\(\int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx\) [25]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 160 \[ \int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx=\frac {209 a^3 \log (1-\sin (c+d x))}{16 d}-\frac {a^3 \log (1+\sin (c+d x))}{16 d}+\frac {7 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {a^6}{6 d (a-a \sin (c+d x))^3}-\frac {13 a^5}{8 d (a-a \sin (c+d x))^2}+\frac {71 a^4}{8 d (a-a \sin (c+d x))} \] Output:

209/16*a^3*ln(1-sin(d*x+c))/d-1/16*a^3*ln(1+sin(d*x+c))/d+7*a^3*sin(d*x+c) 
/d+3/2*a^3*sin(d*x+c)^2/d+1/3*a^3*sin(d*x+c)^3/d+1/6*a^6/d/(a-a*sin(d*x+c) 
)^3-13/8*a^5/d/(a-a*sin(d*x+c))^2+71/8*a^4/d/(a-a*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.62 \[ \int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx=\frac {a^3 \left (627 \log (1-\sin (c+d x))-3 \log (1+\sin (c+d x))-\frac {8}{(-1+\sin (c+d x))^3}-\frac {78}{(-1+\sin (c+d x))^2}-\frac {426}{-1+\sin (c+d x)}+336 \sin (c+d x)+72 \sin ^2(c+d x)+16 \sin ^3(c+d x)\right )}{48 d} \] Input:

Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^7,x]
 

Output:

(a^3*(627*Log[1 - Sin[c + d*x]] - 3*Log[1 + Sin[c + d*x]] - 8/(-1 + Sin[c 
+ d*x])^3 - 78/(-1 + Sin[c + d*x])^2 - 426/(-1 + Sin[c + d*x]) + 336*Sin[c 
 + d*x] + 72*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3))/(48*d)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 99, 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.

\(\displaystyle \int \tan ^7(c+d x) (a \sin (c+d x)+a)^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^7 (a \sin (c+d x)+a)^3dx\)

\(\Big \downarrow \) 3186

\(\displaystyle \frac {\int \frac {a^7 \sin ^7(c+d x)}{(a-a \sin (c+d x))^4 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {a^6}{2 (a-a \sin (c+d x))^4}-\frac {13 a^5}{4 (a-a \sin (c+d x))^3}+\frac {71 a^4}{8 (a-a \sin (c+d x))^2}-\frac {209 a^3}{16 (a-a \sin (c+d x))}-\frac {a^3}{16 (\sin (c+d x) a+a)}+\sin ^2(c+d x) a^2+3 \sin (c+d x) a^2+7 a^2\right )d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a^6}{6 (a-a \sin (c+d x))^3}-\frac {13 a^5}{8 (a-a \sin (c+d x))^2}+\frac {71 a^4}{8 (a-a \sin (c+d x))}+\frac {1}{3} a^3 \sin ^3(c+d x)+\frac {3}{2} a^3 \sin ^2(c+d x)+7 a^3 \sin (c+d x)+\frac {209}{16} a^3 \log (a-a \sin (c+d x))-\frac {1}{16} a^3 \log (a \sin (c+d x)+a)}{d}\)

Input:

Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^7,x]
 

Output:

((209*a^3*Log[a - a*Sin[c + d*x]])/16 - (a^3*Log[a + a*Sin[c + d*x]])/16 + 
 7*a^3*Sin[c + d*x] + (3*a^3*Sin[c + d*x]^2)/2 + (a^3*Sin[c + d*x]^3)/3 + 
a^6/(6*(a - a*Sin[c + d*x])^3) - (13*a^5)/(8*(a - a*Sin[c + d*x])^2) + (71 
*a^4)/(8*(a - a*Sin[c + d*x])))/d
 

Defintions of rubi rules used

rule 99
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3186
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) 
^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E 
qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 18.80 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.62

