Integrand size = 23, antiderivative size = 113 \[ \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {\left (8 a^2-4 a b+b^2\right ) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\left (8 a^2-4 a b+b^2\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(12 a-7 b) b \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {b^2 \sec ^5(c+d x) \tan (c+d x)}{6 d} \] Output:
1/16*(8*a^2-4*a*b+b^2)*arctanh(sin(d*x+c))/d+1/16*(8*a^2-4*a*b+b^2)*sec(d* x+c)*tan(d*x+c)/d+1/24*(12*a-7*b)*b*sec(d*x+c)^3*tan(d*x+c)/d+1/6*b^2*sec( d*x+c)^5*tan(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 9.19 (sec) , antiderivative size = 875, normalized size of antiderivative = 7.74 \[ \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx =\text {Too large to display} \] Input:
Integrate[Sec[c + d*x]^3*(a + b*Tan[c + d*x]^2)^2,x]
Output:
(Sin[c + d*x]*(65625*a^2*ArcTanh[Sqrt[Sin[c + d*x]^2]] - 36855*a^2*ArcTanh [Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^2 - 91875*a*(a - b)*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^2 + 1680*a^2*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^4 + 54180*a*(a - b)*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^4 + 32970*(a - b)^2*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^4 - 1365*a*(a - b)*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^6 - 19845*(a - b)^2*ArcTanh [Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^6 + 525*(a - b)^2*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^8 - 65625*a^2*Sqrt[Sin[c + d*x]^2] - 23555*a*(a - b )*Sin[c + d*x]^4*Sqrt[Sin[c + d*x]^2] - 32970*(a - b)^2*Sin[c + d*x]^4*Sqr t[Sin[c + d*x]^2] + 8855*(a - b)^2*Sin[c + d*x]^6*Sqrt[Sin[c + d*x]^2] + 6 20*a^2*HypergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, Sin[c + d*x]^2]*Sin[ c + d*x]^6*Sqrt[Sin[c + d*x]^2] + 160*a^2*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^6*Sqrt[Sin[c + d*x]^2] + 16*a^2*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^6*Sqrt[Sin[c + d*x]^2] - 968*a*(a - b)*Hypergeometri cPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^8*Sqrt[Sin[ c + d*x]^2] - 288*a*(a - b)*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^8*Sqrt[Sin[c + d*x]^2] - 32*a*(a - b)* HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 9/2}, Sin[c + d*x]^2] *Sin[c + d*x]^8*Sqrt[Sin[c + d*x]^2] + 380*(a - b)^2*HypergeometricPFQ[...
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4159, 315, 25, 298, 215, 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.
| \(\displaystyle \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^3 \left (a+b \tan (c+d x)^2\right )^2dx\) |
| \(\Big \downarrow \) 4159 |
| \(\displaystyle \frac {\int \frac {\left (a-(a-b) \sin ^2(c+d x)\right )^2}{\left (1-\sin ^2(c+d x)\right )^4}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\frac {b \sin (c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{6 \left (1-\sin ^2(c+d x)\right )^3}-\frac {1}{6} \int -\frac {a (6 a-b)-3 (a-b) (2 a-b) \sin ^2(c+d x)}{\left (1-\sin ^2(c+d x)\right )^3}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {a (6 a-b)-3 (a-b) (2 a-b) \sin ^2(c+d x)}{\left (1-\sin ^2(c+d x)\right )^3}d\sin (c+d x)+\frac {b \sin (c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{6 \left (1-\sin ^2(c+d x)\right )^3}}{d}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} \left (8 a^2-4 a b+b^2\right ) \int \frac {1}{\left (1-\sin ^2(c+d x)\right )^2}d\sin (c+d x)+\frac {b (8 a-3 b) \sin (c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}\right )+\frac {b \sin (c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{6 \left (1-\sin ^2(c+d x)\right )^3}}{d}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} \left (8 a^2-4 a b+b^2\right ) \left (\frac {1}{2} \int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)+\frac {\sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )+\frac {b (8 a-3 b) \sin (c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}\right )+\frac {b \sin (c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{6 \left (1-\sin ^2(c+d x)\right )^3}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} \left (8 a^2-4 a b+b^2\right ) \left (\frac {1}{2} \text {arctanh}(\sin (c+d x))+\frac {\sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )+\frac {b (8 a-3 b) \sin (c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}\right )+\frac {b \sin (c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{6 \left (1-\sin ^2(c+d x)\right )^3}}{d}\) |
