Integrand size = 21, antiderivative size = 148 \[ \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {3 a \text {arctanh}(\sin (c+d x))}{b^4 d}-\frac {3 \left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^4 \sqrt {a^2+b^2} d}-\frac {\sec ^3(c+d x)}{2 b d (a+b \tan (c+d x))^2}+\frac {3 \sec (c+d x) (2 a+b \tan (c+d x))}{2 b^3 d (a+b \tan (c+d x))} \]
-3*a*arctanh(sin(d*x+c))/b^4/d-3/2*(2*a^2+b^2)*arctanh((b*cos(d*x+c)-a*sin (d*x+c))/(a^2+b^2)^(1/2))/b^4/d/(a^2+b^2)^(1/2)-1/2*sec(d*x+c)^3/b/d/(a+b* tan(d*x+c))^2+3/2*sec(d*x+c)*(2*a+b*tan(d*x+c))/b^3/d/(a+b*tan(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(396\) vs. \(2(148)=296\).
Time = 3.68 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.68 \[ \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac {b^2 \left (a^2+b^2\right ) \sin (c+d x)}{a}+\frac {(2 a-b) b (2 a+b) (a \cos (c+d x)+b \sin (c+d x))}{a}+2 b (a \cos (c+d x)+b \sin (c+d x))^2+\frac {6 \left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\sqrt {a^2+b^2}}+6 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2-6 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2+\frac {2 b \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {2 b \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{2 b^4 d (a+b \tan (c+d x))^3} \]
(Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])*((b^2*(a^2 + b^2)*Sin[c + d*x])/a + ((2*a - b)*b*(2*a + b)*(a*Cos[c + d*x] + b*Sin[c + d*x]))/a + 2*b*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 + (6*(2*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])^2)/S qrt[a^2 + b^2] + 6*a*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a*Cos[c + d *x] + b*Sin[c + d*x])^2 - 6*a*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a* Cos[c + d*x] + b*Sin[c + d*x])^2 + (2*b*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (2*b*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(2*b^4*d*(a + b*Tan[c + d*x])^3)
Time = 0.38 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3992, 492, 590, 25, 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.
\(\displaystyle \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^5}{(a+b \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3992 |
\(\displaystyle \frac {\sec (c+d x) \int \frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{(a+b \tan (c+d x))^3}d(b \tan (c+d x))}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 492 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {3 \int \frac {b \tan (c+d x) \sqrt {\tan ^2(c+d x)+1}}{(a+b \tan (c+d x))^2}d(b \tan (c+d x))}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 590 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {3 \left (\frac {\sqrt {\tan ^2(c+d x)+1} (2 a+b \tan (c+d x))}{a+b \tan (c+d x)}-\int -\frac {b^2-2 a b \tan (c+d x)}{b^2 (a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {3 \left (\int \frac {b^2-2 a b \tan (c+d x)}{b^2 (a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))+\frac {\sqrt {\tan ^2(c+d x)+1} (2 a+b \tan (c+d x))}{a+b \tan (c+d x)}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {3 \left (\frac {\int \frac {b^2-2 a b \tan (c+d x)}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{b^2}+\frac {\sqrt {\tan ^2(c+d x)+1} (2 a+b \tan (c+d x))}{a+b \tan (c+d x)}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 719 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {3 \left (\frac {\left (2 a^2+b^2\right ) \int \frac {1}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))-2 a \int \frac {1}{\sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{b^2}+\frac {\sqrt {\tan ^2(c+d x)+1} (2 a+b \tan (c+d x))}{a+b \tan (c+d x)}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {3 \left (\frac {\left (2 a^2+b^2\right ) \int \frac {1}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))-2 a b \text {arcsinh}(\tan (c+d x))}{b^2}+\frac {\sqrt {\tan ^2(c+d x)+1} (2 a+b \tan (c+d x))}{a+b \tan (c+d x)}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {3 \left (\frac {-\left (2 a^2+b^2\right ) \int \frac {1}{\frac {a^2}{b^2}-b^2 \tan ^2(c+d x)+1}d\frac {1-\frac {a \tan (c+d x)}{b}}{\sqrt {\tan ^2(c+d x)+1}}-2 a b \text {arcsinh}(\tan (c+d x))}{b^2}+\frac {\sqrt {\tan ^2(c+d x)+1} (2 a+b \tan (c+d x))}{a+b \tan (c+d x)}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sec (c+d x) \left (\frac {3 \left (\frac {-\frac {b \left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {b^2 \tan (c+d x)}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-2 a b \text {arcsinh}(\tan (c+d x))}{b^2}+\frac {\sqrt {\tan ^2(c+d x)+1} (2 a+b \tan (c+d x))}{a+b \tan (c+d x)}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{3/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
