Integrand size = 23, antiderivative size = 167 \[ \int \frac {\sec ^7(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=-\frac {(4 a-5 b) \text {arctanh}(\sin (c+d x))}{2 b^3 d}+\frac {(a-b)^{3/2} (4 a+b) \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^3 d}+\frac {(a-b) (2 a-b) \sin (c+d x)}{2 a b^2 d \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac {\sec (c+d x) \tan (c+d x)}{2 b d \left (a-(a-b) \sin ^2(c+d x)\right )} \]
-1/2*(4*a-5*b)*arctanh(sin(d*x+c))/b^3/d+1/2*(a-b)^(3/2)*(4*a+b)*arctanh(s in(d*x+c)*(a-b)^(1/2)/a^(1/2))/a^(3/2)/b^3/d+1/2*(a-b)*(2*a-b)*sin(d*x+c)/ a/b^2/d/(a-(a-b)*sin(d*x+c)^2)+1/2*sec(d*x+c)*tan(d*x+c)/b/d/(a-(a-b)*sin( d*x+c)^2)
Time = 4.14 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.52 \[ \int \frac {\sec ^7(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {2 (4 a-5 b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (-4 a+5 b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {(a-b)^{3/2} (4 a+b) \log \left (\sqrt {a}-\sqrt {a-b} \sin (c+d x)\right )}{a^{3/2}}+\frac {(a-b)^{3/2} (4 a+b) \log \left (\sqrt {a}+\sqrt {a-b} \sin (c+d x)\right )}{a^{3/2}}+\frac {b}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (a-b)^2 b \sin (c+d x)}{a (a+b+(a-b) \cos (2 (c+d x)))}}{4 b^3 d} \]
(2*(4*a - 5*b)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(-4*a + 5*b)*L og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - ((a - b)^(3/2)*(4*a + b)*Log[Sqr t[a] - Sqrt[a - b]*Sin[c + d*x]])/a^(3/2) + ((a - b)^(3/2)*(4*a + b)*Log[S qrt[a] + Sqrt[a - b]*Sin[c + d*x]])/a^(3/2) + b/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - b/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*(a - b)^2*b *Sin[c + d*x])/(a*(a + b + (a - b)*Cos[2*(c + d*x)])))/(4*b^3*d)
Time = 0.41 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4159, 316, 25, 402, 27, 397, 219, 221}
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.
| \(\displaystyle \int \frac {\sec ^7(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^7}{\left (a+b \tan (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4159 |
| \(\displaystyle \frac {\int \frac {1}{\left (1-\sin ^2(c+d x)\right )^2 \left (a-(a-b) \sin ^2(c+d x)\right )^2}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int -\frac {3 (a-b) \sin ^2(c+d x)+a-2 b}{\left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )^2}d\sin (c+d x)}{2 b}+\frac {\sin (c+d x)}{2 b \left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\sin (c+d x)}{2 b \left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {\int \frac {3 (a-b) \sin ^2(c+d x)+a-2 b}{\left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )^2}d\sin (c+d x)}{2 b}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\sin (c+d x)}{2 b \left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {-\frac {\int -\frac {2 \left (2 a^2-2 b a-b^2+(a-b) (2 a-b) \sin ^2(c+d x)\right )}{\left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}d\sin (c+d x)}{2 a b}-\frac {(a-b) (2 a-b) \sin (c+d x)}{a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{2 b}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\sin (c+d x)}{2 b \left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {\frac {\int \frac {2 a^2-2 b a-b^2+(a-b) (2 a-b) \sin ^2(c+d x)}{\left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}d\sin (c+d x)}{a b}-\frac {(a-b) (2 a-b) \sin (c+d x)}{a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{2 b}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\sin (c+d x)}{2 b \left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {\frac {\frac {a (4 a-5 b) \int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)}{b}-\frac {(a-b)^2 (4 a+b) \int \frac {1}{a-(a-b) \sin ^2(c+d x)}d\sin (c+d x)}{b}}{a b}-\frac {(a-b) (2 a-b) \sin (c+d x)}{a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{2 b}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\sin (c+d x)}{2 b \left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {\frac {\frac {a (4 a-5 b) \text {arctanh}(\sin (c+d x))}{b}-\frac {(a-b)^2 (4 a+b) \int \frac {1}{a-(a-b) \sin ^2(c+d x)}d\sin (c+d x)}{b}}{a b}-\frac {(a-b) (2 a-b) \sin (c+d x)}{a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{2 b}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\sin (c+d x)}{2 b \left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {\frac {\frac {a (4 a-5 b) \text {arctanh}(\sin (c+d x))}{b}-\frac {(a-b)^{3/2} (4 a+b) \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b}}{a b}-\frac {(a-b) (2 a-b) \sin (c+d x)}{a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{2 b}}{d}\) |
(Sin[c + d*x]/(2*b*(1 - Sin[c + d*x]^2)*(a - (a - b)*Sin[c + d*x]^2)) - (( (a*(4*a - 5*b)*ArcTanh[Sin[c + d*x]])/b - ((a - b)^(3/2)*(4*a + b)*ArcTanh [(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(Sqrt[a]*b))/(a*b) - ((a - b)*(2*a - b)*Sin[c + d*x])/(a*b*(a - (a - b)*Sin[c + d*x]^2)))/(2*b))/d
3.5.62.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 61.65 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{4 b^{2} \left (\sin \left (d x +c \right )+1\right )}+\frac {\left (-4 a +5 b \right ) \ln \left (\sin \left (d x +c \right )+1\right )}{4 b^{3}}-\frac {1}{4 b^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (4 a -5 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 b^{3}}-\frac {\left (a^{2}-2 a b +b^{2}\right ) \left (\frac {b \sin \left (d x +c \right )}{2 a \left (\sin \left (d x +c \right )^{2} a -b \sin \left (d x +c \right )^{2}-a \right )}-\frac {\left (4 a +b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}\right )}{b^{3}}}{d}\) | \(175\) |
| default | \(\frac {-\frac {1}{4 b^{2} \left (\sin \left (d x +c \right )+1\right )}+\frac {\left (-4 a +5 b \right ) \ln \left (\sin \left (d x +c \right )+1\right )}{4 b^{3}}-\frac {1}{4 b^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (4 a -5 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 b^{3}}-\frac {\left (a^{2}-2 a b +b^{2}\right ) \left (\frac {b \sin \left (d x +c \right )}{2 a \left (\sin \left (d x +c \right )^{2} a -b \sin \left (d x +c \right )^{2}-a \right )}-\frac {\left (4 a +b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}\right )}{b^{3}}}{d}\) | \(175\) |
