3.451 \(\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=414 \[ -\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{4 \sqrt {2} \sqrt {b} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}+1\right )}{4 \sqrt {2} \sqrt {b} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{8 \sqrt {2} \sqrt {b} f}-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{8 \sqrt {2} \sqrt {b} f}-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}} \]

[Out]

-1/2*a*b*(a*sin(f*x+e))^(3/2)/f/(b*sec(f*x+e))^(1/2)-3/8*a^(5/2)*arctan(1-2^(1/2)*b^(1/2)*(a*sin(f*x+e))^(1/2)
/a^(1/2)/(b*cos(f*x+e))^(1/2))*(b*cos(f*x+e))^(1/2)*(b*sec(f*x+e))^(1/2)/f*2^(1/2)/b^(1/2)+3/8*a^(5/2)*arctan(
1+2^(1/2)*b^(1/2)*(a*sin(f*x+e))^(1/2)/a^(1/2)/(b*cos(f*x+e))^(1/2))*(b*cos(f*x+e))^(1/2)*(b*sec(f*x+e))^(1/2)
/f*2^(1/2)/b^(1/2)+3/16*a^(5/2)*ln(a^(1/2)-2^(1/2)*b^(1/2)*(a*sin(f*x+e))^(1/2)/(b*cos(f*x+e))^(1/2)+a^(1/2)*t
an(f*x+e))*(b*cos(f*x+e))^(1/2)*(b*sec(f*x+e))^(1/2)/f*2^(1/2)/b^(1/2)-3/16*a^(5/2)*ln(a^(1/2)+2^(1/2)*b^(1/2)
*(a*sin(f*x+e))^(1/2)/(b*cos(f*x+e))^(1/2)+a^(1/2)*tan(f*x+e))*(b*cos(f*x+e))^(1/2)*(b*sec(f*x+e))^(1/2)/f*2^(
1/2)/b^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 414, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {2583, 2585, 2574, 297, 1162, 617, 204, 1165, 628} \[ -\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{4 \sqrt {2} \sqrt {b} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}+1\right )}{4 \sqrt {2} \sqrt {b} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{8 \sqrt {2} \sqrt {b} f}-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{8 \sqrt {2} \sqrt {b} f}-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(5/2),x]

[Out]

(-3*a^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/(Sqrt[a]*Sqrt[b*Cos[e + f*x]])]*Sqrt[b*Cos[e + f
*x]]*Sqrt[b*Sec[e + f*x]])/(4*Sqrt[2]*Sqrt[b]*f) + (3*a^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]]
)/(Sqrt[a]*Sqrt[b*Cos[e + f*x]])]*Sqrt[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]])/(4*Sqrt[2]*Sqrt[b]*f) + (3*a^(5/2
)*Sqrt[b*Cos[e + f*x]]*Log[Sqrt[a] - (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/Sqrt[b*Cos[e + f*x]] + Sqrt[a]*Tan
[e + f*x]]*Sqrt[b*Sec[e + f*x]])/(8*Sqrt[2]*Sqrt[b]*f) - (3*a^(5/2)*Sqrt[b*Cos[e + f*x]]*Log[Sqrt[a] + (Sqrt[2
]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/Sqrt[b*Cos[e + f*x]] + Sqrt[a]*Tan[e + f*x]]*Sqrt[b*Sec[e + f*x]])/(8*Sqrt[2]*
Sqrt[b]*f) - (a*b*(a*Sin[e + f*x])^(3/2))/(2*f*Sqrt[b*Sec[e + f*x]])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 2583

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*b*(a*Sin[
e + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(m - n)), x] + Dist[(a^2*(m - 1))/(m - n), Int[(a*Sin[e + f*x])
^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m - n, 0] && IntegersQ[2*
m, 2*n]

Rule 2585

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(b*Cos[e + f*
x])^n*(b*Sec[e + f*x])^n, Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&
IntegerQ[m - 1/2] && IntegerQ[n - 1/2]

Rubi steps

\begin {align*} \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{5/2} \, dx &=-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}}+\frac {1}{4} \left (3 a^2\right ) \int \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)} \, dx\\ &=-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}}+\frac {1}{4} \left (3 a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}} \, dx\\ &=-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}}+\frac {\left (3 a^3 b \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {x^2}{a^2+b^2 x^4} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{2 f}\\ &=-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}}-\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {a-b x^2}{a^2+b^2 x^4} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{4 f}+\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {a+b x^2}{a^2+b^2 x^4} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{4 f}\\ &=-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}}+\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {a}{b}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {b}}+x^2} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{8 b f}+\frac {\left (3 a^3 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {b}}+x^2} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{8 b f}+\frac {\left (3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {b}}+2 x}{-\frac {a}{b}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {b}}-x^2} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{8 \sqrt {2} \sqrt {b} f}+\frac {\left (3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {b}}-2 x}{-\frac {a}{b}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {b}}-x^2} \, dx,x,\frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}\right )}{8 \sqrt {2} \sqrt {b} f}\\ &=\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{8 \sqrt {2} \sqrt {b} f}-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{8 \sqrt {2} \sqrt {b} f}-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}}+\frac {\left (3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{4 \sqrt {2} \sqrt {b} f}-\frac {\left (3 a^{5/2} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{4 \sqrt {2} \sqrt {b} f}\\ &=-\frac {3 a^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{4 \sqrt {2} \sqrt {b} f}+\frac {3 a^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{4 \sqrt {2} \sqrt {b} f}+\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{8 \sqrt {2} \sqrt {b} f}-\frac {3 a^{5/2} \sqrt {b \cos (e+f x)} \log \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)\right ) \sqrt {b \sec (e+f x)}}{8 \sqrt {2} \sqrt {b} f}-\frac {a b (a \sin (e+f x))^{3/2}}{2 f \sqrt {b \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 65, normalized size = 0.16 \[ -\frac {a (a \sin (e+f x))^{3/2} (b \sec (e+f x))^{3/2} \left (-2 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right )+\cos (2 (e+f x))+1\right )}{4 b f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(5/2),x]

[Out]

-1/4*(a*(1 + Cos[2*(e + f*x)] - 2*Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2])*(b*Sec[e + f*x])^(3/2)*(a*S
in[e + f*x])^(3/2))/(b*f)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec \left (f x + e\right )} \left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e))*(a*sin(f*x + e))^(5/2), x)

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maple [C]  time = 0.18, size = 512, normalized size = 1.24 \[ -\frac {\left (-3 i \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+3 i \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}-2 \cos \left (f x +e \right ) \sqrt {2}\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {\frac {b}{\cos \left (f x +e \right )}}\, \sqrt {2}}{8 f \left (-1+\cos \left (f x +e \right )\right ) \sin \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^(5/2)*(b*sec(f*x+e))^(1/2),x)

[Out]

-1/8/f*(-3*I*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+c
os(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))+3*
I*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/
sin(f*x+e))^(1/2)*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))-3*((1-cos(f*x
+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(
1/2)*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))-3*((1-cos(f*x+e)+sin(f*x+e
))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*Elliptic
Pi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))+2*cos(f*x+e)^2*2^(1/2)-2*cos(f*x+e)*2^(
1/2))*(a*sin(f*x+e))^(5/2)*(b/cos(f*x+e))^(1/2)/(-1+cos(f*x+e))/sin(f*x+e)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec \left (f x + e\right )} \left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e))*(a*sin(f*x + e))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(e + f*x))^(5/2)*(b/cos(e + f*x))^(1/2),x)

[Out]

int((a*sin(e + f*x))^(5/2)*(b/cos(e + f*x))^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))**(5/2)*(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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