Integrand size = 21, antiderivative size = 255 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {5 a \left (a^2+b^2\right ) \sec (c+d x)}{b^5 d}-\frac {5 a \sec ^3(c+d x)}{3 b^3 d}+\frac {5 \left (8 a^4+12 a^2 b^2+3 b^4\right ) \text {arcsinh}(\tan (c+d x)) \sec (c+d x)}{8 b^6 d \sqrt {\sec ^2(c+d x)}}+\frac {5 a \left (a^2+b^2\right )^{3/2} \text {arctanh}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right ) \sec (c+d x)}{b^6 d \sqrt {\sec ^2(c+d x)}}+\frac {5 \left (4 a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 b^4 d}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{4 b^2 d}-\frac {\sec ^5(c+d x)}{b d (a+b \tan (c+d x))} \] Output:
-5*a*(a^2+b^2)*sec(d*x+c)/b^5/d-5/3*a*sec(d*x+c)^3/b^3/d+5/8*(8*a^4+12*a^2 *b^2+3*b^4)*arcsinh(tan(d*x+c))*sec(d*x+c)/b^6/d/(sec(d*x+c)^2)^(1/2)+5*a* (a^2+b^2)^(3/2)*arctanh((b-a*tan(d*x+c))/(a^2+b^2)^(1/2)/(sec(d*x+c)^2)^(1 /2))*sec(d*x+c)/b^6/d/(sec(d*x+c)^2)^(1/2)+5/8*(4*a^2+3*b^2)*sec(d*x+c)*ta n(d*x+c)/b^4/d+5/4*sec(d*x+c)^3*tan(d*x+c)/b^2/d-sec(d*x+c)^5/b/d/(a+b*tan (d*x+c))
Result contains complex when optimal does not.
Time = 6.21 (sec) , antiderivative size = 747, normalized size of antiderivative = 2.93 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:
Integrate[Sec[c + d*x]^7/(a + b*Tan[c + d*x])^2,x]
Output:
(Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])*(-48*b*(a^2 + b^2)^2 - 1 6*a*b*(12*a^2 + 13*b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x]) - 480*a*(a - I*b )*(a + I*b)*Sqrt[a^2 + b^2]*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b ^2]]*(a*Cos[c + d*x] + b*Sin[c + d*x]) - 30*(8*a^4 + 12*a^2*b^2 + 3*b^4)*L og[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c + d*x]) + 30*(8*a^4 + 12*a^2*b^2 + 3*b^4)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] *(a*Cos[c + d*x] + b*Sin[c + d*x]) + (3*b^4*(a*Cos[c + d*x] + b*Sin[c + d* x]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4 + (b^2*(36*a^2 - 8*a*b + 21*b ^2)*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2 ])^2 - (16*a*b^3*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(Cos[ (c + d*x)/2] - Sin[(c + d*x)/2])^3 - (16*a*b*(12*a^2 + 13*b^2)*Sin[(c + d* x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(Cos[(c + d*x)/2] - Sin[(c + d*x) /2]) - (3*b^4*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(Cos[(c + d*x)/2] + Sin[( c + d*x)/2])^4 + (16*a*b^3*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d* x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (b^2*(36*a^2 + 8*a*b + 21*b ^2)*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2 ])^2 + (16*a*b*(12*a^2 + 13*b^2)*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[ c + d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(48*b^6*d*(a + b*Tan[c + d*x])^2)
Time = 0.52 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.87, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3992, 492, 591, 25, 682, 27, 719, 222, 488, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^7}{(a+b \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 3992 |
\(\displaystyle \frac {\sec (c+d x) \int \frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{(a+b \tan (c+d x))^2}d(b \tan (c+d x))}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 492 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \int \frac {b \tan (c+d x) \left (\tan ^2(c+d x)+1\right )^{3/2}}{a+b \tan (c+d x)}d(b \tan (c+d x))}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 591 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {1}{4} \int -\frac {\left (a-\left (\frac {4 a^2}{b^2}+3\right ) b \tan (c+d x)\right ) \sqrt {\tan ^2(c+d x)+1}}{a+b \tan (c+d x)}d(b \tan (c+d x))-\frac {1}{12} \left (\tan ^2(c+d x)+1\right )^{3/2} (4 a-3 b \tan (c+d x))\right )}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (-\frac {1}{4} \int \frac {\left (a-\left (\frac {4 a^2}{b^2}+3\right ) b \tan (c+d x)\right ) \sqrt {\tan ^2(c+d x)+1}}{a+b \tan (c+d x)}d(b \tan (c+d x))-\frac {1}{12} \left (\tan ^2(c+d x)+1\right )^{3/2} (4 