Integrand size = 19, antiderivative size = 90 \[ \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {a \left (2 a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b \left (3 a^2-b^2\right ) \sec (c+d x)}{d}+\frac {b^3 \sec ^3(c+d x)}{3 d}+\frac {3 a b^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*a*(2*a^2-3*b^2)*arctanh(sin(d*x+c))/d+b*(3*a^2-b^2)*sec(d*x+c)/d+1/3*b ^3*sec(d*x+c)^3/d+3/2*a*b^2*sec(d*x+c)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(293\) vs. \(2(90)=180\).
Time = 1.51 (sec) , antiderivative size = 293, normalized size of antiderivative = 3.26 \[ \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {36 a^2 b-10 b^3-6 a \left (2 a^2-3 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-18 a b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {9 a b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+2 b \left (18 a^2-b^2+2 b^2 \cos (c+d x)+\left (18 a^2-5 b^2\right ) \cos (2 (c+d x))\right ) \sec ^3(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )-\frac {9 a b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}}{12 d} \] Input:
Integrate[Sec[c + d*x]*(a + b*Tan[c + d*x])^3,x]
Output:
(36*a^2*b - 10*b^3 - 6*a*(2*a^2 - 3*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d *x)/2]] + 12*a^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 18*a*b^2*Log[C os[(c + d*x)/2] + Sin[(c + d*x)/2]] + (9*a*b^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + b^3/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 2*b*(18*a^2 - b^2 + 2*b^2*Cos[c + d*x] + (18*a^2 - 5*b^2)*Cos[2*(c + d*x)])*Sec[c + d* x]^3*Sin[(c + d*x)/2]^2 - (9*a*b^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^ 2 + b^3/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(12*d)
Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {3042, 3991, 3042, 4159, 27, 298, 219, 4861, 27, 244, 2009}
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.
| \(\displaystyle \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x) (a+b \tan (c+d x))^3dx\) |
| \(\Big \downarrow \) 3991 |
| \(\displaystyle \int \sec (c+d x) \left (a^3+3 b^2 \tan ^2(c+d x) a\right )dx+\int \sec (c+d x) \tan (c+d x) \left (\tan ^2(c+d x) b^3+3 a^2 b\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x) \left (a^3+3 b^2 \tan (c+d x)^2 a\right )dx+\int \sec (c+d x) \tan (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )dx\) |
| \(\Big \downarrow \) 4159 |
| \(\displaystyle \int \sec (c+d x) \tan (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )dx+\frac {\int \frac {a \left (a^2-\left (a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{\left (1-\sin ^2(c+d x)\right )^2}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \sec (c+d x) \tan (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )dx+\frac {a \int \frac {a^2-\left (a^2-3 b^2\right ) \sin ^2(c+d x)}{\left (1-\sin ^2(c+d x)\right )^2}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \int \sec (c+d x) \tan (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )dx+\frac {a \left (\frac {1}{2} \left (2 a^2-3 b^2\right ) \int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)+\frac {3 b^2 \sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \int \sec (c+d x) \tan (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )dx+\frac {a \left (\frac {1}{2} \left (2 a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {3 b^2 \sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )}{d}\) |
| \(\Big \downarrow \) 4861 |
| \(\displaystyle \frac {a \left (\frac {1}{2} \left (2 a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {3 b^2 \sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )}{d}-\frac {\int b \left (b^2+\left (3 a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \left (\frac {1}{2} \left (2 a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {3 b^2 \sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )}{d}-\frac {b \int \left (b^2+\left (3 a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \frac {a \left (\frac {1}{2} \left (2 a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {3 b^2 \sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )}{d}-\frac {b \int \left (b^2 \sec ^4(c+d x)+\left (3 a^2-b^2\right ) \sec ^2(c+d x)\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {1}{2} \left (2 a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {3 b^2 \sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )}{d}-\frac {b \left (-\left (3 a^2-b^2\right ) \sec (c+d x)-\frac {1}{3} b^2 \sec ^3(c+d x)\right )}{d}\) |
Input:
Int[Sec[c + d*x]*(a + b*Tan[c + d*x])^3,x]
Output:
-((b*(-((3*a^2 - b^2)*Sec[c + d*x]) - (b^2*Sec[c + d*x]^3)/3))/d) + (a*((( 2*a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/2 + (3*b^2*Sin[c + d*x])/(2*(1 - Sin [c + d*x]^2))))/d
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 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[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Module[{k}, Int[Sec[e + f*x]^m*Sum[Binomial[n, 2*k]*a^(n - 2*k)*b^(2*k)*Tan[e + f*x]^(2*k), {k, 0, n/2}], x] + Int[Sec[e + f*x]^m*Tan [e + f*x]*Sum[Binomial[n, 2*k + 1]*a^(n - 2*k - 1)*b^(2*k + 1)*Tan[e + f*x] ^(2*k), {k, 0, (n - 1)/2}], x]] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]
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]
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Cos[c*(a + b*x)], x]}, Simp[-(b*c)^(-1) Subst[Int[SubstFor[1/x, Cos[c* (a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])
