\(\int \sin (e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [106]

Optimal result

Integrand size = 23, antiderivative size = 113 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {3 (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {3 b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f}-\frac {\cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{f} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.15 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.50 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\left (6 \sqrt {2} (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right ) \cos ^2(e+f x)-2 (a-2 b+(a-b) \cos (2 (e+f x))) \sqrt {a+b+(a-b) \cos (2 (e+f x))}\right ) \sec (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{4 \sqrt {2} f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \] Input:

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

Output:

((6*Sqrt[2]*(a - b)*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]] 
/(Sqrt[2]*Sqrt[b])]*Cos[e + f*x]^2 - 2*(a - 2*b + (a - b)*Cos[2*(e + f*x)] 
)*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])*Sec[e + f*x]*Sqrt[(a + b + (a - 
b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(4*Sqrt[2]*f*Sqrt[a + b + (a - b)*Co 
s[2*(e + f*x)]])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4147, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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. \(683\) vs. \(2(99)=198\).

Time = 8.50 (sec) , antiderivative size = 684, normalized size of antiderivative = 6.05

method	result	size
default	\(-\frac {\left (-3 b^{\frac {7}{2}} \ln \left (\frac {4 b \cot \left (f x +e \right )^{2}-8 b \cot \left (f x +e \right ) \csc \left (f x +e \right )+4 b \csc \left (f x +e \right )^{2}+8 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 b}{\cot \left (f x +e \right )^{2}-2 \csc \left (f x +e \right ) \cot \left (f x +e \right )+\csc \left (f x +e \right )^{2}-1}\right ) \cos \left (f x +e \right )^{2}+6 b^{\frac {5}{2}} \ln \left (\frac {4 b \cot \left (f x +e \right )^{2}-8 b \cot \left (f x +e \right ) \csc \left (f x +e \right )+4 b \csc \left (f x +e \right )^{2}+8 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 b}{\cot \left (f x +e \right )^{2}-2 \csc \left (f x +e \right ) \cot \left (f x +e \right )+\csc \left (f x +e \right )^{2}-1}\right ) a \cos \left (f x +e \right )^{2}-3 b^{\frac {3}{2}} \ln \left (\frac {4 b \cot \left (f x +e \right )^{2}-8 b \cot \left (f x +e \right ) \csc \left (f x +e \right )+4 b \csc \left (f x +e \right )^{2}+8 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 b}{\cot \left (f x +e \right )^{2}-2 \csc \left (f x +e \right ) \cot \left (f x +e \right )+\csc \left (f x +e \right )^{2}-1}\right ) a^{2} \cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )^{2} \left (2 \cos \left (f x +e \right )+2\right ) a^{2} \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b +\left (-4 \cos \left (f x +e \right )^{3}-4 \cos \left (f x +e \right )^{2}-\cos \left (f x +e \right )-1\right ) a \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b^{2}+\left (2 \cos \left (f x +e \right )^{3}+2 \cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )+1\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b^{3}\right ) \cos \left (f x +e \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{2 f \left (a -b \right ) b \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cos \left (f x +e \right )^{2} \left (\cos \left (f x +e \right )+1\right ) a +\left (\cos \left (f x +e \right )+1\right ) \sin \left (f x +e \right )^{2} b \right )}\)	\(684\)

Input:

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

Output:

-1/2/f/(a-b)/b*(-3*b^(7/2)*ln(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+ 
b*csc(f*x+e)^2+2*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2 
)^(1/2)+b)/(cot(f*x+e)^2-2*csc(f*x+e)*cot(f*x+e)+csc(f*x+e)^2-1))*cos(f*x+ 
e)^2+6*b^(5/2)*ln(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*csc(f*x+e) 
^2+2*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)+b)/( 
cot(f*x+e)^2-2*csc(f*x+e)*cot(f*x+e)+csc(f*x+e)^2-1))*a*cos(f*x+e)^2-3*b^( 
3/2)*ln(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*csc(f*x+e)^2+2*b^(1/ 
2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)+b)/(cot(f*x+e) 
^2-2*csc(f*x+e)*cot(f*x+e)+csc(f*x+e)^2-1))*a^2*cos(f*x+e)^2+cos(f*x+e)^2* 
(2*cos(f*x+e)+2)*a^2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1 
/2)*b+(-4*cos(f*x+e)^3-4*cos(f*x+e)^2-cos(f*x+e)-1)*a*((a*cos(f*x+e)^2+b*s 
in(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*b^2+(2*cos(f*x+e)^3+2*cos(f*x+e)^2+co 
s(f*x+e)+1)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*b^3)* 
cos(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2)/((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos( 
f*x+e)+1)^2)^(1/2)/(cos(f*x+e)^2*(cos(f*x+e)+1)*a+(cos(f*x+e)+1)*sin(f*x+e 
)^2*b)
 

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.52 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (a - b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right )}, \frac {3 \, {\left (a - b\right )} \sqrt {-b} \arctan \left (-\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) \cos \left (f x + e\right ) - {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, f \cos \left (f x + e\right )}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.56 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {4 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} {\left (a - b\right )} \cos \left (f x + e\right ) - \frac {2 \, {\left (a b - b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} + \frac {3 \, {\left (a b - b^{2}\right )} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{\sqrt {b}}}{4 \, f} \] Input:

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

Output:

-1/4*(4*sqrt(a - b + b/cos(f*x + e)^2)*(a - b)*cos(f*x + e) - 2*(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) + 3*(a*b - b^2)*log((sqrt(a - b + b/cos(f*x + e)^2)*co 
s(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt( 
b)))/sqrt(b))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1366 vs. \(2 (99) = 198\).

Time = 1.98 (sec) , antiderivative size = 1366, normalized size of antiderivative = 12.09 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

(3*(a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 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) - sq 
rt(a))/sqrt(-b))/sqrt(-b) + 4*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta 
n(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*sgn(tan(1/2*f*x + 1/2*e)^2 - 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))*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))*b^2*sgn(tan(1/2*f*x + 1/2 
*e)^2 - 1) - a^(5/2)*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 2*a^(3/2)*b*sgn(tan 
(1/2*f*x + 1/2*e)^2 - 1) - sqrt(a)*b^2*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))^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))*sqrt(a) - 3*a + 4*b) + 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*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*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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