3.474 \(\int \frac {\sqrt {a \sin (e+f x)}}{(b \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=418 \[ -\frac {\sqrt {a} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{4 \sqrt {2} b^{5/2} f}+\frac {\sqrt {a} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}+1\right )}{4 \sqrt {2} b^{5/2} f}+\frac {\sqrt {a} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{8 \sqrt {2} b^{5/2} f}-\frac {\sqrt {a} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{8 \sqrt {2} b^{5/2} f}+\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 418, 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 = {2582, 2585, 2574, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\sqrt {a} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right )}{4 \sqrt {2} b^{5/2} f}+\frac {\sqrt {a} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}+1\right )}{4 \sqrt {2} b^{5/2} f}+\frac {\sqrt {a} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{8 \sqrt {2} b^{5/2} f}-\frac {\sqrt {a} \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \log \left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)}}+\sqrt {a} \tan (e+f x)+\sqrt {a}\right )}{8 \sqrt {2} b^{5/2} f}+\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a]*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]*b^(5/2)*f) + (Sqrt[a]*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/(S
qrt[a]*Sqrt[b*Cos[e + f*x]])]*Sqrt[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]])/(4*Sqrt[2]*b^(5/2)*f) + (Sqrt[a]*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]*b^(5/2)*f) - (Sqrt[a]*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]*b^(5/2)*
f) + (a*Sin[e + f*x])^(3/2)/(2*a*b*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 2582

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((a*Sin[e + f
*x])^(m + 1)*(b*Sec[e + f*x])^(n + 1))/(a*b*f*(m - n)), x] - Dist[(n + 1)/(b^2*(m - n)), Int[(a*Sin[e + f*x])^
m*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -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 \frac {\sqrt {a \sin (e+f x)}}{(b \sec (e+f x))^{3/2}} \, dx &=\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}}+\frac {\int \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)} \, dx}{4 b^2}\\ &=\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (\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}{4 b^2}\\ &=\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (a \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 b f}\\ &=\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}}-\frac {\left (a \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 b^2 f}+\frac {\left (a \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 b^2 f}\\ &=\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (a \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^3 f}+\frac {\left (a \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^3 f}+\frac {\left (\sqrt {a} \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} b^{5/2} f}+\frac {\left (\sqrt {a} \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} b^{5/2} f}\\ &=\frac {\sqrt {a} \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} b^{5/2} f}-\frac {\sqrt {a} \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} b^{5/2} f}+\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}}+\frac {\left (\sqrt {a} \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} b^{5/2} f}-\frac {\left (\sqrt {a} \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} b^{5/2} f}\\ &=-\frac {\sqrt {a} \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} b^{5/2} f}+\frac {\sqrt {a} \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} b^{5/2} f}+\frac {\sqrt {a} \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} b^{5/2} f}-\frac {\sqrt {a} \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} b^{5/2} f}+\frac {(a \sin (e+f x))^{3/2}}{2 a b f \sqrt {b \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 76, normalized size = 0.18 \[ \frac {\sec ^2(e+f x) \sqrt {a \sin (e+f x)} \left (2 \tan (e+f x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right )+3 \sin (2 (e+f x))\right )}{12 f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*Sqrt[a*Sin[e + f*x]]*(3*Sin[2*(e + f*x)] + 2*Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2]*T
an[e + f*x]))/(12*f*(b*Sec[e + f*x])^(3/2))

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sin \left (f x + e\right )}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.21, size = 516, normalized size = 1.23 \[ \frac {\left (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 )-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 )+\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 )+\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 ) \sqrt {a \sin \left (f x +e \right )}\, \sin \left (f x +e \right ) \sqrt {2}}{8 f \left (-1+\cos \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f*(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))-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))+((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)*Ell
ipticPi(((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))+((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))+2*cos(f*x+e)^2*2^(1/2)-2*cos(f*x+e)*2^(1/2))*(a*s
in(f*x+e))^(1/2)*sin(f*x+e)/(-1+cos(f*x+e))/cos(f*x+e)^2/(b/cos(f*x+e))^(3/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sin \left (f x + e\right )}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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