Input:
Int[Sec[c + d*x]^3*(a + b*Tan[c + d*x]^2)^2,x]
Output:
((b*Sin[c + d*x]*(a - (a - b)*Sin[c + d*x]^2))/(6*(1 - Sin[c + d*x]^2)^3) + (((8*a - 3*b)*b*Sin[c + d*x])/(4*(1 - Sin[c + d*x]^2)^2) + (3*(8*a^2 - 4 *a*b + b^2)*(ArcTanh[Sin[c + d*x]]/2 + Sin[c + d*x]/(2*(1 - Sin[c + d*x]^2 ))))/4)/6)/d
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
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 = 4.17 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.76
| method | result | size |
| derivativedivides | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{5}}{24 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{48 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{48}-\frac {\sin \left (d x +c \right )}{16}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(199\) |
| default | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{5}}{24 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{48 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{48}-\frac {\sin \left (d x +c \right )}{16}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(199\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (24 a^{2} {\mathrm e}^{10 i \left (d x +c \right )}-12 a b \,{\mathrm e}^{10 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+72 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+60 a b \,{\mathrm e}^{8 i \left (d x +c \right )}-47 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+48 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+72 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+78 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-48 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-72 \,{\mathrm e}^{4 i \left (d x +c \right )} a b -78 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-72 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-60 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+47 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-24 a^{2}+12 a b -3 b^{2}\right )}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{4 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{16 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{4 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{16 d}\) | \(406\) |
Input:
int(sec(d*x+c)^3*(a+b*tan(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(b^2*(1/6*sin(d*x+c)^5/cos(d*x+c)^6+1/24*sin(d*x+c)^5/cos(d*x+c)^4-1/4 8*sin(d*x+c)^5/cos(d*x+c)^2-1/48*sin(d*x+c)^3-1/16*sin(d*x+c)+1/16*ln(sec( d*x+c)+tan(d*x+c)))+2*a*b*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/ cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c)))+a^2*(1/2*sec(d* x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21 \[ \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {3 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (12 \, a b - 7 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \] Input:
integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")
Output:
1/96*(3*(8*a^2 - 4*a*b + b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(8* a^2 - 4*a*b + b^2)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(3*(8*a^2 - 4 *a*b + b^2)*cos(d*x + c)^4 + 2*(12*a*b - 7*b^2)*cos(d*x + c)^2 + 8*b^2)*si n(d*x + c))/(d*cos(d*x + c)^6)
\[ \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**3*(a+b*tan(d*x+c)**2)**2,x)
Output:
Integral((a + b*tan(c + d*x)**2)**2*sec(c + d*x)**3, x)
Time = 0.04 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.38 \[ \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {3 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{5} - 8 \, {\left (6 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} + 4 \, a b - b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/96*(3*(8*a^2 - 4*a*b + b^2)*log(sin(d*x + c) + 1) - 3*(8*a^2 - 4*a*b + b ^2)*log(sin(d*x + c) - 1) - 2*(3*(8*a^2 - 4*a*b + b^2)*sin(d*x + c)^5 - 8* (6*a^2 - b^2)*sin(d*x + c)^3 + 3*(8*a^2 + 4*a*b - b^2)*sin(d*x + c))/(sin( d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1))/d