a-3 b \tan (c+d x))\right )}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 682 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {1}{4} \left (-\frac {1}{2} b^2 \int \frac {a b^2 \left (4 a^2+5 b^2\right )-b \left (8 a^4+12 b^2 a^2+3 b^4\right ) \tan (c+d x)}{b^6 (a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))-\frac {\sqrt {\tan ^2(c+d x)+1} \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{2 b^2}\right )-\frac {1}{12} \left (\tan ^2(c+d x)+1\right )^{3/2} (4 a-3 b \tan (c+d x))\right )}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {1}{4} \left (-\frac {\int \frac {a b^2 \left (4 a^2+5 b^2\right )-b \left (8 a^4+12 b^2 a^2+3 b^4\right ) \tan (c+d x)}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{2 b^4}-\frac {\sqrt {\tan ^2(c+d x)+1} \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{2 b^2}\right )-\frac {1}{12} \left (\tan ^2(c+d x)+1\right )^{3/2} (4 a-3 b \tan (c+d x))\right )}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 719 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {1}{4} \left (-\frac {8 a \left (a^2+b^2\right )^2 \int \frac {1}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))-\left (8 a^4+12 a^2 b^2+3 b^4\right ) \int \frac {1}{\sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{2 b^4}-\frac {\sqrt {\tan ^2(c+d x)+1} \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{2 b^2}\right )-\frac {1}{12} \left (\tan ^2(c+d x)+1\right )^{3/2} (4 a-3 b \tan (c+d x))\right )}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {1}{4} \left (-\frac {8 a \left (a^2+b^2\right )^2 \int \frac {1}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))-b \left (8 a^4+12 a^2 b^2+3 b^4\right ) \text {arcsinh}(\tan (c+d x))}{2 b^4}-\frac {\sqrt {\tan ^2(c+d x)+1} \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{2 b^2}\right )-\frac {1}{12} \left (\tan ^2(c+d x)+1\right )^{3/2} (4 a-3 b \tan (c+d x))\right )}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {1}{4} \left (-\frac {-8 a \left (a^2+b^2\right )^2 \int \frac {1}{\frac {a^2}{b^2}-b^2 \tan ^2(c+d x)+1}d\frac {1-\frac {a \tan (c+d x)}{b}}{\sqrt {\tan ^2(c+d x)+1}}-b \left (8 a^4+12 a^2 b^2+3 b^4\right ) \text {arcsinh}(\tan (c+d x))}{2 b^4}-\frac {\sqrt {\tan ^2(c+d x)+1} \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{2 b^2}\right )-\frac {1}{12} \left (\tan ^2(c+d x)+1\right )^{3/2} (4 a-3 b \tan (c+d x))\right )}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {1}{4} \left (-\frac {\sqrt {\tan ^2(c+d x)+1} \left (8 a \left (a^2+b^2\right )-b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{2 b^2}-\frac {-8 a b \left (a^2+b^2\right )^{3/2} \text {arctanh}\left (\frac {b^2 \tan (c+d x)}{\sqrt {a^2+b^2}}\right )-b \left (8 a^4+12 a^2 b^2+3 b^4\right ) \text {arcsinh}(\tan (c+d x))}{2 b^4}\right )-\frac {1}{12} \left (\tan ^2(c+d x)+1\right )^{3/2} (4 a-3 b \tan (c+d x))\right )}{b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{a+b \tan (c+d x)}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
Input:
Int[Sec[c + d*x]^7/(a + b*Tan[c + d*x])^2,x]
Output:
(Sec[c + d*x]*(-((1 + Tan[c + d*x]^2)^(5/2)/(a + b*Tan[c + d*x])) + (5*(-1 /12*((4*a - 3*b*Tan[c + d*x])*(1 + Tan[c + d*x]^2)^(3/2)) + (-1/2*(-(b*(8* a^4 + 12*a^2*b^2 + 3*b^4)*ArcSinh[Tan[c + d*x]]) - 8*a*b*(a^2 + b^2)^(3/2) *ArcTanh[(b^2*Tan[c + d*x])/Sqrt[a^2 + b^2]])/b^4 - ((8*a*(a^2 + b^2) - b* (4*a^2 + 3*b^2)*Tan[c + d*x])*Sqrt[1 + Tan[c + d*x]^2])/(2*b^2))/4))/b^2)) /(b*d*Sqrt[Sec[c + d*x]^2])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 1))), x] - Simp[2*b*(p/(d*(n + 1)) ) Int[x*(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && GtQ[p, 0] && (IntegerQ[p] || LtQ[n, -1]) && NeQ[n, -1] && !IL tQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c + d*x)^(n + 1))*(a + b*x^2)^p*((c*(2*p + 1) - d*(n + 2*p + 1)*x)/ (d^2*(n + 2*p + 1)*(n + 2*p + 2))), x] + Simp[2*(p/(d^2*(n + 2*p + 1)*(n + 2*p + 2))) Int[(c + d*x)^n*(a + b*x^2)^(p - 1)*Simp[a*c*d*n + (b*c^2*(2*p + 1) + a*d^2*(n + 