(Sec[c + d*x]*(-1/2*(1 + Tan[c + d*x]^2)^(3/2)/(a + b*Tan[c + d*x])^2 + (3 *((-2*a*b*ArcSinh[Tan[c + d*x]] - (b*(2*a^2 + b^2)*ArcTanh[(b^2*Tan[c + d* x])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2])/b^2 + ((2*a + b*Tan[c + d*x])*Sqrt[ 1 + Tan[c + d*x]^2])/(a + b*Tan[c + d*x])))/(2*b^2)))/(b*d*Sqrt[Sec[c + d* x]^2])
3.6.73.3.1 Defintions of rubi rules used
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 + 1)*x)/(d^2*( n + 1)*(n + 2*p + 2))), x] + Simp[2*(p/(d^2*(n + 1)*(n + 2*p + 2))) Int[( c + d*x)^(n + 1)*(a + b*x^2)^(p - 1)*(a*d*(n + 1) + b*c*(2*p + 1)*x), x], x ] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && LtQ[n, -1] && !ILtQ[n + 2*p + 1, 0]
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 = 94.52 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.82
method | result | size |
derivativedivides | \(\frac {\frac {1}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}-\frac {2 \left (\frac {\frac {b^{2} \left (3 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {b \left (4 a^{4}-9 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}-\frac {b^{2} \left (13 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-2 a^{2} b +\frac {b^{3}}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}^{2}}-\frac {3 \left (2 a^{2}+b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{4}}-\frac {1}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}}{d}\) | \(269\) |
default | \(\frac {\frac {1}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}-\frac {2 \left (\frac {\frac {b^{2} \left (3 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {b \left (4 a^{4}-9 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}-\frac {b^{2} \left (13 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-2 a^{2} b +\frac {b^{3}}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}^{2}}-\frac {3 \left (2 a^{2}+b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{4}}-\frac {1}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}}{d}\) | \(269\) |
risch | \(\frac {-9 i a b \,{\mathrm e}^{5 i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+9 i a b \,{\mathrm e}^{i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )^{2} b^{3} d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,b^{4}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \,b^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {a^{2}+b^{2}}\, d \,b^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}\, d \,b^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {a^{2}+b^{2}}\, d \,b^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}\, d \,b^{2}}\) | \(403\) |
1/d*(1/b^3/(tan(1/2*d*x+1/2*c)+1)-3*a/b^4*ln(tan(1/2*d*x+1/2*c)+1)-2/b^4*( (1/2*b^2*(3*a^2-2*b^2)/a*tan(1/2*d*x+1/2*c)^3+1/2*b*(4*a^4-9*a^2*b^2+2*b^4 )/a^2*tan(1/2*d*x+1/2*c)^2-1/2*b^2*(13*a^2-2*b^2)/a*tan(1/2*d*x+1/2*c)-2*a ^2*b+1/2*b^3)/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2-3/2*(2*a ^2+b^2)/(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)))-1/b^3/(tan(1/2*d*x+1/2*c)-1)+3*a/b^4*ln(tan(1/2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (138) = 276\).
Time = 0.34 (sec) , antiderivative size = 513, normalized size of antiderivative = 3.47 \[ \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {4 \, a^{2} b^{3} + 4 \, b^{5} + 6 \, {\left (2 \, a^{4} b + a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 18 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left ({\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 6 \, {\left ({\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left ({\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \, {\left ({\left (a^{4} b^{4} - b^{8}\right )} d \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )\right )}} \]
1/4*(4*a^2*b^3 + 4*b^5 + 6*(2*a^4*b + a^2*b^3 - b^5)*cos(d*x + c)^2 + 18*( a^3*b^2 + a*b^4)*cos(d*x + c)*sin(d*x + c) + 3*((2*a^4 - a^2*b^2 - b^4)*co s(d*x + c)^3 + 2*(2*a^3*b + a*b^3)*cos(d*x + c)^2*sin(d*x + c) + (2*a^2*b^ 2 + b^4)*cos(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)) - 6*((a^5 - a*b^4)*cos(d*x + c)^3 + 2*(a^4*b + a^2* b^3)*cos(d*x + c)^2*sin(d*x + c) + (a^3*b^2 + a*b^4)*cos(d*x + c))*log(sin (d*x + c) + 1) + 6*((a^5 - a*b^4)*cos(d*x + c)^3 + 2*(a^4*b + a^2*b^3)*cos (d*x + c)^2*sin(d*x + c) + (a^3*b^2 + a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1))/((a^4*b^4 - b^8)*d*cos(d*x + c)^3 + 2*(a^3*b^5 + a*b^7)*d*cos(d*x + c)^2*sin(d*x + c) + (a^2*b^6 + b^8)*d*cos(d*x + c))
\[ \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (138) = 276\).
