\(\int (a+b \sec ^2(e+f x))^{3/2} \sin ^3(e+f x) \, dx\) [81]

Optimal result

Integrand size = 25, antiderivative size = 162 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\frac {(3 a-2 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {(3 a-2 b) b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 a f}-\frac {(3 a-2 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f} \] Output:

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.01 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\frac {\sqrt {2} \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (3 \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )^{5/2}-(3 a-2 b) \left (-3 b^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right )+\sqrt {a+b-a \sin ^2(e+f x)} \left (a+4 b-a \sin ^2(e+f x)\right )\right )\right )}{3 b f (a+2 b+a \cos (2 (e+f x)))^{3/2}} \] Input:

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

Output:

(Sqrt[2]*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2)*(3*Sec[e + f*x]^2*(a 
+ b - a*Sin[e + f*x]^2)^(5/2) - (3*a - 2*b)*(-3*b^(3/2)*ArcTanh[Sqrt[a + b 
 - a*Sin[e + f*x]^2]/Sqrt[b]] + Sqrt[a + b - a*Sin[e + f*x]^2]*(a + 4*b - 
a*Sin[e + f*x]^2))))/(3*b*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4622, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

\(\displaystyle \frac {\int -\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 3042

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

rule 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(398\) vs. \(2(142)=284\).

Time = 9.30 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.46

method	result	size
default	\(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (-6 b^{\frac {5}{2}} \ln \left (-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sec \left (f x +e \right )-4 b \sec \left (f x +e \right )\right ) \cos \left (f x +e \right )^{3}+9 b^{\frac {3}{2}} \ln \left (-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sec \left (f x +e \right )-4 b \sec \left (f x +e \right )\right ) a \cos \left (f x +e \right )^{3}+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \left (2 \cos \left (f x +e \right )^{6}+2 \cos \left (f x +e \right )^{5}-6 \cos \left (f x +e \right )^{4}-6 \cos \left (f x +e \right )^{3}\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \left (8 \cos \left (f x +e \right )^{4}+8 \cos \left (f x +e \right )^{3}+3 \cos \left (f x +e \right )^{2}+3 \cos \left (f x +e \right )\right )\right )}{6 f b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (a \cos \left (f x +e \right )^{3}+a \cos \left (f x +e \right )^{2}+\cos \left (f x +e \right ) b +b \right )}\)	\(399\)

Input:

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

Output:

1/6/f/b*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 
2)/(a*cos(f*x+e)^3+a*cos(f*x+e)^2+cos(f*x+e)*b+b)*(-6*b^(5/2)*ln(-4*b^(1/2 
)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*b^(1/2)*((b+a*cos(f*x+e)^2 
)/(1+cos(f*x+e))^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*cos(f*x+e)^3+9*b^(3/2 
)*ln(-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*b^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*a*cos(f 
*x+e)^3+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b*(2*cos(f*x+e)^6+2* 
cos(f*x+e)^5-6*cos(f*x+e)^4-6*cos(f*x+e)^3)+((b+a*cos(f*x+e)^2)/(1+cos(f*x 
+e))^2)^(1/2)*b^2*(8*cos(f*x+e)^4+8*cos(f*x+e)^3+3*cos(f*x+e)^2+3*cos(f*x+ 
e)))
 

Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.78 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\left [-\frac {3 \, {\left (3 \, a - 2 \, b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, a \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, f \cos \left (f x + e\right )}, -\frac {3 \, {\left (3 \, a - 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) \cos \left (f x + e\right ) - {\left (2 \, a \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, f \cos \left (f x + e\right )}\right ] \] Input:

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

Output:

[-1/12*(3*(3*a - 2*b)*sqrt(b)*cos(f*x + e)*log((a*cos(f*x + e)^2 - 2*sqrt( 
b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x 
 + e)^2) - 2*(2*a*cos(f*x + e)^4 - 2*(3*a - 4*b)*cos(f*x + e)^2 + 3*b)*sqr 
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/6*(3*(3*a - 
 2*b)*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2) 
*cos(f*x + e)/(a*cos(f*x + e)^2 + b))*cos(f*x + e) - (2*a*cos(f*x + e)^4 - 
 2*(3*a - 4*b)*cos(f*x + e)^2 + 3*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + 
 e)^2))/(f*cos(f*x + e))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.54 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\frac {4 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 12 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a \cos \left (f x + e\right ) + 12 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right ) + \frac {6 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a b \cos \left (f x + e\right )}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} - 9 \, a \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) + 6 \, b^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{12 \, f} \] Input:

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

Output:

1/12*(4*(a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 12*sqrt(a + b/cos(f* 
x + e)^2)*a*cos(f*x + e) + 12*sqrt(a + b/cos(f*x + e)^2)*b*cos(f*x + e) + 
6*sqrt(a + b/cos(f*x + e)^2)*a*b*cos(f*x + e)/((a + b/cos(f*x + e)^2)*cos( 
f*x + e)^2 - b) - 9*a*sqrt(b)*log((sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) 
 - sqrt(b))/(sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b))) + 6*b^(3/ 
2)*log((sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a + b/cos 
(f*x + e)^2)*cos(f*x + e) + sqrt(b))))/f
 

Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.90 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=-\frac {a {\left (\frac {3 \, {\left (3 \, a b - 2 \, b^{2}\right )} \arctan \left (\frac {\sqrt {a \cos \left (f x + e\right )^{2} + b}}{\sqrt {-b}}\right )}{a \sqrt {-b}} - \frac {3 \, \sqrt {a \cos \left (f x + e\right )^{2} + b} b}{a \cos \left (f x + e\right )^{2}} - \frac {2 \, {\left ({\left (a \cos \left (f x + e\right )^{2} + b\right )}^{\frac {3}{2}} a^{2} - 3 \, \sqrt {a \cos \left (f x + e\right )^{2} + b} a^{3} + 3 \, \sqrt {a \cos \left (f x + e\right )^{2} + b} a^{2} b\right )}}{a^{3}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{6 \, f} \] Input:

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

Output:

-1/6*a*(3*(3*a*b - 2*b^2)*arctan(sqrt(a*cos(f*x + e)^2 + b)/sqrt(-b))/(a*s 
qrt(-b)) - 3*sqrt(a*cos(f*x + e)^2 + b)*b/(a*cos(f*x + e)^2) - 2*((a*cos(f 
*x + e)^2 + b)^(3/2)*a^2 - 3*sqrt(a*cos(f*x + e)^2 + b)*a^3 + 3*sqrt(a*cos 
(f*x + e)^2 + b)*a^2*b)/a^3)*sgn(cos(f*x + e))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*sin(e + f*x)**3,x)*b + int 
(sqrt(sec(e + f*x)**2*b + a)*sin(e + f*x)**3,x)*a