\(\int \sin ^5(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [104]

Optimal result

Integrand size = 25, antiderivative size = 227 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {(3 a-7 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {(3 a-7 b) b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 (a-b) f}-\frac {(3 a-7 b) \cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac {2 \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{5/2}}{3 (a-b) f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{5/2}}{5 (a-b) f} \] Output:

1/2*(3*a-7*b)*b^(1/2)*arctanh(b^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2 
))/f+1/2*(3*a-7*b)*b*sec(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/(a-b)/f-1/3*(3* 
a-7*b)*cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)/f+2/3*cos(f*x+e)^3*(a-b 
+b*sec(f*x+e)^2)^(5/2)/(a-b)/f-1/5*cos(f*x+e)^5*(a-b+b*sec(f*x+e)^2)^(5/2) 
/(a-b)/f
 

Mathematica [A] (warning: unable to verify)

Time = 3.96 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.03 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\cos (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (120 \sqrt {2} \sqrt {b} \left (3 a^2-10 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right )+2 \sqrt {a+b+(a-b) \cos (2 (e+f x))} \left (-89 a^2+474 a b-409 b^2+4 \left (7 a^2-20 a b+13 b^2\right ) \cos (2 (e+f x))-3 (a-b)^2 \cos (4 (e+f x))+60 (a-b) b \sec ^2(e+f x)\right )\right )}{240 \sqrt {2} (a-b) f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \] Input:

Integrate[Sin[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

(Cos[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(120 
*Sqrt[2]*Sqrt[b]*(3*a^2 - 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b + (a - b)*Cos 
[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])] + 2*Sqrt[a + b + (a - b)*Cos[2*(e + f*x) 
]]*(-89*a^2 + 474*a*b - 409*b^2 + 4*(7*a^2 - 20*a*b + 13*b^2)*Cos[2*(e + f 
*x)] - 3*(a - b)^2*Cos[4*(e + f*x)] + 60*(a - b)*b*Sec[e + f*x]^2)))/(240* 
Sqrt[2]*(a - b)*f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.91, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4147, 365, 27, 359, 247, 211, 224, 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.

\(\displaystyle \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x)^5 \left (a+b \tan (e+f x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 4147

\(\displaystyle \frac {\int \cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )^{3/2}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 365

\(\displaystyle \frac {\frac {\int -5 (a-b) \cos ^4(e+f x) \left (2-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{5 (a-b)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\int \cos ^4(e+f x) \left (2-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}d\sec (e+f x)-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{5 (a-b)}}{f}\)

\(\Big \downarrow \) 359

\(\displaystyle \frac {\frac {(3 a-7 b) \int \cos ^2(e+f x) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}d\sec (e+f x)}{3 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{5 (a-b)}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {\frac {(3 a-7 b) \left (3 b \int \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}\right )}{3 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{5 (a-b)}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {(3 a-7 b) \left (3 b \left (\frac {1}{2} (a-b) \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}\right )}{3 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{5 (a-b)}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {(3 a-7 b) \left (3 b \left (\frac {1}{2} (a-b) \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}\right )}{3 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{5 (a-b)}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {(3 a-7 b) \left (3 b \left (\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}\right )}{3 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{5 (a-b)}+\frac {2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 (a-b)}}{f}\)

Input:

Int[Sin[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

((2*Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(5/2))/(3*(a - b)) - (Cos[e 
+ f*x]^5*(a - b + b*Sec[e + f*x]^2)^(5/2))/(5*(a - b)) + ((3*a - 7*b)*(-(C 
os[e + f*x]*(a - b + b*Sec[e + f*x]^2)^(3/2)) + 3*b*(((a - b)*ArcTanh[(Sqr 
t[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*Sqrt[b]) + (Sec[e + 
 f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/2)))/(3*(a - b)))/f
 

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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 219

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])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 365

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x 
_Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] 
- Simp[1/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p 
+ 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, 
 b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
 

rule 3042

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

rule 4147

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ 
m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 
)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( 
m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1636\) vs. \(2(203)=406\).

Time = 10.86 (sec) , antiderivative size = 1637, normalized size of antiderivative = 7.21

method	result	size
default	\(\text {Expression too large to display}\)	\(1637\)

Input:

int(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/30/f/(a-b)^5/b*(-105*b^(15/2)*ln(-4*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+ 
e)^2)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/( 
cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*cos(f*x+e)^2+570*b^(13/2 
)*ln(-4*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)-4 
*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+ 
e)-4*b*sec(f*x+e))*a*cos(f*x+e)^2-1275*b^(11/2)*ln(-4*b^(1/2)*((a*cos(f*x+ 
e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2+b* 
sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*a^2*cos(f 
*x+e)^2+1500*b^(9/2)*ln(-4*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f 
*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1 
)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*a^3*cos(f*x+e)^2-975*b^(7/2)*ln(-4*b 
^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)* 
((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*se 
c(f*x+e))*a^4*cos(f*x+e)^2+330*b^(5/2)*ln(-4*b^(1/2)*((a*cos(f*x+e)^2+b*si 
n(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e 
)^2)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*a^5*cos(f*x+e)^2-4 
5*b^(3/2)*ln(-4*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2) 
^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)* 
sec(f*x+e)-4*b*sec(f*x+e))*a^6*cos(f*x+e)^2+2*cos(f*x+e)^2*(3*cos(f*x+e)^5 
+3*cos(f*x+e)^4-10*cos(f*x+e)^3-10*cos(f*x+e)^2+15*cos(f*x+e)+15)*((a*c...
 

