\(\int (a+b \sin ^2(e+f x))^{3/2} \tan ^3(e+f x) \, dx\) [431]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 148 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=-\frac {\sqrt {a+b} (2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}+\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f} \] Output:

-1/2*(a+b)^(1/2)*(2*a+5*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/f 
+1/2*(2*a+5*b)*(a+b*sin(f*x+e)^2)^(1/2)/f+1/6*(2*a+5*b)*(a+b*sin(f*x+e)^2) 
^(3/2)/(a+b)/f+1/2*sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(5/2)/(a+b)/f
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.78 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {3 \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}+(2 a+5 b) \left (-3 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b \sin ^2(e+f x)} \left (4 a+3 b+b \sin ^2(e+f x)\right )\right )}{6 (a+b) f} \] Input:

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

Output:

(3*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2) + (2*a + 5*b)*(-3*(a + b)^( 
3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] + Sqrt[a + b*Sin[e + 
f*x]^2]*(4*a + 3*b + b*Sin[e + f*x]^2)))/(6*(a + b)*f)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3673, 87, 60, 60, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {(2 a+5 b) \left ((a+b) \left (\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{2 f}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

((a + b*Sin[e + f*x]^2)^(5/2)/((a + b)*(1 - Sin[e + f*x]^2)) - ((2*a + 5*b 
)*((-2*(a + b*Sin[e + f*x]^2)^(3/2))/3 + (a + b)*(2*Sqrt[a + b]*ArcTanh[Sq 
rt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] - 2*Sqrt[a + b*Sin[e + f*x]^2])))/(2 
*(a + b)))/(2*f)
 

Defintions of rubi rules used

rule 60
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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

rule 3673
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ 
(m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m 
 + 1)/2)/(2*f)   Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 
)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ 
erQ[(m - 1)/2]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(425\) vs. \(2(128)=256\).

Time = 0.52 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.88

method result size
default \(\frac {-\frac {\sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (b \cos \left (f x +e \right )^{2}+2 a -b \right )}{3}+2 b \sqrt {a +b \sin \left (f x +e \right )^{2}}+\left (-\frac {1}{4} a^{2}-\frac {1}{2} a b -\frac {1}{4} b^{2}\right ) \left (-\frac {\sqrt {a +b -b \cos \left (f x +e \right )^{2}}}{\left (a +b \right ) \left (1+\sin \left (f x +e \right )\right )}-\frac {b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\left (a +b \right )^{\frac {3}{2}}}\right )+\left (\frac {1}{4} a^{2}+\frac {1}{2} a b +\frac {1}{4} b^{2}\right ) \left (-\frac {\sqrt {a +b -b \cos \left (f x +e \right )^{2}}}{\left (a +b \right ) \left (\sin \left (f x +e \right )-1\right )}+\frac {b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\left (a +b \right )^{\frac {3}{2}}}\right )-\frac {\left (\frac {1}{2} a^{2}+2 a b +\frac {3}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\sqrt {a +b}}-\frac {\left (\frac {1}{2} a^{2}+2 a b +\frac {3}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\sqrt {a +b}}+2 a \sqrt {a +b \sin \left (f x +e \right )^{2}}}{f}\) \(426\)

Input:

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

Output:

(-1/3*(a+b-b*cos(f*x+e)^2)^(1/2)*(b*cos(f*x+e)^2+2*a-b)+2*b*(a+b*sin(f*x+e 
)^2)^(1/2)+(-1/4*a^2-1/2*a*b-1/4*b^2)*(-1/(a+b)/(1+sin(f*x+e))*(a+b-b*cos( 
f*x+e)^2)^(1/2)-b/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2) 
-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e))))+(1/4*a^2+1/2*a*b+1/4*b^2)*(-1/(a+b)/ 
(sin(f*x+e)-1)*(a+b-b*cos(f*x+e)^2)^(1/2)+b/(a+b)^(3/2)*ln((2*(a+b)^(1/2)* 
(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e)-1)))-(1/2*a^2+2 
*a*b+3/2*b^2)/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b 
*sin(f*x+e)+2*a)/(sin(f*x+e)-1))-(1/2*a^2+2*a*b+3/2*b^2)/(a+b)^(1/2)*ln((2 
*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)) 
)+2*a*(a+b*sin(f*x+e)^2)^(1/2))/f
 

Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.88 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\left [\frac {3 \, {\left (2 \, a + 5 \, b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, f \cos \left (f x + e\right )^{2}}, -\frac {3 \, {\left (2 \, a + 5 \, b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) \cos \left (f x + e\right )^{2} + {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f \cos \left (f x + e\right )^{2}}\right ] \] Input:

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

Output:

[1/12*(3*(2*a + 5*b)*sqrt(a + b)*cos(f*x + e)^2*log((b*cos(f*x + e)^2 + 2* 
sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) - 
 2*(2*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b)*sqrt(-b 
*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2), -1/6*(3*(2*a + 5*b)*sqrt(-a 
- b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(b*cos(f*x + e)^2 
 - a - b))*cos(f*x + e)^2 + (2*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + 
e)^2 - 3*a - 3*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.14 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {4 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{2} + 12 \, {\left (a b^{2} + 2 \, b^{3}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {3 \, {\left (2 \, a^{2} b^{2} + 7 \, a b^{3} + 5 \, b^{4}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{\sqrt {a + b}} - \frac {6 \, {\left (a b^{3} + b^{4}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}}{b \sin \left (f x + e\right )^{2} - b}}{12 \, b^{2} f} \] Input:

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

Output:

1/12*(4*(b*sin(f*x + e)^2 + a)^(3/2)*b^2 + 12*(a*b^2 + 2*b^3)*sqrt(b*sin(f 
*x + e)^2 + a) + 3*(2*a^2*b^2 + 7*a*b^3 + 5*b^4)*log((sqrt(b*sin(f*x + e)^ 
2 + a) - sqrt(a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/sqrt(a + 
 b) - 6*(a*b^3 + b^4)*sqrt(b*sin(f*x + e)^2 + a)/(b*sin(f*x + e)^2 - b))/( 
b^2*f)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2073 vs. \(2 (128) = 256\).

Time = 1.31 (sec) , antiderivative size = 2073, normalized size of antiderivative = 14.01 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/3*(3*(2*a^2 + 7*a*b + 5*b^2)*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(-a - b))/sqrt(-a - b) - 6*(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))^3*a^2 + 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))^3*a*b + (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 + 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))^2*a^(5/2) + 7*(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 + 5*(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*sqrt(a)*b^2 - 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 
))*a^3 - 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 + 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*b^2 + 4*(sqrt(...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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