method result size
risch \(-13 i a^{3} x +\frac {i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {3 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {29 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {29 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {3 a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {26 i a^{3} c}{d}-\frac {i \left (213 a^{3} {\mathrm e}^{i \left (d x +c \right )}+774 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-1138 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-774 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+213 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}\right )}{12 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {209 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) \(260\)
derivativedivides \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{11}}{6 \cos \left (d x +c \right )^{6}}-\frac {5 \sin \left (d x +c \right )^{11}}{24 \cos \left (d x +c \right )^{4}}+\frac {35 \sin \left (d x +c \right )^{11}}{48 \cos \left (d x +c \right )^{2}}+\frac {35 \sin \left (d x +c \right )^{9}}{48}+\frac {15 \sin \left (d x +c \right )^{7}}{16}+\frac {21 \sin \left (d x +c \right )^{5}}{16}+\frac {35 \sin \left (d x +c \right )^{3}}{16}+\frac {105 \sin \left (d x +c \right )}{16}-\frac {105 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 a^{3} \left (\frac {\sin \left (d x +c \right )^{10}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{10}}{6 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{10}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{8}}{2}+\frac {2 \sin \left (d x +c \right )^{6}}{3}+\sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{2}+4 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin \left (d x +c \right )^{9}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{4}}+\frac {5 \sin \left (d x +c \right )^{9}}{16 \cos \left (d x +c \right )^{2}}+\frac {5 \sin \left (d x +c \right )^{7}}{16}+\frac {7 \sin \left (d x +c \right )^{5}}{16}+\frac {35 \sin \left (d x +c \right )^{3}}{48}+\frac {35 \sin \left (d x +c \right )}{16}-\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a^{3} \left (\frac {\tan \left (d x +c \right )^{6}}{6}-\frac {\tan \left (d x +c \right )^{4}}{4}+\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) \(392\)
default \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{11}}{6 \cos \left (d x +c \right )^{6}}-\frac {5 \sin \left (d x +c \right )^{11}}{24 \cos \left (d x +c \right )^{4}}+\frac {35 \sin \left (d x +c \right )^{11}}{48 \cos \left (d x +c \right )^{2}}+\frac {35 \sin \left (d x +c \right )^{9}}{48}+\frac {15 \sin \left (d x +c \right )^{7}}{16}+\frac {21 \sin \left (d x +c \right )^{5}}{16}+\frac {35 \sin \left (d x +c \right )^{3}}{16}+\frac {105 \sin \left (d x +c \right )}{16}-\frac {105 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 a^{3} \left (\frac {\sin \left (d x +c \right )^{10}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{10}}{6 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{10}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{8}}{2}+\frac {2 \sin \left (d x +c \right )^{6}}{3}+\sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{2}+4 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin \left (d x +c \right )^{9}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{4}}+\frac {5 \sin \left (d x +c \right )^{9}}{16 \cos \left (d x +c \right )^{2}}+\frac {5 \sin \left (d x +c \right )^{7}}{16}+\frac {7 \sin \left (d x +c \right )^{5}}{16}+\frac {35 \sin \left (d x +c \right )^{3}}{48}+\frac {35 \sin \left (d x +c \right )}{16}-\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a^{3} \left (\frac {\tan \left (d x +c \right )^{6}}{6}-\frac {\tan \left (d x +c \right )^{4}}{4}+\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) \(392\)
parts \(\frac {a^{3} \left (\frac {\tan \left (d x +c \right )^{6}}{6}-\frac {\tan \left (d x +c \right )^{4}}{4}+\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{11}}{6 \cos \left (d x +c \right )^{6}}-\frac {5 \sin \left (d x +c \right )^{11}}{24 \cos \left (d x +c \right )^{4}}+\frac {35 \sin \left (d x +c \right )^{11}}{48 \cos \left (d x +c \right )^{2}}+\frac {35 \sin \left (d x +c \right )^{9}}{48}+\frac {15 \sin \left (d x +c \right )^{7}}{16}+\frac {21 \sin \left (d x +c \right )^{5}}{16}+\frac {35 \sin \left (d x +c \right )^{3}}{16}+\frac {105 \sin \left (d x +c \right )}{16}-\frac {105 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}+\frac {3 a^{3} \left (\frac {\sin \left (d x +c \right )^{9}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{4}}+\frac {5 \sin \left (d x +c \right )^{9}}{16 \cos \left (d x +c \right )^{2}}+\frac {5 \sin \left (d x +c \right )^{7}}{16}+\frac {7 \sin \left (d x +c \right )^{5}}{16}+\frac {35 \sin \left (d x +c \right )^{3}}{48}+\frac {35 \sin \left (d x +c \right )}{16}-\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}+\frac {3 a^{3} \left (\frac {\sin \left (d x +c \right )^{10}}{6 \cos \left (d x +c \right )^{6}}-\frac {\sin \left (d x +c \right )^{10}}{6 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{10}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{8}}{2}+\frac {2 \sin \left (d x +c \right )^{6}}{3}+\sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{2}+4 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) \(406\)

Input:

int((a+a*sin(d*x+c))^3*tan(d*x+c)^7,x,method=_RETURNVERBOSE)
 

Output:

-13*I*a^3*x+1/24*I*a^3/d*exp(3*I*(d*x+c))-3/8*a^3/d*exp(2*I*(d*x+c))-29/8* 
I*a^3/d*exp(I*(d*x+c))+29/8*I*a^3/d*exp(-I*(d*x+c))-3/8*a^3/d*exp(-2*I*(d* 
x+c))-1/24*I*a^3/d*exp(-3*I*(d*x+c))-26*I/d*a^3*c-1/12*I*(213*a^3*exp(I*(d 
*x+c))+774*I*a^3*exp(2*I*(d*x+c))-1138*a^3*exp(3*I*(d*x+c))-774*I*a^3*exp( 
4*I*(d*x+c))+213*a^3*exp(5*I*(d*x+c)))/(exp(I*(d*x+c))-I)^6/d-1/8*a^3/d*ln 
(exp(I*(d*x+c))+I)+209/8*a^3/d*ln(exp(I*(d*x+c))-I)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.50 \[ \int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx=-\frac {16 \, a^{3} \cos \left (d x + c\right )^{6} - 216 \, a^{3} \cos \left (d x + c\right )^{4} + 1002 \, a^{3} \cos \left (d x + c\right )^{2} - 482 \, a^{3} + 3 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 627 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (12 \, a^{3} \cos \left (d x + c\right )^{4} + 398 \, a^{3} \cos \left (d x + c\right )^{2} - 245 \, a^{3}\right )} \sin \left (d x + c\right )}{48 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \] Input:

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^7,x, algorithm="fricas")
 

Output:

-1/48*(16*a^3*cos(d*x + c)^6 - 216*a^3*cos(d*x + c)^4 + 1002*a^3*cos(d*x + 
 c)^2 - 482*a^3 + 3*(3*a^3*cos(d*x + c)^2 - 4*a^3 - (a^3*cos(d*x + c)^2 - 
4*a^3)*sin(d*x + c))*log(sin(d*x + c) + 1) - 627*(3*a^3*cos(d*x + c)^2 - 4 
*a^3 - (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 
 2*(12*a^3*cos(d*x + c)^4 + 398*a^3*cos(d*x + c)^2 - 245*a^3)*sin(d*x + c) 
)/(3*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 4*d)*sin(d*x + c) - 4*d)
 

Sympy [F]

\[ \int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx=a^{3} \left (\int 3 \sin {\left (c + d x \right )} \tan ^{7}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan ^{7}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{7}{\left (c + d x \right )}\, dx + \int \tan ^{7}{\left (c + d x \right )}\, dx\right ) \] Input:

integrate((a+a*sin(d*x+c))**3*tan(d*x+c)**7,x)
 

Output:

a**3*(Integral(3*sin(c + d*x)*tan(c + d*x)**7, x) + Integral(3*sin(c + d*x 
)**2*tan(c + d*x)**7, x) + Integral(sin(c + d*x)**3*tan(c + d*x)**7, x) + 
Integral(tan(c + d*x)**7, x))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.83 \[ \int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx=\frac {16 \, a^{3} \sin \left (d x + c\right )^{3} + 72 \, a^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 627 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 336 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, {\left (213 \, a^{3} \sin \left (d x + c\right )^{2} - 387 \, a^{3} \sin \left (d x + c\right ) + 178 \, a^{3}\right )}}{\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \] Input:

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^7,x, algorithm="maxima")
 

Output:

1/48*(16*a^3*sin(d*x + c)^3 + 72*a^3*sin(d*x + c)^2 - 3*a^3*log(sin(d*x + 
c) + 1) + 627*a^3*log(sin(d*x + c) - 1) + 336*a^3*sin(d*x + c) - 2*(213*a^ 
3*sin(d*x + c)^2 - 387*a^3*sin(d*x + c) + 178*a^3)/(sin(d*x + c)^3 - 3*sin 
(d*x + c)^2 + 3*sin(d*x + c) - 1))/d
 

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.71 \[ \int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx=-\frac {1}{48} \, a^{3} {\left (\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} - \frac {627 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {8 \, {\left (2 \, d^{2} \sin \left (d x + c\right )^{3} + 9 \, d^{2} \sin \left (d x + c\right )^{2} + 42 \, d^{2} \sin \left (d x + c\right )\right )}}{d^{3}} + \frac {2 \, {\left (213 \, \sin \left (d x + c\right )^{2} - 387 \, \sin \left (d x + c\right ) + 178\right )}}{d {\left (\sin \left (d x + c\right ) - 1\right )}^{3}}\right )} \] Input:

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^7,x, algorithm="giac")
 

Output:

-1/48*a^3*(3*log(abs(sin(d*x + c) + 1))/d - 627*log(abs(sin(d*x + c) - 1)) 
/d - 8*(2*d^2*sin(d*x + c)^3 + 9*d^2*sin(d*x + c)^2 + 42*d^2*sin(d*x + c)) 
/d^3 + 2*(213*sin(d*x + c)^2 - 387*sin(d*x + c) + 178)/(d*(sin(d*x + c) - 
1)^3))
 