Time = 2.41 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62
| method | result | size |
| derivativedivides | \(\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {3 a^{2} b}{\cos \left (d x +c \right )}+a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(146\) |
| default | \(\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {3 a^{2} b}{\cos \left (d x +c \right )}+a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(146\) |
| risch | \(-\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (9 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-18 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-36 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 i a b -18 a^{2}+6 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{2 d}\) | \(213\) |
Input:
int(sec(d*x+c)*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(b^3*(1/3*sin(d*x+c)^4/cos(d*x+c)^3-1/3*sin(d*x+c)^4/cos(d*x+c)-1/3*(2 +sin(d*x+c)^2)*cos(d*x+c))+3*a*b^2*(1/2*sin(d*x+c)^3/cos(d*x+c)^2+1/2*sin( d*x+c)-1/2*ln(sec(d*x+c)+tan(d*x+c)))+3*a^2*b/cos(d*x+c)+a^3*ln(sec(d*x+c) +tan(d*x+c)))
Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.37 \[ \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 18 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, b^{3} + 12 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(sec(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/12*(3*(2*a^3 - 3*a*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*a^3 - 3*a*b^2)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 18*a*b^2*cos(d*x + c)*s in(d*x + c) + 4*b^3 + 12*(3*a^2*b - b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)^3 )
\[ \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(a+b*tan(d*x+c))**3,x)
Output:
Integral((a + b*tan(c + d*x))**3*sec(c + d*x), x)
Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.23 \[ \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {9 \, a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - \frac {36 \, a^{2} b}{\cos \left (d x + c\right )} + \frac {4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{3}}{\cos \left (d x + c\right )^{3}}}{12 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/12*(9*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1 ) - log(sin(d*x + c) - 1)) - 12*a^3*log(sec(d*x + c) + tan(d*x + c)) - 36* a^2*b/cos(d*x + c) + 4*(3*cos(d*x + c)^2 - 1)*b^3/cos(d*x + c)^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (84) = 168\).
Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.90 \[ \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, a^{2} b + 4 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/6*(3*(2*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*a^3 - 3 *a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(9*a*b^2*tan(1/2*d*x + 1/2* c)^5 - 18*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 36*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 12*b^3*tan(1/2*d*x + 1/2*c)^2 - 9*a*b^2*tan(1/2*d*x + 1/2*c) - 18*a^2*b + 4*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 2.64 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.78 \[ \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,a\,b^2-2\,a^3\right )}{d}-\frac {6\,a^2\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a^2\,b-4\,b^3\right )-\frac {4\,b^3}{3}+3\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + b*tan(c + d*x))^3/cos(c + d*x),x)
Output:
- (atanh(tan(c/2 + (d*x)/2))*(3*a*b^2 - 2*a^3))/d - (6*a^2*b - tan(c/2 + ( d*x)/2)^2*(12*a^2*b - 4*b^3) - (4*b^3)/3 + 3*a*b^2*tan(c/2 + (d*x)/2) + 6* a^2*b*tan(c/2 + (d*x)/2)^4 - 3*a*b^2*tan(c/2 + (d*x)/2)^5)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
Time = 0.18 (sec) , antiderivative size = 364, normalized size of antiderivative = 4.04 \[ \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {-6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3}+9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}+6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}-9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3}-9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}-6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}+9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-18 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b +4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{3}-9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}+18 \cos \left (d x +c \right ) a^{2} b -4 \cos \left (d x +c \right ) b^{3}+18 \sin \left (d x +c \right )^{2} a^{2} b -6 \sin \left (d x +c \right )^{2} b^{3}-18 a^{2} b +4 b^{3}}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)*(a+b*tan(d*x+c))^3,x)
Output:
( - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 + 9*cos( c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2 + 6*cos(c + d*x) *log(tan((c + d*x)/2) - 1)*a**3 - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1) *a*b**2 + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 - 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**2 - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3 + 9*cos(c + d*x)*log(tan((c + d*x)/2 ) + 1)*a*b**2 - 18*cos(c + d*x)*sin(c + d*x)**2*a**2*b + 4*cos(c + d*x)*si n(c + d*x)**2*b**3 - 9*cos(c + d*x)*sin(c + d*x)*a*b**2 + 18*cos(c + d*x)* a**2*b - 4*cos(c + d*x)*b**3 + 18*sin(c + d*x)**2*a**2*b - 6*sin(c + d*x)* *2*b**3 - 18*a**2*b + 4*b**3)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))