\(\int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [198]

Optimal result

Integrand size = 21, antiderivative size = 474 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {8 a^4 b^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {12 a^2 b^2 \left (a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {a^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))} \] Output:

8*a^4*b^2*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(9/2) 
/d+12*a^2*b^2*(a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/( 
a^2-b^2)^(9/2)/d+a^4*(2*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2) 
^(1/2))/(a^2-b^2)^(9/2)/d+1/12*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))^2-3/4*a 
*cos(d*x+c)/(a+b)^4/d/(1-sin(d*x+c))+1/12*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+ 
c))-1/12*cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))^2+3/4*a*cos(d*x+c)/(a-b)^4/d/ 
(1+sin(d*x+c))-1/12*cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))+1/2*a^4*b*cos(d*x+ 
c)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))^2+3/2*a^5*b*cos(d*x+c)/(a^2-b^2)^4/d/(a+ 
b*sin(d*x+c))+4*a^3*b^3*cos(d*x+c)/(a^2-b^2)^4/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 351, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {96 a^2 \left (2 a^4+21 a^2 b^2+12 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2}}-\frac {\sec ^3(c+d x) \left (-264 a^6 b-358 a^4 b^3+8 a^2 b^5-16 b^7-8 \left (44 a^6 b+55 a^4 b^3+8 a^2 b^5-2 b^7\right ) \cos (2 (c+d x))-2 \left (28 a^6 b+89 a^4 b^3-12 a^2 b^5\right ) \cos (4 (c+d x))+22 a^5 b^2 \sin (c+d x)-264 a^3 b^4 \sin (c+d x)+32 a b^6 \sin (c+d x)+32 a^7 \sin (3 (c+d x))-91 a^5 b^2 \sin (3 (c+d x))-244 a^3 b^4 \sin (3 (c+d x))-12 a b^6 \sin (3 (c+d x))-17 a^5 b^2 \sin (5 (c+d x))-76 a^3 b^4 \sin (5 (c+d x))-12 a b^6 \sin (5 (c+d x))\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))^2}}{96 d} \] Input:

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

Output:

((96*a^2*(2*a^4 + 21*a^2*b^2 + 12*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqr 
t[a^2 - b^2]])/(a^2 - b^2)^(9/2) - (Sec[c + d*x]^3*(-264*a^6*b - 358*a^4*b 
^3 + 8*a^2*b^5 - 16*b^7 - 8*(44*a^6*b + 55*a^4*b^3 + 8*a^2*b^5 - 2*b^7)*Co 
s[2*(c + d*x)] - 2*(28*a^6*b + 89*a^4*b^3 - 12*a^2*b^5)*Cos[4*(c + d*x)] + 
 22*a^5*b^2*Sin[c + d*x] - 264*a^3*b^4*Sin[c + d*x] + 32*a*b^6*Sin[c + d*x 
] + 32*a^7*Sin[3*(c + d*x)] - 91*a^5*b^2*Sin[3*(c + d*x)] - 244*a^3*b^4*Si 
n[3*(c + d*x)] - 12*a*b^6*Sin[3*(c + d*x)] - 17*a^5*b^2*Sin[5*(c + d*x)] - 
 76*a^3*b^4*Sin[5*(c + d*x)] - 12*a*b^6*Sin[5*(c + d*x)]))/((a^2 - b^2)^4* 
(a + b*Sin[c + d*x])^2))/(96*d)
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3210, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^4}{(a+b \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3210

\(\displaystyle \int \left (\frac {6 a^2 b^2 \left (a^2+b^2\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac {a^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^3}+\frac {4 a^3 b^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac {3 a}{4 (a+b)^4 (\sin (c+d x)-1)}-\frac {3 a}{4 (a-b)^4 (\sin (c+d x)+1)}+\frac {1}{4 (a+b)^3 (\sin (c+d x)-1)^2}+\frac {1}{4 (a-b)^3 (\sin (c+d x)+1)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {12 a^2 b^2 \left (a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {3 a^5 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac {a^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {8 a^4 b^2 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {a^4 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac {4 a^3 b^3 \cos (c+d x)}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac {3 a \cos (c+d x)}{4 d (a+b)^4 (1-\sin (c+d x))}+\frac {3 a \cos (c+d x)}{4 d (a-b)^4 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)^2}\)