Mupad [B] (verification not implemented)

Time = 17.78 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.49 \[ \int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx=\frac {\frac {105\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{4}-\frac {263\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{2}+\frac {1301\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}-582\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {1657\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-\frac {2767\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {1657\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-582\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {1301\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {263\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {105\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{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 )}^{11}+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-38\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+92\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-38\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}+\frac {209\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}-\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}-\frac {13\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:

int(tan(c + d*x)^7*(a + a*sin(c + d*x))^3,x)
 

Output:

((1301*a^3*tan(c/2 + (d*x)/2)^3)/4 - (263*a^3*tan(c/2 + (d*x)/2)^2)/2 - 58 
2*a^3*tan(c/2 + (d*x)/2)^4 + (1657*a^3*tan(c/2 + (d*x)/2)^5)/2 - (2767*a^3 
*tan(c/2 + (d*x)/2)^6)/3 + (1657*a^3*tan(c/2 + (d*x)/2)^7)/2 - 582*a^3*tan 
(c/2 + (d*x)/2)^8 + (1301*a^3*tan(c/2 + (d*x)/2)^9)/4 - (263*a^3*tan(c/2 + 
 (d*x)/2)^10)/2 + (105*a^3*tan(c/2 + (d*x)/2)^11)/4 + (105*a^3*tan(c/2 + ( 
d*x)/2))/4)/(d*(18*tan(c/2 + (d*x)/2)^2 - 6*tan(c/2 + (d*x)/2) - 38*tan(c/ 
2 + (d*x)/2)^3 + 63*tan(c/2 + (d*x)/2)^4 - 84*tan(c/2 + (d*x)/2)^5 + 92*ta 
n(c/2 + (d*x)/2)^6 - 84*tan(c/2 + (d*x)/2)^7 + 63*tan(c/2 + (d*x)/2)^8 - 3 
8*tan(c/2 + (d*x)/2)^9 + 18*tan(c/2 + (d*x)/2)^10 - 6*tan(c/2 + (d*x)/2)^1 
1 + tan(c/2 + (d*x)/2)^12 + 1)) + (209*a^3*log(tan(c/2 + (d*x)/2) - 1))/(8 
*d) - (a^3*log(tan(c/2 + (d*x)/2) + 1))/(8*d) - (13*a^3*log(tan(c/2 + (d*x 
)/2)^2 + 1))/d
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 645, normalized size of antiderivative = 4.03 \[ \int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+a*sin(d*x+c))^3*tan(d*x+c)^7,x)
 

Output:

(a**3*( - 12*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**6 + 36*log(tan(c + d*x 
)**2 + 1)*sin(c + d*x)**4 - 36*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**2 + 
12*log(tan(c + d*x)**2 + 1) - 288*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x 
)**6 + 864*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4 - 864*log(tan((c + 
 d*x)/2)**2 + 1)*sin(c + d*x)**2 + 288*log(tan((c + d*x)/2)**2 + 1) + 603* 
log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 1809*log(tan((c + d*x)/2) - 1) 
*sin(c + d*x)**4 + 1809*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 603*lo 
g(tan((c + d*x)/2) - 1) - 27*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 + 8 
1*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 81*log(tan((c + d*x)/2) + 1) 
*sin(c + d*x)**2 + 27*log(tan((c + d*x)/2) + 1) + 8*sin(c + d*x)**9 + 36*s 
in(c + d*x)**8 + 144*sin(c + d*x)**7 + 4*sin(c + d*x)**6*tan(c + d*x)**6 - 
 6*sin(c + d*x)**6*tan(c + d*x)**4 + 12*sin(c + d*x)**6*tan(c + d*x)**2 - 
264*sin(c + d*x)**6 - 693*sin(c + d*x)**5 - 12*sin(c + d*x)**4*tan(c + d*x 
)**6 + 18*sin(c + d*x)**4*tan(c + d*x)**4 - 36*sin(c + d*x)**4*tan(c + d*x 
)**2 + 360*sin(c + d*x)**4 + 840*sin(c + d*x)**3 + 12*sin(c + d*x)**2*tan( 
c + d*x)**6 - 18*sin(c + d*x)**2*tan(c + d*x)**4 + 36*sin(c + d*x)**2*tan( 
c + d*x)**2 - 144*sin(c + d*x)**2 - 315*sin(c + d*x) - 4*tan(c + d*x)**6 + 
 6*tan(c + d*x)**4 - 12*tan(c + d*x)**2))/(24*d*(sin(c + d*x)**6 - 3*sin(c 
 + d*x)**4 + 3*sin(c + d*x)**2 - 1))