| risch | \(\frac {i \left (2 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+2 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+a b \,{\mathrm e}^{5 i \left (d x +c \right )}+b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-2 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-a b \,{\mathrm e}^{3 i \left (d x +c \right )}-b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-2 a^{2} {\mathrm e}^{i \left (d x +c \right )}+3 a b \,{\mathrm e}^{i \left (d x +c \right )}-b^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} a \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +b \right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{d \,b^{3}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d \,b^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{d \,b^{3}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d \,b^{2}}+\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{d \,b^{3}}-\frac {3 \sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{4 a d \,b^{2}}-\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{4 a^{2} d b}-\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{d \,b^{3}}+\frac {3 \sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{4 a d \,b^{2}}+\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{4 a^{2} d b}\) | \(682\) |
1/d*(-1/4/b^2/(sin(d*x+c)+1)+1/4/b^3*(-4*a+5*b)*ln(sin(d*x+c)+1)-1/4/b^2/( sin(d*x+c)-1)+1/4*(4*a-5*b)/b^3*ln(sin(d*x+c)-1)-(a^2-2*a*b+b^2)/b^3*(1/2* b/a*sin(d*x+c)/(sin(d*x+c)^2*a-b*sin(d*x+c)^2-a)-1/2*(4*a+b)/a/(a*(a-b))^( 1/2)*arctanh((a-b)*sin(d*x+c)/(a*(a-b))^(1/2))))
Time = 0.37 (sec) , antiderivative size = 635, normalized size of antiderivative = 3.80 \[ \int \frac {\sec ^7(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [-\frac {{\left ({\left (4 \, a^{3} - 7 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a - b}{a}} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, a \sqrt {\frac {a - b}{a}} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) + {\left ({\left (4 \, a^{3} - 9 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (4 \, a^{3} - 9 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a b^{2} + {\left (2 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a b^{4} d \cos \left (d x + c\right )^{2} + {\left (a^{2} b^{3} - a b^{4}\right )} d \cos \left (d x + c\right )^{4}\right )}}, -\frac {2 \, {\left ({\left (4 \, a^{3} - 7 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {a - b}{a}} \arctan \left (\sqrt {-\frac {a - b}{a}} \sin \left (d x + c\right )\right ) + {\left ({\left (4 \, a^{3} - 9 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (4 \, a^{3} - 9 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a b^{2} + {\left (2 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a b^{4} d \cos \left (d x + c\right )^{2} + {\left (a^{2} b^{3} - a b^{4}\right )} d \cos \left (d x + c\right )^{4}\right )}}\right ] \]
[-1/4*(((4*a^3 - 7*a^2*b + 2*a*b^2 + b^3)*cos(d*x + c)^4 + (4*a^2*b - 3*a* b^2 - b^3)*cos(d*x + c)^2)*sqrt((a - b)/a)*log(-((a - b)*cos(d*x + c)^2 + 2*a*sqrt((a - b)/a)*sin(d*x + c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b)) + ((4*a^3 - 9*a^2*b + 5*a*b^2)*cos(d*x + c)^4 + (4*a^2*b - 5*a*b^2)*cos(d* x + c)^2)*log(sin(d*x + c) + 1) - ((4*a^3 - 9*a^2*b + 5*a*b^2)*cos(d*x + c )^4 + (4*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a*b^ 2 + (2*a^2*b - 3*a*b^2 + b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^4*d*cos(d *x + c)^2 + (a^2*b^3 - a*b^4)*d*cos(d*x + c)^4), -1/4*(2*((4*a^3 - 7*a^2*b + 2*a*b^2 + b^3)*cos(d*x + c)^4 + (4*a^2*b - 3*a*b^2 - b^3)*cos(d*x + c)^ 2)*sqrt(-(a - b)/a)*arctan(sqrt(-(a - b)/a)*sin(d*x + c)) + ((4*a^3 - 9*a^ 2*b + 5*a*b^2)*cos(d*x + c)^4 + (4*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*log(si n(d*x + c) + 1) - ((4*a^3 - 9*a^2*b + 5*a*b^2)*cos(d*x + c)^4 + (4*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a*b^2 + (2*a^2*b - 3 *a*b^2 + b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^4*d*cos(d*x + c)^2 + (a^2 *b^3 - a*b^4)*d*cos(d*x + c)^4)]