Fricas [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.95 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (3 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (6 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 4 \, {\left (5 \, a^{2} - 13 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (15 \, a^{2} - 70 \, a b + 58 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, a b + 15 \, b^{2}\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, {\left (a - b\right )} f \cos \left (f x + e\right )}, \frac {15 \, {\left (3 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) \cos \left (f x + e\right ) - {\left (6 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 4 \, {\left (5 \, a^{2} - 13 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (15 \, a^{2} - 70 \, a b + 58 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, a b + 15 \, b^{2}\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, {\left (a - b\right )} f \cos \left (f x + e\right )}\right ] \] Input:

integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/60*(15*(3*a^2 - 10*a*b + 7*b^2)*sqrt(b)*cos(f*x + e)*log(-((a - b)*cos 
(f*x + e)^2 - 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)* 
cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*(6*(a^2 - 2*a*b + b^2)*cos(f*x + e 
)^6 - 4*(5*a^2 - 13*a*b + 8*b^2)*cos(f*x + e)^4 + 2*(15*a^2 - 70*a*b + 58* 
b^2)*cos(f*x + e)^2 - 15*a*b + 15*b^2)*sqrt(((a - b)*cos(f*x + e)^2 + b)/c 
os(f*x + e)^2))/((a - b)*f*cos(f*x + e)), 1/30*(15*(3*a^2 - 10*a*b + 7*b^2 
)*sqrt(-b)*arctan(-sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e) 
^2)*cos(f*x + e)/((a - b)*cos(f*x + e)^2 + b))*cos(f*x + e) - (6*(a^2 - 2* 
a*b + b^2)*cos(f*x + e)^6 - 4*(5*a^2 - 13*a*b + 8*b^2)*cos(f*x + e)^4 + 2* 
(15*a^2 - 70*a*b + 58*b^2)*cos(f*x + e)^2 - 15*a*b + 15*b^2)*sqrt(((a - b) 
*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a - b)*f*cos(f*x + e))]
 

Sympy [F(-1)]

Timed out. \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**5*(a+b*tan(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.45 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {\frac {12 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5}}{a - b} - 40 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} + 60 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} {\left (a - b\right )} \cos \left (f x + e\right ) - 120 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right ) - 60 \, b^{\frac {3}{2}} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) - \frac {30 \, {\left (a b - b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} + \frac {45 \, {\left (a b - b^{2}\right )} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{\sqrt {b}}}{60 \, f} \] Input:

integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

-1/60*(12*(a - b + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5/(a - b) - 40*(a 
- b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 + 60*sqrt(a - b + b/cos(f*x + 
 e)^2)*(a - b)*cos(f*x + e) - 120*sqrt(a - b + b/cos(f*x + e)^2)*b*cos(f*x 
 + e) - 60*b^(3/2)*log((sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt 
(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b))) - 30*(a*b - 
b^2)*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/((a - b + b/cos(f*x + e)^ 
2)*cos(f*x + e)^2 - b) + 45*(a*b - b^2)*log((sqrt(a - b + b/cos(f*x + e)^2 
)*cos(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) + s 
qrt(b)))/sqrt(b))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4702 vs. \(2 (203) = 406\).

Time = 5.14 (sec) , antiderivative size = 4702, normalized size of antiderivative = 20.71 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

1/15*(15*(3*a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 7*b^2*sgn(tan(1/2*f*x + 
1/2*e)^2 - 1))*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/ 
2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 
 + a) - sqrt(a))/sqrt(-b))/sqrt(-b) + 30*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 
- sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2 
*f*x + 1/2*e)^2 + a))^3*a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + (sqrt(a)*tan 
(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/ 
2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 
 - 1) - 3*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 
- 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(3/2)* 
b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 5*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sq 
rt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x 
 + 1/2*e)^2 + a))^2*sqrt(a)*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 3*(sqrt( 
a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f* 
x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^2*b*sgn(tan(1/2*f*x + 1/ 
2*e)^2 - 1) - 9*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2 
*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a*b^ 
2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 4*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sq 
rt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x 
 + 1/2*e)^2 + a))*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - a^(5/2)*b*sgn(t...
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int {\sin \left (e+f\,x\right )}^5\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \] Input:

int(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)^(3/2),x)
 

Output:

int(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\tan \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{5} \tan \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\tan \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{5}d x \right ) a \] Input:

int(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x)
 

Output:

int(sqrt(tan(e + f*x)**2*b + a)*sin(e + f*x)**5*tan(e + f*x)**2,x)*b + int 
(sqrt(tan(e + f*x)**2*b + a)*sin(e + f*x)**5,x)*a