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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4219, 464, 283, 201, 223, 212} \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\frac {\sqrt {b} (3 a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {b (3 a-2 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 a f}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {(3 a-2 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f} \]

[In]

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

[Out]

((3*a - 2*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) + ((3*a - 2*b)*b*Sec[e
+ f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(2*a*f) - ((3*a - 2*b)*Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2))/(3*a*f) +
(Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(5/2))/(3*a*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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 4219

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]}, Dist[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] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right ) \left (a+b x^2\right )^{3/2}}{x^4} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}+\frac {(3 a-2 b) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{x^2} \, dx,x,\sec (e+f x)\right )}{3 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}+\frac {((3 a-2 b) b) \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\sec (e+f x)\right )}{a 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}+\frac {((3 a-2 b) b) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sec (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}+\frac {((3 a-2 b) b) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.01 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\frac {\sqrt {2} \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (3 \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )^{5/2}-(3 a-2 b) \left (-3 b^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right )+\sqrt {a+b-a \sin ^2(e+f x)} \left (a+4 b-a \sin ^2(e+f x)\right )\right )\right )}{3 b f (a+2 b+a \cos (2 (e+f x)))^{3/2}} \]

[In]

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

[Out]

(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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(551\) vs. \(2(142)=284\).

Time = 6.93 (sec) , antiderivative size = 552, normalized size of antiderivative = 3.41

method	result	size
default	\(-\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (6 \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 \sec \left (f x +e \right ) b \right ) \cos \left (f x +e \right )^{3} b^{\frac {5}{2}}-2 \cos \left (f x +e \right )^{6} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -9 \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 \sec \left (f x +e \right ) b \right ) \cos \left (f x +e \right )^{3} b^{\frac {3}{2}} a -2 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \cos \left (f x +e \right )^{5}+6 \cos \left (f x +e \right )^{4} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -8 \cos \left (f x +e \right )^{4} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}+6 \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 -8 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \cos \left (f x +e \right )^{3}-3 \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}-3 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \cos \left (f x +e \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 (b +a \cos \left (f x +e \right )^{2}\right ) \left (1+\cos \left (f x +e \right )\right )}\)	\(552\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.72 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\left [-\frac {3 \, {\left (3 \, a - 2 \, b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, a \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, f \cos \left (f x + e\right )}, -\frac {3 \, {\left (3 \, a - 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) - {\left (2 \, a \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, f \cos \left (f x + e\right )}\right ] \]

[In]

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

[Out]

[-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)*s
qrt((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)*sqr
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)*cos(f*x + e) - (2*a*cos(f*x + e)^4 - 2*(3*a - 4*b)*co
s(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. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.54 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\frac {4 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 12 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a \cos \left (f x + e\right ) + 12 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right ) + \frac {6 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a b \cos \left (f x + e\right )}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} - 9 \, a \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) + 6 \, b^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{12 \, f} \]

[In]

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

[Out]

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)*co
s(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1388 vs. \(2 (142) = 284\).

Time = 1.24 (sec) , antiderivative size = 1388, normalized size of antiderivative = 8.57 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/3*(3*(3*a*b - 2*b^2)*arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*ta
n(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) - sqrt(a + b))/sqrt(-b
))*sgn(cos(f*x + e))/sqrt(-b) + 6*((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan
(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^3*(a*b + 2*b^2)*sgn(co
s(f*x + e)) - (sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 -
2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2*(3*a*b - 2*b^2)*sqrt(a + b)*sgn(cos(f*x +
e)) + (3*a^2*b - 3*a*b^2 - 2*b^3)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan
(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sgn(cos(f*x + e)) - (a
^2*b - a*b^2 + 2*b^3)*sqrt(a + b)*sgn(cos(f*x + e)))/((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x
+ 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2 -
2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/
2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sqrt(a + b) + a - 3*b)^2 - 8*(3*(sqrt(a + b)*tan(1/2*f
*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*ta
n(1/2*f*x + 1/2*e)^2 + a + b))^5*a*b*sgn(cos(f*x + e)) - 3*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/
2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)
)^4*(2*a^2 - 3*a*b)*sqrt(a + b)*sgn(cos(f*x + e)) + 2*(4*a^3 - 9*a^2*b + 3*a*b^2)*(sqrt(a + b)*tan(1/2*f*x + 1
/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*
f*x + 1/2*e)^2 + a + b))^3*sgn(cos(f*x + e)) + 6*(2*a^3 - 3*a^2*b - a*b^2)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2
- sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1
/2*e)^2 + a + b))^2*sqrt(a + b)*sgn(cos(f*x + e)) - 3*(8*a^4 - 13*a^3*b - 2*a^2*b^2 + 3*a*b^3)*(sqrt(a + b)*ta
n(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 +
2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sgn(cos(f*x + e)) + (10*a^4 - 23*a^3*b + 12*a^2*b^2 - 3*a*b^3)*sqrt(a +
b)*sgn(cos(f*x + e)))/((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1
/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2 + 2*(sqrt(a + b)*tan(1/2*f*x + 1
/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*
f*x + 1/2*e)^2 + a + b))*sqrt(a + b) - 3*a + b)^3)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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