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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3745, 283, 201, 223, 212} \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {b} (a-b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 f}+\frac {3 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 f}-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f} \]

[In]

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

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 283

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

Rule 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a-b+b x^2\right )^{3/2}}{x^2} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{f}+\frac {(3 b) \text {Subst}\left (\int \sqrt {a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{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}+\frac {(3 (a-b) b) \text {Subst}\left (\int \frac {1}{\sqrt {a-b+b x^2}} \, dx,x,\sec (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}+\frac {(3 (a-b) b) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.72 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.50 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\left (6 \sqrt {2} (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right ) \cos ^2(e+f x)-2 (a-2 b+(a-b) \cos (2 (e+f x))) \sqrt {a+b+(a-b) \cos (2 (e+f x))}\right ) \sec (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{4 \sqrt {2} f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \]

[In]

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

[Out]

((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)*Cos[2*(e + f*x)]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(919\) vs. \(2(99)=198\).

Time = 2.49 (sec) , antiderivative size = 920, normalized size of antiderivative = 8.14

method	result	size
default	\(\text {Expression too large to display}\)	\(920\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.37 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (a - b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right )}, -\frac {3 \, {\left (a - b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) + {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, f \cos \left (f x + e\right )}\right ] \]

[In]

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

[Out]

[-1/4*(3*(a - b)*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*(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)*arctan(sqrt(-b)*sqrt(((a - b)*cos(f*
x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.56 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {4 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} {\left (a - b\right )} \cos \left (f x + e\right ) - \frac {2 \, {\left (a b - b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} + \frac {3 \, {\left (a b - b^{2}\right )} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{\sqrt {b}}}{4 \, f} \]

[In]

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

[Out]

-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)*
cos(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 = 1.88 (sec) , antiderivative size = 1366, normalized size of antiderivative = 12.09 \[ \int \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(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) -
sqrt(a))/sqrt(-b))/sqrt(-b) + 4*((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*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 - s
qrt(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 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*t
an(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*(sq
rt(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 - 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^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - a^(5/2)*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 3*a
^(3/2)*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 4*sqrt(a)*b^3*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) + a - 4*b)^2)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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