\(\int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx\) [1501]

Optimal result

Integrand size = 27, antiderivative size = 88 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {3 b \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {3 b \sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d} \] Output:

-3/8*b*(a^2-b^2)*arctanh(sin(d*x+c))/d-3/8*b*sec(d*x+c)^2*(b+a*sin(d*x+c)) 
*(a+b*sin(d*x+c))/d+1/4*sec(d*x+c)^4*(a+b*sin(d*x+c))^3/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(370\) vs. \(2(88)=176\).

Time = 1.69 (sec) , antiderivative size = 370, normalized size of antiderivative = 4.20 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {2 a \left (a^2-b^2\right )^2 \sec ^4(c+d x) (a+b \sin (c+d x))^4+2 b \left (a^2-b^2\right ) \sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^5+b \sec ^2(c+d x) (a+b \sin (c+d x))^5 \left (5 a^2 b+b^3-3 a \left (a^2+b^2\right ) \sin (c+d x)\right )+\frac {1}{2} \left (5 a^4 b+10 a^2 b^3+b^5\right ) \left (3 \left ((a+b)^4 \log (1-\sin (c+d x))-(a-b)^4 \log (1+\sin (c+d x))\right )+6 b^2 \left (6 a^2+b^2\right ) \sin (c+d x)+12 a b^3 \sin ^2(c+d x)+2 b^4 \sin ^3(c+d x)\right )-a b \left (a^2+b^2\right ) \left (6 (a+b)^5 \log (1-\sin (c+d x))-6 (a-b)^5 \log (1+\sin (c+d x))+60 a b^2 \left (2 a^2+b^2\right ) \sin (c+d x)+6 b^3 \left (10 a^2+b^2\right ) \sin ^2(c+d x)+20 a b^4 \sin ^3(c+d x)+3 b^5 \sin ^4(c+d x)\right )}{8 \left (a^2-b^2\right )^3 d} \] Input:

Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3*Tan[c + d*x],x]
 

Output:

(2*a*(a^2 - b^2)^2*Sec[c + d*x]^4*(a + b*Sin[c + d*x])^4 + 2*b*(a^2 - b^2) 
*Sec[c + d*x]^4*(b - a*Sin[c + d*x])*(a + b*Sin[c + d*x])^5 + b*Sec[c + d* 
x]^2*(a + b*Sin[c + d*x])^5*(5*a^2*b + b^3 - 3*a*(a^2 + b^2)*Sin[c + d*x]) 
 + ((5*a^4*b + 10*a^2*b^3 + b^5)*(3*((a + b)^4*Log[1 - Sin[c + d*x]] - (a 
- b)^4*Log[1 + Sin[c + d*x]]) + 6*b^2*(6*a^2 + b^2)*Sin[c + d*x] + 12*a*b^ 
3*Sin[c + d*x]^2 + 2*b^4*Sin[c + d*x]^3))/2 - a*b*(a^2 + b^2)*(6*(a + b)^5 
*Log[1 - Sin[c + d*x]] - 6*(a - b)^5*Log[1 + Sin[c + d*x]] + 60*a*b^2*(2*a 
^2 + b^2)*Sin[c + d*x] + 6*b^3*(10*a^2 + b^2)*Sin[c + d*x]^2 + 20*a*b^4*Si 
n[c + d*x]^3 + 3*b^5*Sin[c + d*x]^4))/(8*(a^2 - b^2)^3*d)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3316, 27, 531, 27, 477, 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.