Input:

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

Output:

(8*a^4*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^ 
(9/2)*d) + (12*a^2*b^2*(a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^ 
2 - b^2]])/((a^2 - b^2)^(9/2)*d) + (a^4*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c 
 + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + Cos[c + d*x]/(12*(a 
+ b)^3*d*(1 - Sin[c + d*x])^2) - (3*a*Cos[c + d*x])/(4*(a + b)^4*d*(1 - Si 
n[c + d*x])) + Cos[c + d*x]/(12*(a + b)^3*d*(1 - Sin[c + d*x])) - Cos[c + 
d*x]/(12*(a - b)^3*d*(1 + Sin[c + d*x])^2) + (3*a*Cos[c + d*x])/(4*(a - b) 
^4*d*(1 + Sin[c + d*x])) - Cos[c + d*x]/(12*(a - b)^3*d*(1 + Sin[c + d*x]) 
) + (a^4*b*Cos[c + d*x])/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^2) + (3*a 
^5*b*Cos[c + d*x])/(2*(a^2 - b^2)^4*d*(a + b*Sin[c + d*x])) + (4*a^3*b^3*C 
os[c + d*x])/((a^2 - b^2)^4*d*(a + b*Sin[c + d*x]))
 

Defintions of rubi rules used

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 3210

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_ 
), x_Symbol] :> Int[ExpandIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^m/ 
(1 - Sin[e + f*x]^2)^(p/2)), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - 
b^2, 0] && IntegersQ[m, p/2]
 

Maple [A] (verified)

Time = 2.81 (sec) , antiderivative size = 354, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-2 a +b}{2 \left (a +b \right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-2 a -b}{2 \left (a -b \right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 a^{2} \left (\frac {\frac {b^{2} a \left (5 a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+\frac {b \left (4 a^{4}+15 b^{2} a^{2}+14 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {11 b^{2} a \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2 b \,a^{4}+\frac {7 b^{3} a^{2}}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{4}+21 b^{2} a^{2}+12 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{4} \left (a +b \right )^{4}}}{d}\)	\(354\)
default	\(\frac {-\frac {1}{3 \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-2 a +b}{2 \left (a +b \right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-2 a -b}{2 \left (a -b \right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 a^{2} \left (\frac {\frac {b^{2} a \left (5 a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+\frac {b \left (4 a^{4}+15 b^{2} a^{2}+14 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {11 b^{2} a \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2 b \,a^{4}+\frac {7 b^{3} a^{2}}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{4}+21 b^{2} a^{2}+12 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{4} \left (a +b \right )^{4}}}{d}\)	\(354\)
risch	\(\text {Expression too large to display}\)	\(1140\)

Input:

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

Output:

1/d*(-1/3/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)^3-1/2/(a+b)^3/(tan(1/2*d*x+1/2*c) 
-1)^2-1/2/(a+b)^4*(-2*a+b)/(tan(1/2*d*x+1/2*c)-1)-1/3/(a-b)^3/(tan(1/2*d*x 
+1/2*c)+1)^3+1/2/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)^2-1/2/(a-b)^4*(-2*a-b)/(ta 
n(1/2*d*x+1/2*c)+1)+2*a^2/(a-b)^4/(a+b)^4*((1/2*b^2*a*(5*a^2+6*b^2)*tan(1/ 
2*d*x+1/2*c)^3+1/2*b*(4*a^4+15*a^2*b^2+14*b^4)*tan(1/2*d*x+1/2*c)^2+11/2*b 
^2*a*(a^2+2*b^2)*tan(1/2*d*x+1/2*c)+2*b*a^4+7/2*b^3*a^2)/(tan(1/2*d*x+1/2* 
c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)^2+1/2*(2*a^4+21*a^2*b^2+12*b^4)/(a^2-b^2) 
^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 1249, normalized size of antiderivative = 2.64 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/12*(4*a^8*b - 16*a^6*b^3 + 24*a^4*b^5 - 16*a^2*b^7 + 4*b^9 - 2*(28*a^8* 
b + 61*a^6*b^3 - 101*a^4*b^5 + 12*a^2*b^7)*cos(d*x + c)^4 - 4*(8*a^8*b - 2 
5*a^6*b^3 + 27*a^4*b^5 - 11*a^2*b^7 + b^9)*cos(d*x + c)^2 - 3*((2*a^6*b^2 
+ 21*a^4*b^4 + 12*a^2*b^6)*cos(d*x + c)^5 - 2*(2*a^7*b + 21*a^5*b^3 + 12*a 
^3*b^5)*cos(d*x + c)^3*sin(d*x + c) - (2*a^8 + 23*a^6*b^2 + 33*a^4*b^4 + 1 
2*a^2*b^6)*cos(d*x + c)^3)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c 
)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b* 
cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - 
 a^2 - b^2)) - 2*(2*a^9 - 8*a^7*b^2 + 12*a^5*b^4 - 8*a^3*b^6 + 2*a*b^8 + ( 
17*a^7*b^2 + 59*a^5*b^4 - 64*a^3*b^6 - 12*a*b^8)*cos(d*x + c)^4 - 2*(4*a^9 
 - 9*a^7*b^2 + 3*a^5*b^4 + 5*a^3*b^6 - 3*a*b^8)*cos(d*x + c)^2)*sin(d*x + 
c))/((a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)* 
d*cos(d*x + c)^5 - 2*(a^11*b - 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3 
*b^9 - a*b^11)*d*cos(d*x + c)^3*sin(d*x + c) - (a^12 - 4*a^10*b^2 + 5*a^8* 
b^4 - 5*a^4*b^8 + 4*a^2*b^10 - b^12)*d*cos(d*x + c)^3), 1/6*(2*a^8*b - 8*a 
^6*b^3 + 12*a^4*b^5 - 8*a^2*b^7 + 2*b^9 - (28*a^8*b + 61*a^6*b^3 - 101*a^4 
*b^5 + 12*a^2*b^7)*cos(d*x + c)^4 - 2*(8*a^8*b - 25*a^6*b^3 + 27*a^4*b^5 - 
 11*a^2*b^7 + b^9)*cos(d*x + c)^2 - 3*((2*a^6*b^2 + 21*a^4*b^4 + 12*a^2*b^ 
6)*cos(d*x + c)^5 - 2*(2*a^7*b + 21*a^5*b^3 + 12*a^3*b^5)*cos(d*x + c)^3*s 
in(d*x + c) - (2*a^8 + 23*a^6*b^2 + 33*a^4*b^4 + 12*a^2*b^6)*cos(d*x + ...
 

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

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

Output:

Integral(tan(c + d*x)**4/(a + b*sin(c + d*x))**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 632, normalized size of antiderivative = 1.33 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

1/3*(3*(2*a^6 + 21*a^4*b^2 + 12*a^2*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2) 
*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^8 - 4* 
a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(a^2 - b^2)) + 3*(5*a^5*b^2*tan 
(1/2*d*x + 1/2*c)^3 + 6*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^6*b*tan(1/2*d 
*x + 1/2*c)^2 + 15*a^4*b^3*tan(1/2*d*x + 1/2*c)^2 + 14*a^2*b^5*tan(1/2*d*x 
 + 1/2*c)^2 + 11*a^5*b^2*tan(1/2*d*x + 1/2*c) + 22*a^3*b^4*tan(1/2*d*x + 1 
/2*c) + 4*a^6*b + 7*a^4*b^3)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b 
^8)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) + 2*(3*a^ 
5*tan(1/2*d*x + 1/2*c)^5 + 24*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*a*b^4*tan 
(1/2*d*x + 1/2*c)^5 - 9*a^4*b*tan(1/2*d*x + 1/2*c)^4 - 24*a^2*b^3*tan(1/2* 
d*x + 1/2*c)^4 - 3*b^5*tan(1/2*d*x + 1/2*c)^4 - 10*a^5*tan(1/2*d*x + 1/2*c 
)^3 - 56*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 
 36*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 36*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a 
^5*tan(1/2*d*x + 1/2*c) + 24*a^3*b^2*tan(1/2*d*x + 1/2*c) + 9*a*b^4*tan(1/ 
2*d*x + 1/2*c) - 15*a^4*b - 20*a^2*b^3 - b^5)/((a^8 - 4*a^6*b^2 + 6*a^4*b^ 
4 - 4*a^2*b^6 + b^8)*(tan(1/2*d*x + 1/2*c)^2 - 1)^3))/d
 