2*p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && GtQ[p, 0] && LeQ[-1, n, 0] && !ILtQ[n + 2*p, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] + Simp[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))) Int[(d + e*x) ^m*(a + c*x^2)^(p - 1)*Simp[f*a*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f* d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))*x, x], x ], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] || ! RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) && !ILtQ[m + 2*p, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[Sec[e + f*x]/(b*f*Sqrt[Sec[e + f*x]^2]) Subst[Int[( a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b , e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[(m - 1)/2]
Time = 150.12 (sec) , antiderivative size = 454, normalized size of antiderivative = 1.78
method | result | size |
derivativedivides | \(\frac {-\frac {1}{4 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {4 a -3 b}{6 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {12 a^{2}-8 a b +11 b^{2}}{8 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (40 a^{4}+60 b^{2} a^{2}+15 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 b^{6}}-\frac {32 a^{3}-12 a^{2} b +40 a \,b^{2}-9 b^{3}}{8 b^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{4 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {-4 a -3 b}{6 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-12 a^{2}-8 a b -11 b^{2}}{8 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-40 a^{4}-60 b^{2} a^{2}-15 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 b^{6}}-\frac {-32 a^{3}-12 a^{2} b -40 a \,b^{2}-9 b^{3}}{8 b^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\frac {2 \left (\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+b \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )\right )}{a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}-\frac {10 a \left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{b^{6}}}{d}\) | \(454\) |
default | \(\frac {-\frac {1}{4 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {4 a -3 b}{6 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {12 a^{2}-8 a b +11 b^{2}}{8 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (40 a^{4}+60 b^{2} a^{2}+15 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 b^{6}}-\frac {32 a^{3}-12 a^{2} b +40 a \,b^{2}-9 b^{3}}{8 b^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{4 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {-4 a -3 b}{6 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-12 a^{2}-8 a b -11 b^{2}}{8 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-40 a^{4}-60 b^{2} a^{2}-15 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 b^{6}}-\frac {-32 a^{3}-12 a^{2} b -40 a \,b^{2}-9 b^{3}}{8 b^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\frac {2 \left (\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+b \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )\right )}{a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}-\frac {10 a \left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{b^{6}}}{d}\) | \(454\) |
risch | \(-\frac {120 i a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+60 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+75 i a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-190 i a \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+190 i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-120 i a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}-60 i a^{3} b \,{\mathrm e}^{9 i \left (d x +c \right )}-75 i a \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}+180 a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}+180 a^{2} b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+640 a^{2} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+920 a^{2} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+640 a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+480 a^{4} {\mathrm e}^{7 i \left (d x +c \right )}+120 b^{4} {\mathrm e}^{7 i \left (d x +c \right )}+720 a^{4} {\mathrm e}^{5 i \left (d x +c \right )}+54 b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+480 a^{4} {\mathrm e}^{3 i \left (d x +c \right )}+120 b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+120 a^{4} {\mathrm e}^{i \left (d x +c \right )}+45 b^{4} {\mathrm e}^{i \left (d x +c \right )}+120 a^{4} {\mathrm e}^{9 i \left (d x +c \right )}+45 b^{4} {\mathrm e}^{9 i \left (d x +c \right )}}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right ) d \,b^{5}}+\frac {5 \left (a^{2}+b^{2}\right )^{\frac {3}{2}} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} \left (i a -b \right )}{a^{4}+2 b^{2} a^{2}+b^{4}}\right )}{d \,b^{6}}-\frac {5 \left (a^{2}+b^{2}\right )^{\frac {3}{2}} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} \left (i a -b \right )}{a^{4}+2 b^{2} a^{2}+b^{4}}\right )}{d \,b^{6}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{4}}{d \,b^{6}}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{2 d \,b^{4}}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d \,b^{2}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{4}}{d \,b^{6}}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{2 d \,b^{4}}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d \,b^{2}}\) | \(676\) |
Input:
int(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/4/b^2/(tan(1/2*d*x+1/2*c)+1)^4-1/6*(4*a-3*b)/b^3/(tan(1/2*d*x+1/2* c)+1)^3-1/8*(12*a^2-8*a*b+11*b^2)/b^4/(tan(1/2*d*x+1/2*c)+1)^2+1/8/b^6*(40 *a^4+60*a^2*b^2+15*b^4)*ln(tan(1/2*d*x+1/2*c)+1)-1/8*(32*a^3-12*a^2*b+40*a *b^2-9*b^3)/b^5/(tan(1/2*d*x+1/2*c)+1)+1/4/b^2/(tan(1/2*d*x+1/2*c)-1)^4-1/ 6*(-4*a-3*b)/b^3/(tan(1/2*d*x+1/2*c)-1)^3-1/8*(-12*a^2-8*a*b-11*b^2)/b^4/( tan(1/2*d*x+1/2*c)-1)^2+1/8/b^6*(-40*a^4-60*a^2*b^2-15*b^4)*ln(tan(1/2*d*x +1/2*c)-1)-1/8*(-32*a^3-12*a^2*b-40*a*b^2-9*b^3)/b^5/(tan(1/2*d*x+1/2*c)-1 )+2/b^6*(((a^4+2*a^2*b^2+b^4)*b^2/a*tan(1/2*d*x+1/2*c)+b*(a^4+2*a^2*b^2+b^ 4))/(a*tan(1/2*d*x+1/2*c)^2-2*b*tan(1/2*d*x+1/2*c)-a)-5*a*(a^4+2*a^2*b^2+b ^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/ 2))))
Time = 0.24 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.85 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {12 \, b^{5} - 30 \, {\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} + 120 \, {\left ({\left (a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{5} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 15 \, {\left ({\left (8 \, a^{5} + 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + {\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (8 \, a^{5} + 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + {\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (2 \, a b^{4} \cos \left (d x + c\right ) + 3 \, {\left (4 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{48 \, {\left (a b^{6} d \cos \left (d x + c\right )^{5} + b^{7} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
1/48*(12*b^5 - 30*(8*a^4*b + 12*a^2*b^3 + 3*b^5)*cos(d*x + c)^4 + 10*(4*a^ 2*b^3 + 3*b^5)*cos(d*x + c)^2 + 120*((a^4 + a^2*b^2)*cos(d*x + c)^5 + (a^3 *b + a*b^3)*cos(d*x + c)^4*sin(d*x + c))*sqrt(a^2 + b^2)*log((2*a*b*cos(d* x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^ 2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) + 15*((8*a^5 + 12*a^3*b^2 + 3*a*b^ 4)*cos(d*x + c)^5 + (8*a^4*b + 12*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*sin(d*x + c))*log(sin(d*x + c) + 1) - 15*((8*a^5 + 12*a^3*b^2 + 3*a*b^4)*cos(d*x + c)^5 + (8*a^4*b + 12*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*sin(d*x + c))*log(-s in(d*x + c) + 1) - 10*(2*a*b^4*cos(d*x + c) + 3*(4*a^3*b^2 + 5*a*b^4)*cos( d*x + c)^3)*sin(d*x + c))/(a*b^6*d*cos(d*x + c)^5 + b^7*d*cos(d*x + c)^4*s in(d*x + c))
\[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\sec ^{7}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sec(d*x+c)**7/(a+b*tan(d*x+c))**2,x)
Output:
Integral(sec(c + d*x)**7/(a + b*tan(c + d*x))**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 827 vs. \(2 (239) = 478\).