Time = 0.34 (sec) , antiderivative size = 518, normalized size of antiderivative = 3.50 \[ \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (6 \, a^{4} - a^{2} b^{2} + \frac {{\left (21 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, {\left (6 \, a^{4} - 9 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (6 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {{\left (6 \, a^{4} - 9 \, a^{2} b^{2} + 2 \, b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{4} b^{3} + \frac {4 \, a^{3} b^{4} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {8 \, a^{3} b^{4} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {4 \, a^{3} b^{4} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{4} b^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {{\left (3 \, a^{4} b^{3} - 4 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (3 \, a^{4} b^{3} - 4 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {6 \, a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{4}} + \frac {6 \, a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{4}} - \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}}}{2 \, d} \]
1/2*(2*(6*a^4 - a^2*b^2 + (21*a^3*b - 2*a*b^3)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*(6*a^4 - 9*a^2*b^2 + b^4)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4 *(6*a^3*b - a*b^3)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + (6*a^4 - 9*a^2*b^ 2 + 2*b^4)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (3*a^3*b - 2*a*b^3)*sin(d *x + c)^5/(cos(d*x + c) + 1)^5)/(a^4*b^3 + 4*a^3*b^4*sin(d*x + c)/(cos(d*x + c) + 1) - 8*a^3*b^4*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 4*a^3*b^4*sin (d*x + c)^5/(cos(d*x + c) + 1)^5 - a^4*b^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - (3*a^4*b^3 - 4*a^2*b^5)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (3*a^ 4*b^3 - 4*a^2*b^5)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 6*a*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b^4 + 6*a*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b^4 - 3*(2*a^2 + b^2)*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + s qrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2))) /(sqrt(a^2 + b^2)*b^4))/d
Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (138) = 276\).
Time = 0.61 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.12 \[ \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {6 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac {6 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} + \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {4}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} b^{3}} + \frac {2 \, {\left (3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 13 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{4} + a^{2} b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{2} a^{2} b^{3}}}{2 \, d} \]
-1/2*(6*a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 6*a*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 + 3*(2*a^2 + b^2)*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^4) + 4/((tan(1/2*d*x + 1/2*c)^2 - 1)*b^3) + 2*(3 *a^3*b*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*tan (1/2*d*x + 1/2*c)^2 - 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 + 2*b^4*tan(1/2*d*x + 1/2*c)^2 - 13*a^3*b*tan(1/2*d*x + 1/2*c) + 2*a*b^3*tan(1/2*d*x + 1/2*c) - 4*a^4 + a^2*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2*a^2*b^3))/d
Time = 6.48 (sec) , antiderivative size = 1311, normalized size of antiderivative = 8.86 \[ \int \frac {\sec ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \]
((6*a^2 - b^2)/b^3 - (2*tan(c/2 + (d*x)/2)^2*(6*a^4 + b^4 - 9*a^2*b^2))/(a ^2*b^3) + (tan(c/2 + (d*x)/2)*(21*a^2 - 2*b^2))/(a*b^2) + (tan(c/2 + (d*x) /2)^4*(6*a^4 + 2*b^4 - 9*a^2*b^2))/(a^2*b^3) - (4*tan(c/2 + (d*x)/2)^3*(6* a^2 - b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)^5*(3*a^2 - 2*b^2))/(a*b^2))/(d*( tan(c/2 + (d*x)/2)^4*(3*a^2 - 4*b^2) - tan(c/2 + (d*x)/2)^2*(3*a^2 - 4*b^2 ) - a^2*tan(c/2 + (d*x)/2)^6 + a^2 - 8*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*ta n(c/2 + (d*x)/2)^5 + 4*a*b*tan(c/2 + (d*x)/2))) - (6*a*atanh(tan(c/2 + (d* x)/2)))/(b^4*d) + (atan((((2*a^2 + b^2)*(a^2 + b^2)^(1/2)*((288*a^4)/b^5 + (8*tan(c/2 + (d*x)/2)*(9*a*b^7 + 108*a^3*b^5 + 72*a^5*b^3))/b^9 - (3*(2*a ^2 + b^2)*(a^2 + b^2)^(1/2)*((8*tan(c/2 + (d*x)/2)*(12*a*b^10 + 24*a^3*b^8 ))/b^9 - 48*a^2 + (3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*(32*a^2*b^3 + (8*tan( c/2 + (d*x)/2)*(12*a*b^13 + 8*a^3*b^11))/b^9))/(2*(b^6 + a^2*b^4))))/(2*(b ^6 + a^2*b^4)))*3i)/(2*(b^6 + a^2*b^4)) + ((2*a^2 + b^2)*(a^2 + b^2)^(1/2) *((288*a^4)/b^5 + (8*tan(c/2 + (d*x)/2)*(9*a*b^7 + 108*a^3*b^5 + 72*a^5*b^ 3))/b^9 - (3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*(48*a^2 - (8*tan(c/2 + (d*x)/ 2)*(12*a*b^10 + 24*a^3*b^8))/b^9 + (3*(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*(32* a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 + 8*a^3*b^11))/b^9))/(2*(b^6 + a^2*b^4))))/(2*(b^6 + a^2*b^4)))*3i)/(2*(b^6 + a^2*b^4)))/((16*(54*a^4 + 2 7*a^2*b^2))/b^8 - (16*tan(c/2 + (d*x)/2)*(216*a^5 + 108*a^3*b^2))/b^9 - (3 *(2*a^2 + b^2)*(a^2 + b^2)^(1/2)*((288*a^4)/b^5 + (8*tan(c/2 + (d*x)/2)...