Mupad [B] (verification not implemented)

Time = 21.01 (sec) , antiderivative size = 1099, normalized size of antiderivative = 2.32 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((3*tan(c/2 + (d*x)/2)^8*(2*a^6*b + 12*a^2*b^5 + 21*a^4*b^3))/(a^8 + b^8 - 
 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2) - (2*tan(c/2 + (d*x)/2)^5*(36*a*b^6 + 
14*a^7 + 242*a^3*b^4 + 23*a^5*b^2))/(3*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 
- 4*a^6*b^2)) - (4*tan(c/2 + (d*x)/2)^7*(12*a*b^4 + 2*a^5 + 21*a^3*b^2))/( 
3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^2*(30*a^4*b 
 + 4*b^5 + 71*a^2*b^3))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (a^2*(42 
*a^4*b + 2*b^5 + 61*a^2*b^3))/(3*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^ 
4*b^2)) + (10*tan(c/2 + (d*x)/2)^4*(4*a^6*b + 43*a^2*b^5 + 16*a^4*b^3))/(3 
*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (4*tan(c/2 + (d*x)/2)^ 
3*(16*a*b^6 - 2*a^7 + 131*a^3*b^4 + 65*a^5*b^2))/(3*(a^2 - b^2)*(a^6 - b^6 
 + 3*a^2*b^4 - 3*a^4*b^2)) - (2*tan(c/2 + (d*x)/2)^6*(22*a^6*b + 12*b^7 + 
153*a^2*b^5 + 233*a^4*b^3))/(3*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4* 
b^2)) + (a^3*tan(c/2 + (d*x)/2)^9*(2*a^4 + 12*b^4 + 21*a^2*b^2))/(a^8 + b^ 
8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2) - (a*tan(c/2 + (d*x)/2)*(8*b^6 - 6* 
a^6 + 208*a^2*b^4 + 105*a^4*b^2))/(3*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 
3*a^4*b^2)))/(d*(tan(c/2 + (d*x)/2)^4*(2*a^2 + 12*b^2) - tan(c/2 + (d*x)/2 
)^6*(2*a^2 + 12*b^2) + a^2*tan(c/2 + (d*x)/2)^10 - a^2 + tan(c/2 + (d*x)/2 
)^2*(a^2 - 4*b^2) - tan(c/2 + (d*x)/2)^8*(a^2 - 4*b^2) + 8*a*b*tan(c/2 + ( 
d*x)/2)^3 - 8*a*b*tan(c/2 + (d*x)/2)^7 + 4*a*b*tan(c/2 + (d*x)/2)^9 - 4*a* 
b*tan(c/2 + (d*x)/2))) + (a^2*atan(((a^2*(2*a^4 + 12*b^4 + 21*a^2*b^2)*...
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 2221, normalized size of antiderivative = 4.69 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos 
(c + d*x)*sin(c + d*x)**4*a**6*b**3 + 252*sqrt(a**2 - b**2)*atan((tan((c + 
 d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**4*a**4*b**5 
+ 144*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*c 
os(c + d*x)*sin(c + d*x)**4*a**2*b**7 + 48*sqrt(a**2 - b**2)*atan((tan((c 
+ d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**3*a**7*b**2 
 + 504*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* 
cos(c + d*x)*sin(c + d*x)**3*a**5*b**4 + 288*sqrt(a**2 - b**2)*atan((tan(( 
c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**3*a**3*b* 
*6 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2)) 
*cos(c + d*x)*sin(c + d*x)**2*a**8*b + 228*sqrt(a**2 - b**2)*atan((tan((c 
+ d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**6*b**3 
 - 108*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* 
cos(c + d*x)*sin(c + d*x)**2*a**4*b**5 - 144*sqrt(a**2 - b**2)*atan((tan(( 
c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**2*b* 
*7 - 48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2)) 
*cos(c + d*x)*sin(c + d*x)*a**7*b**2 - 504*sqrt(a**2 - b**2)*atan((tan((c 
+ d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)*a**5*b**4 - 
288*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos 
(c + d*x)*sin(c + d*x)*a**3*b**6 - 24*sqrt(a**2 - b**2)*atan((tan((c + ...