Time = 0.14 (sec) , antiderivative size = 827, normalized size of antiderivative = 3.24 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
-1/24*(2*(120*a^5 + 160*a^3*b^2 + 24*a*b^4 + (180*a^4*b + 245*a^2*b^3 + 24 *b^5)*sin(d*x + c)/(cos(d*x + c) + 1) - 10*(48*a^5 + 68*a^3*b^2 + 15*a*b^4 )*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(300*a^4*b + 385*a^2*b^3 + 48*b^ 5)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 10*(72*a^5 + 100*a^3*b^2 + 15*a*b ^4)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 48*(15*a^4*b + 20*a^2*b^3 + 3*b^ 5)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 30*(16*a^5 + 20*a^3*b^2 + 3*a*b^4 )*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6*(60*a^4*b + 85*a^2*b^3 + 16*b^5) *sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 30*(4*a^5 + 4*a^3*b^2 - a*b^4)*sin( d*x + c)^8/(cos(d*x + c) + 1)^8 + 3*(20*a^4*b + 25*a^2*b^3 + 8*b^5)*sin(d* x + c)^9/(cos(d*x + c) + 1)^9)/(a^2*b^5 + 2*a*b^6*sin(d*x + c)/(cos(d*x + c) + 1) - 5*a^2*b^5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*a*b^6*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 10*a^2*b^5*sin(d*x + c)^4/(cos(d*x + c) + 1) ^4 + 12*a*b^6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 10*a^2*b^5*sin(d*x + c )^6/(cos(d*x + c) + 1)^6 - 8*a*b^6*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 5 *a^2*b^5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 2*a*b^6*sin(d*x + c)^9/(cos (d*x + c) + 1)^9 - a^2*b^5*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) - 120*(a ^4 + 2*a^2*b^2 + b^4)*a*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt( a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2)))/(sq rt(a^2 + b^2)*b^6) - 15*(8*a^4 + 12*a^2*b^2 + 3*b^4)*log(sin(d*x + c)/(cos (d*x + c) + 1) + 1)/b^6 + 15*(8*a^4 + 12*a^2*b^2 + 3*b^4)*log(sin(d*x +...
Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (239) = 478\).
Time = 0.30 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.08 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
1/24*(15*(8*a^4 + 12*a^2*b^2 + 3*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b ^6 - 15*(8*a^4 + 12*a^2*b^2 + 3*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^ 6 + 120*(a^5 + 2*a^3*b^2 + a*b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2) ))/(sqrt(a^2 + b^2)*b^6) + 48*(a^4*b*tan(1/2*d*x + 1/2*c) + 2*a^2*b^3*tan( 1/2*d*x + 1/2*c) + b^5*tan(1/2*d*x + 1/2*c) + a^5 + 2*a^3*b^2 + a*b^4)/((a *tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)*a*b^5) + 2*(36*a^2 *b*tan(1/2*d*x + 1/2*c)^7 + 27*b^3*tan(1/2*d*x + 1/2*c)^7 + 96*a^3*tan(1/2 *d*x + 1/2*c)^6 + 144*a*b^2*tan(1/2*d*x + 1/2*c)^6 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 3*b^3*tan(1/2*d*x + 1/2*c)^5 - 288*a^3*tan(1/2*d*x + 1/2*c)^4 - 336*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 3* b^3*tan(1/2*d*x + 1/2*c)^3 + 288*a^3*tan(1/2*d*x + 1/2*c)^2 + 304*a*b^2*ta n(1/2*d*x + 1/2*c)^2 + 36*a^2*b*tan(1/2*d*x + 1/2*c) + 27*b^3*tan(1/2*d*x + 1/2*c) - 96*a^3 - 112*a*b^2)/((tan(1/2*d*x + 1/2*c)^2 - 1)^4*b^5))/d
Time = 3.70 (sec) , antiderivative size = 2654, normalized size of antiderivative = 10.41 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^7*(a + b*tan(c + d*x))^2),x)