\[ \int \frac {\sec ^7(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\int \frac {\sec ^{7}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec ^7(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Time = 0.64 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^7(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (4 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{b^{3}} - \frac {{\left (4 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{b^{3}} - \frac {2 \, {\left (4 \, a^{3} - 7 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{\sqrt {-a^{2} + a b} a b^{3}} + \frac {2 \, {\left (2 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{3} + b^{2} \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right ) + 2 \, a b \sin \left (d x + c\right ) - b^{2} \sin \left (d x + c\right )\right )}}{{\left (a \sin \left (d x + c\right )^{4} - b \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{2} + b \sin \left (d x + c\right )^{2} + a\right )} a b^{2}}}{4 \, d} \]
-1/4*((4*a - 5*b)*log(abs(sin(d*x + c) + 1))/b^3 - (4*a - 5*b)*log(abs(sin (d*x + c) - 1))/b^3 - 2*(4*a^3 - 7*a^2*b + 2*a*b^2 + b^3)*arctan(-(a*sin(d *x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*a*b^3) + 2*( 2*a^2*sin(d*x + c)^3 - 3*a*b*sin(d*x + c)^3 + b^2*sin(d*x + c)^3 - 2*a^2*s in(d*x + c) + 2*a*b*sin(d*x + c) - b^2*sin(d*x + c))/((a*sin(d*x + c)^4 - b*sin(d*x + c)^4 - 2*a*sin(d*x + c)^2 + b*sin(d*x + c)^2 + a)*a*b^2))/d
Time = 15.51 (sec) , antiderivative size = 4304, normalized size of antiderivative = 25.77 \[ \int \frac {\sec ^7(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)*(2*a^2 - 2*a*b + b^2))/(a*b^2) - (tan(c/2 + (d*x)/2)^ 3*(2*a^2 - 6*a*b + b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)^7*(2*a^2 - 2*a*b + b^2))/(a*b^2) - (tan(c/2 + (d*x)/2)^5*(2*a^2 - 6*a*b + b^2))/(a*b^2))/(d*( a - tan(c/2 + (d*x)/2)^2*(4*a - 4*b) - tan(c/2 + (d*x)/2)^6*(4*a - 4*b) + tan(c/2 + (d*x)/2)^4*(6*a - 8*b) + a*tan(c/2 + (d*x)/2)^8)) - (atan((((4*a - 5*b)*((((256*(16*a*b^15 + 92*a^2*b^14 - 8*a^3*b^13 - 2236*a^4*b^12 + 76 8*a^5*b^11 + 18228*a^6*b^10 - 41560*a^7*b^9 + 37420*a^8*b^8 - 13552*a^9*b^ 7 + 64*a^10*b^6 + 768*a^11*b^5))/(a^3*b^10) + (((((256*(256*a^4*b^16 + 192 *a^5*b^15 - 1088*a^6*b^14 - 192*a^7*b^13 + 1600*a^8*b^12 - 768*a^9*b^11))/ (a^3*b^10) - (256*tan(c/2 + (d*x)/2)*(4*a - 5*b)*(1024*a^5*b^15 - 2304*a^6 *b^14 + 1664*a^7*b^13 - 384*a^8*b^12))/(a^3*b^11))*(4*a - 5*b))/(2*b^3) - (512*tan(c/2 + (d*x)/2)*(64*a^2*b^14 + 160*a^3*b^13 - 984*a^4*b^12 - 6560* a^5*b^11 + 28720*a^6*b^10 - 42400*a^7*b^9 + 29512*a^8*b^8 - 9664*a^9*b^7 + 1152*a^10*b^6))/(a^3*b^8))*(4*a - 5*b))/(2*b^3))*(4*a - 5*b))/(2*b^3) + ( 512*tan(c/2 + (d*x)/2)*(8*a*b^11 - 8960*a^11*b + 768*a^12 + b^12 + 396*a^2 *b^10 + 440*a^3*b^9 - 7144*a^4*b^8 + 6656*a^5*b^7 + 34712*a^6*b^6 - 106784 *a^7*b^5 + 138675*a^8*b^4 - 100016*a^9*b^3 + 41248*a^10*b^2))/(a^3*b^8))*1 i)/(2*b^3) - ((4*a - 5*b)*((((256*(16*a*b^15 + 92*a^2*b^14 - 8*a^3*b^13 - 2236*a^4*b^12 + 768*a^5*b^11 + 18228*a^6*b^10 - 41560*a^7*b^9 + 37420*a^8* b^8 - 13552*a^9*b^7 + 64*a^10*b^6 + 768*a^11*b^5))/(a^3*b^10) + (((((25...