Time = 0.41 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.48 \[ \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {3 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (24 \, a^{2} \sin \left (d x + c\right )^{5} - 12 \, a b \sin \left (d x + c\right )^{5} + 3 \, b^{2} \sin \left (d x + c\right )^{5} - 48 \, a^{2} \sin \left (d x + c\right )^{3} + 8 \, b^{2} \sin \left (d x + c\right )^{3} + 24 \, a^{2} \sin \left (d x + c\right ) + 12 \, a b \sin \left (d x + c\right ) - 3 \, b^{2} \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/96*(3*(8*a^2 - 4*a*b + b^2)*log(abs(sin(d*x + c) + 1)) - 3*(8*a^2 - 4*a* b + b^2)*log(abs(sin(d*x + c) - 1)) - 2*(24*a^2*sin(d*x + c)^5 - 12*a*b*si n(d*x + c)^5 + 3*b^2*sin(d*x + c)^5 - 48*a^2*sin(d*x + c)^3 + 8*b^2*sin(d* x + c)^3 + 24*a^2*sin(d*x + c) + 12*a*b*sin(d*x + c) - 3*b^2*sin(d*x + c)) /(sin(d*x + c)^2 - 1)^3)/d
Time = 11.91 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.38 \[ \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {\left (a^2+\frac {a\,b}{2}-\frac {b^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-3\,a^2+\frac {5\,a\,b}{2}+\frac {17\,b^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,a^2-3\,a\,b+\frac {19\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (2\,a^2-3\,a\,b+\frac {19\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-3\,a^2+\frac {5\,a\,b}{2}+\frac {17\,b^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a^2+\frac {a\,b}{2}-\frac {b^2}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-\frac {a\,b}{2}+\frac {b^2}{8}\right )}{d} \] Input:
int((a + b*tan(c + d*x)^2)^2/cos(c + d*x)^3,x)
Output:
(tan(c/2 + (d*x)/2)^5*(2*a^2 - 3*a*b + (19*b^2)/4) + tan(c/2 + (d*x)/2)^7* (2*a^2 - 3*a*b + (19*b^2)/4) + tan(c/2 + (d*x)/2)^3*((5*a*b)/2 - 3*a^2 + ( 17*b^2)/24) + tan(c/2 + (d*x)/2)^9*((5*a*b)/2 - 3*a^2 + (17*b^2)/24) + tan (c/2 + (d*x)/2)*((a*b)/2 + a^2 - b^2/8) + tan(c/2 + (d*x)/2)^11*((a*b)/2 + a^2 - b^2/8))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*t an(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (atanh(tan(c/2 + (d*x)/2))*(a^2 - (a*b)/2 + b^2/8))/d
Time = 0.21 (sec) , antiderivative size = 678, normalized size of antiderivative = 6.00 \[ \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3*(a+b*tan(d*x+c)^2)^2,x)
Output:
( - 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a**2 + 12*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**6*a*b - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)* *6*b**2 + 72*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 - 36*log(tan(( c + d*x)/2) - 1)*sin(c + d*x)**4*a*b + 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**2 - 72*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + 36*log (tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b - 9*log(tan((c + d*x)/2) - 1)*s in(c + d*x)**2*b**2 + 24*log(tan((c + d*x)/2) - 1)*a**2 - 12*log(tan((c + d*x)/2) - 1)*a*b + 3*log(tan((c + d*x)/2) - 1)*b**2 + 24*log(tan((c + d*x) /2) + 1)*sin(c + d*x)**6*a**2 - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* *6*a*b + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*b**2 - 72*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2 + 36*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b - 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**2 + 72*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 - 36*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**2*a*b + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 - 24 *log(tan((c + d*x)/2) + 1)*a**2 + 12*log(tan((c + d*x)/2) + 1)*a*b - 3*log (tan((c + d*x)/2) + 1)*b**2 - 24*sin(c + d*x)**5*a**2 + 12*sin(c + d*x)**5 *a*b - 3*sin(c + d*x)**5*b**2 + 48*sin(c + d*x)**3*a**2 - 8*sin(c + d*x)** 3*b**2 - 24*sin(c + d*x)*a**2 - 12*sin(c + d*x)*a*b + 3*sin(c + d*x)*b**2) /(48*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin(c + d*x)**2 - 1))