Output:
-((9*a*b^5)/64 + (15*a^5*b)/8 + (b^6*sin(c + d*x))/8 + (115*a^3*b^3)/48 + (3*b^6*sin(3*c + 3*d*x))/16 + (b^6*sin(5*c + 5*d*x))/16 + (a^6*cos(c + d*x )*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*25i)/4 + (5*a*b^5*cos(2 *c + 2*d*x))/8 + (5*a^5*b*cos(2*c + 2*d*x))/2 + (5*a*b^5*cos(3*c + 3*d*x)) /16 + (25*a^5*b*cos(3*c + 3*d*x))/16 + (15*a*b^5*cos(4*c + 4*d*x))/64 + (5 *a^5*b*cos(4*c + 4*d*x))/8 + (a*b^5*cos(5*c + 5*d*x))/16 + (5*a^5*b*cos(5* c + 5*d*x))/16 + (25*a^3*b^3*cos(c + d*x))/6 + (5*a^2*b^4*sin(c + d*x))/6 + (5*a^4*b^2*sin(c + d*x))/8 + (a^6*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*25i)/8 + (a^6*atan((sin(c/2 + (d*x)/2)*1i)/cos (c/2 + (d*x)/2))*cos(5*c + 5*d*x)*5i)/8 + (10*a^3*b^3*cos(2*c + 2*d*x))/3 + (25*a^3*b^3*cos(3*c + 3*d*x))/12 + (15*a^3*b^3*cos(4*c + 4*d*x))/16 + (5 *a^3*b^3*cos(5*c + 5*d*x))/12 + (95*a^2*b^4*sin(2*c + 2*d*x))/96 + (5*a^4* b^2*sin(2*c + 2*d*x))/8 + (5*a^2*b^4*sin(3*c + 3*d*x))/4 + (15*a^4*b^2*sin (3*c + 3*d*x))/16 + (25*a^2*b^4*sin(4*c + 4*d*x))/64 + (5*a^4*b^2*sin(4*c + 4*d*x))/16 + (5*a^2*b^4*sin(5*c + 5*d*x))/12 + (5*a^4*b^2*sin(5*c + 5*d* x))/16 + (5*a*b^5*cos(c + d*x))/8 + (25*a^5*b*cos(c + d*x))/8 + (a*b^5*ata n((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x)*45i)/64 + ( a^5*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x)*15 i)/8 + (a*b^5*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*sin(5*c + 5 *d*x)*15i)/64 + (a^5*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))...
Time = 0.25 (sec) , antiderivative size = 6643, normalized size of antiderivative = 26.05 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^7/(a+b*tan(d*x+c))^2,x)
Output:
(240*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2) )*cos(c + d*x)*sin(c + d*x)**6*tan(c + d*x)*a**4*b*i + 240*sqrt(a**2 + b** 2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**6*tan(c + d*x)*a**2*b**3*i + 240*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**6*a**5*i + 240*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b** 2))*cos(c + d*x)*sin(c + d*x)**6*a**3*b**2*i - 720*sqrt(a**2 + b**2)*atan( (tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)* *4*tan(c + d*x)*a**4*b*i - 720*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a* i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)*a**2 *b**3*i - 720*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a** 2 + b**2))*cos(c + d*x)*sin(c + d*x)**4*a**5*i - 720*sqrt(a**2 + b**2)*ata n((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x )**4*a**3*b**2*i + 720*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i) /sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x)*a**4*b*i + 7 20*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))* cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x)*a**2*b**3*i + 720*sqrt(a**2 + b* *2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin( c + d*x)**2*a**5*i + 720*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b* i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**2*a**3*b**2*i - 240*sq...