\(\displaystyle \int \tan (c+d x) \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x) (a+b \sin (c+d x))^3}{\cos (c+d x)^5}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {b^5 \int \frac {\sin (c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^4 \int \frac {b \sin (c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 531

\(\displaystyle \frac {b^4 \left (\frac {\int -\frac {3 b^2 (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {(a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^4 \left (\frac {(a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {3}{4} \int \frac {(a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))\right )}{d}\)

\(\Big \downarrow \) 477

\(\displaystyle \frac {b^4 \left (\frac {(a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {3 \int \left (\frac {\left (a^2-b^2\right ) b^2}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {(a+b)^2 b^2}{4 (b-b \sin (c+d x))^2}+\frac {(a-b)^2 b^2}{4 (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{4 b^4}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^4 \left (\frac {(a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {3 \left (\frac {1}{2} b \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {b^2 (a+b)^2}{4 (b-b \sin (c+d x))}-\frac {b^2 (a-b)^2}{4 (b \sin (c+d x)+b)}\right )}{4 b^4}\right )}{d}\)

Input:

Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3*Tan[c + d*x],x]
 

Output:

(b^4*((a + b*Sin[c + d*x])^3/(4*(b^2 - b^2*Sin[c + d*x]^2)^2) - (3*((b*(a^ 
2 - b^2)*ArcTanh[Sin[c + d*x]])/2 + (b^2*(a + b)^2)/(4*(b - b*Sin[c + d*x] 
)) - ((a - b)^2*b^2)/(4*(b + b*Sin[c + d*x]))))/(4*b^4)))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 477

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
a^p   Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 
]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & 
& NiceSqrtQ[-b/a] &&  !FractionalPowerFactorQ[Rt[-b/a, 2]]
 

rule 531

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi 
alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
 + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 
2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b 
*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p 
 + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt 
Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
 

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 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(182\) vs. \(2(82)=164\).

Time = 4.88 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.08

method	result	size
derivativedivides	\(\frac {\frac {a^{3}}{4 \cos \left (d x +c \right )^{4}}+3 a^{2} b \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a \,b^{2} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+b^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(183\)
default	\(\frac {\frac {a^{3}}{4 \cos \left (d x +c \right )^{4}}+3 a^{2} b \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a \,b^{2} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+b^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(183\)
risch	\(\frac {{\mathrm e}^{i \left (d x +c \right )} \left (-24 a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+3 i a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+5 i b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+16 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-21 i a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 i b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-24 \,{\mathrm e}^{i \left (d x +c \right )} a \,b^{2}+21 i a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 i a^{2} b -5 i b^{3}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{8 d}\)	\(284\)

Input:

int(sec(d*x+c)^4*(a+b*sin(d*x+c))^3*tan(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/4*a^3/cos(d*x+c)^4+3*a^2*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)))+3/4*a*b 
^2*sin(d*x+c)^4/cos(d*x+c)^4+b^3*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d* 
x+c)^5/cos(d*x+c)^2-1/8*sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan( 
d*x+c))))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.62 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {3 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 24 \, a b^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - 12 \, a b^{2} - 2 \, {\left (6 \, a^{2} b + 2 \, b^{3} - {\left (3 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^3*tan(d*x+c),x, algorithm="fricas" 
)
 

Output:

-1/16*(3*(a^2*b - b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(a^2*b - b 
^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 24*a*b^2*cos(d*x + c)^2 - 4*a^ 
3 - 12*a*b^2 - 2*(6*a^2*b + 2*b^3 - (3*a^2*b + 5*b^3)*cos(d*x + c)^2)*sin( 
d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**4*(a+b*sin(d*x+c))**3*tan(d*x+c),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.59 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {3 \, {\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (12 \, a b^{2} \sin \left (d x + c\right )^{2} + {\left (3 \, a^{2} b + 5 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, a^{3} - 6 \, a b^{2} + 3 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^3*tan(d*x+c),x, algorithm="maxima" 
)
 

Output:

-1/16*(3*(a^2*b - b^3)*log(sin(d*x + c) + 1) - 3*(a^2*b - b^3)*log(sin(d*x 
 + c) - 1) - 2*(12*a*b^2*sin(d*x + c)^2 + (3*a^2*b + 5*b^3)*sin(d*x + c)^3 
 + 2*a^3 - 6*a*b^2 + 3*(a^2*b - b^3)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin 
(d*x + c)^2 + 1))/d
 

Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.66 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {3 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} + \frac {3 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} + \frac {3 \, a^{2} b \sin \left (d x + c\right )^{3} + 5 \, b^{3} \sin \left (d x + c\right )^{3} + 12 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{2} b \sin \left (d x + c\right ) - 3 \, b^{3} \sin \left (d x + c\right ) + 2 \, a^{3} - 6 \, a b^{2}}{8 \, {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2} d} \] Input:

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^3*tan(d*x+c),x, algorithm="giac")
 

Output:

-3/16*(a^2*b - b^3)*log(abs(sin(d*x + c) + 1))/d + 3/16*(a^2*b - b^3)*log( 
abs(sin(d*x + c) - 1))/d + 1/8*(3*a^2*b*sin(d*x + c)^3 + 5*b^3*sin(d*x + c 
)^3 + 12*a*b^2*sin(d*x + c)^2 + 3*a^2*b*sin(d*x + c) - 3*b^3*sin(d*x + c) 
+ 2*a^3 - 6*a*b^2)/((sin(d*x + c)^2 - 1)^2*d)
 

Mupad [B] (verification not implemented)

Time = 21.14 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.59 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^2\,b}{4}-\frac {3\,b^3}{4}\right )+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {3\,a^2\,b}{4}-\frac {3\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {21\,a^2\,b}{4}+\frac {11\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {21\,a^2\,b}{4}+\frac {11\,b^3}{4}\right )+12\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-b^2\right )}{4\,d} \] Input:

int((tan(c + d*x)*(a + b*sin(c + d*x))^3)/cos(c + d*x)^4,x)
 

Output:

(tan(c/2 + (d*x)/2)*((3*a^2*b)/4 - (3*b^3)/4) + 2*a^3*tan(c/2 + (d*x)/2)^2 
 + 2*a^3*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^7*((3*a^2*b)/4 - (3*b^3 
)/4) + tan(c/2 + (d*x)/2)^3*((21*a^2*b)/4 + (11*b^3)/4) + tan(c/2 + (d*x)/ 
2)^5*((21*a^2*b)/4 + (11*b^3)/4) + 12*a*b^2*tan(c/2 + (d*x)/2)^4)/(d*(6*ta 
n(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan 
(c/2 + (d*x)/2)^8 + 1)) - (3*b*atanh(tan(c/2 + (d*x)/2))*(a^2 - b^2))/(4*d 
)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 392, normalized size of antiderivative = 4.45 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{2} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}-2 \sin \left (d x +c \right )^{4} a^{3}+6 \sin \left (d x +c \right )^{4} a \,b^{2}+3 \sin \left (d x +c \right )^{3} a^{2} b +5 \sin \left (d x +c \right )^{3} b^{3}+4 \sin \left (d x +c \right )^{2} a^{3}+3 \sin \left (d x +c \right ) a^{2} b -3 \sin \left (d x +c \right ) b^{3}}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:

int(sec(d*x+c)^4*(a+b*sin(d*x+c))^3*tan(d*x+c),x)
 

Output:

(3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b - 3*log(tan((c + d*x)/ 
2) - 1)*sin(c + d*x)**4*b**3 - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 
*a**2*b + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**3 + 3*log(tan((c 
+ d*x)/2) - 1)*a**2*b - 3*log(tan((c + d*x)/2) - 1)*b**3 - 3*log(tan((c + 
d*x)/2) + 1)*sin(c + d*x)**4*a**2*b + 3*log(tan((c + d*x)/2) + 1)*sin(c + 
d*x)**4*b**3 + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b - 6*log( 
tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 - 3*log(tan((c + d*x)/2) + 1)*a 
**2*b + 3*log(tan((c + d*x)/2) + 1)*b**3 - 2*sin(c + d*x)**4*a**3 + 6*sin( 
c + d*x)**4*a*b**2 + 3*sin(c + d*x)**3*a**2*b + 5*sin(c + d*x)**3*b**3 + 4 
*sin(c + d*x)**2*a**3 + 3*sin(c + d*x)*a**2*b - 3*sin(c + d*x)*b**3)/(8*d* 
(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))