3.5.66 \(\int \frac {\sin ^{\frac {3}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) [466]

Optimal. Leaf size=363 \[ \frac {\sqrt {b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {b} \sqrt {\sin (e+f x)}}\right )}{4 \sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {\sqrt {b} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {b} \sqrt {\sin (e+f x)}}\right )}{4 \sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {\sqrt {b} \log \left (\sqrt {b}+\sqrt {b} \cot (e+f x)-\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {\sin (e+f x)}}\right )}{8 \sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {\sqrt {b} \log \left (\sqrt {b}+\sqrt {b} \cot (e+f x)+\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {\sin (e+f x)}}\right )}{8 \sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sqrt {\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}} \]

[Out]

1/8*arctan(1-2^(1/2)*(b*cos(f*x+e))^(1/2)/b^(1/2)/sin(f*x+e)^(1/2))*b^(1/2)/f*2^(1/2)/(b*cos(f*x+e))^(1/2)/(b*
sec(f*x+e))^(1/2)-1/8*arctan(1+2^(1/2)*(b*cos(f*x+e))^(1/2)/b^(1/2)/sin(f*x+e)^(1/2))*b^(1/2)/f*2^(1/2)/(b*cos
(f*x+e))^(1/2)/(b*sec(f*x+e))^(1/2)-1/16*ln(b^(1/2)+cot(f*x+e)*b^(1/2)-2^(1/2)*(b*cos(f*x+e))^(1/2)/sin(f*x+e)
^(1/2))*b^(1/2)/f*2^(1/2)/(b*cos(f*x+e))^(1/2)/(b*sec(f*x+e))^(1/2)+1/16*ln(b^(1/2)+cot(f*x+e)*b^(1/2)+2^(1/2)
*(b*cos(f*x+e))^(1/2)/sin(f*x+e)^(1/2))*b^(1/2)/f*2^(1/2)/(b*cos(f*x+e))^(1/2)/(b*sec(f*x+e))^(1/2)-1/2*b*sin(
f*x+e)^(1/2)/f/(b*sec(f*x+e))^(3/2)

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Rubi [A]
time = 0.18, antiderivative size = 363, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {2663, 2665, 2655, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\sqrt {b} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {b} \sqrt {\sin (e+f x)}}\right )}{4 \sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {b} \sqrt {\sin (e+f x)}}+1\right )}{4 \sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sqrt {\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}-\frac {\sqrt {b} \log \left (\sqrt {b} \cot (e+f x)-\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {\sin (e+f x)}}+\sqrt {b}\right )}{8 \sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {\sqrt {b} \log \left (\sqrt {b} \cot (e+f x)+\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {\sin (e+f x)}}+\sqrt {b}\right )}{8 \sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 210

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

Rule 303

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 631

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 642

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 1176

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 1179

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 2655

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 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*Cos[e + f*x])^(1/k)/(b*
Sin[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 2663

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 2665

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

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Mathematica [A]
time = 0.99, size = 145, normalized size = 0.40 \begin {gather*} \frac {b \left (-4 \sin ^2(e+f x)+\sqrt {2} \tan ^{-1}\left (\frac {-1+\sqrt {\tan ^2(e+f x)}}{\sqrt {2} \sqrt [4]{\tan ^2(e+f x)}}\right ) \tan ^2(e+f x)^{3/4}+\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{\tan ^2(e+f x)}}{1+\sqrt {\tan ^2(e+f x)}}\right ) \tan ^2(e+f x)^{3/4}\right )}{8 f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.20, size = 648, normalized size = 1.79

method	result	size
default	\(\frac {\left (i \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 {\sin \left (f x +e \right )-1+\cos \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 )-i \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 {\sin \left (f x +e \right )-1+\cos \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 )-\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 {\sin \left (f x +e \right )-1+\cos \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 )-\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 {\sin \left (f x +e \right )-1+\cos \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 \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {\sin \left (f x +e \right )-1+\cos \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 )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right )-2 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}+2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}\right ) \left (\sqrt {\sin }\left (f x +e \right )\right ) \sqrt {2}}{8 f \left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {b}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )}\)	\(648\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^(3/2)/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8/f*(I*sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((sin(f*x+e)-1+cos(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))-I*sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((sin(f*x+e)-1+cos(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))
-sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((sin(f*x+e)-1+cos(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))-sin(
f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((sin(f*x+e)-1+cos(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))+2*((1-cos
(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e)
)^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*sin(f*x+e)-2*cos(f*x+e)^3*2^(1/2)+
2*cos(f*x+e)^2*2^(1/2))*sin(f*x+e)^(1/2)/(-1+cos(f*x+e))/(b/cos(f*x+e))^(1/2)/cos(f*x+e)*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^(3/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^(3/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{\frac {3}{2}}{\left (e + f x \right )}}{\sqrt {b \sec {\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**(3/2)/(b*sec(f*x+e))**(1/2),x)

[Out]

Integral(sin(e + f*x)**(3/2)/sqrt(b*sec(e + f*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^(3/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\sin \left (e+f\,x\right )}^{3/2}}{\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^(3/2)/(b/cos(e + f*x))^(1/2),x)

[Out]

int(sin(e + f*x)^(3/2)/(b/cos(e + f*x))^(1/2), x)

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