3.2.20 \(\int (a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2} \, dx\) [120]

Optimal. Leaf size=126 \[ -\frac {24 a^2 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{5 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {12 a^2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{5 f}-\frac {2 b (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{5 f} \]

[Out]

-24/5*a^2*b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*(a*sin(f*x
+e))^(1/2)/f/cos(f*x+e)^(1/2)/(b*tan(f*x+e))^(1/2)-2/5*b*(a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2)/f+12/5*a^2*
b*(a*sin(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.11, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2678, 2674, 2681, 2719} \begin {gather*} -\frac {24 a^2 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{5 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {12 a^2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{5 f}-\frac {2 b (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2),x]

[Out]

(-24*a^2*b^2*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) + (
12*a^2*b*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(5*f) - (2*b*(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]])/
(5*f)

Rule 2674

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

Rule 2678

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

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin {align*} \int (a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2} \, dx &=-\frac {2 b (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{5 f}+\frac {1}{5} \left (6 a^2\right ) \int \sqrt {a \sin (e+f x)} (b \tan (e+f x))^{3/2} \, dx\\ &=\frac {12 a^2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{5 f}-\frac {2 b (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{5 f}-\frac {1}{5} \left (12 a^2 b^2\right ) \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx\\ &=\frac {12 a^2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{5 f}-\frac {2 b (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{5 f}-\frac {\left (12 a^2 b^2 \sqrt {a \sin (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{5 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\\ &=-\frac {24 a^2 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{5 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {12 a^2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{5 f}-\frac {2 b (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{5 f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.69, size = 99, normalized size = 0.79 \begin {gather*} \frac {a^2 b \left (\cos ^2(e+f x)^{3/4} (11+\cos (2 (e+f x)))-12 \cos ^2(e+f x) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right )\right ) \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{5 f \cos ^2(e+f x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2),x]

[Out]

(a^2*b*((Cos[e + f*x]^2)^(3/4)*(11 + Cos[2*(e + f*x)]) - 12*Cos[e + f*x]^2*Hypergeometric2F1[1/4, 1/2, 3/2, Si
n[e + f*x]^2])*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(5*f*(Cos[e + f*x]^2)^(3/4))

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Maple [C] Result contains complex when optimal does not.
time = 4.17, size = 338, normalized size = 2.68

method	result	size
default	\(\frac {2 \left (12 i \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )-12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-\left (\cos ^{4}\left (f x +e \right )\right )+8 \left (\cos ^{2}\left (f x +e \right )\right )-12 \cos \left (f x +e \right )+5\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right )}{5 f \sin \left (f x +e \right )^{5}}\)	\(338\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/5/f*(12*I*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+
e)+1))^(1/2)*cos(f*x+e)-12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e
)-1)/sin(f*x+e),I)*sin(f*x+e)*cos(f*x+e)+12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ellip
ticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)-12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)
*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)-cos(f*x+e)^4+8*cos(f*x+e)^2-12*cos(f*x+e)+5)*(a*sin(f*x+e
))^(5/2)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)*cos(f*x+e)/sin(f*x+e)^5

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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((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(5/2)*(b*tan(f*x + e))^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 135, normalized size = 1.07 \begin {gather*} \frac {2 \, {\left (6 \, \sqrt {2} \sqrt {-a b} a^{2} b {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 6 \, \sqrt {2} \sqrt {-a b} a^{2} b {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + {\left (a^{2} b \cos \left (f x + e\right )^{2} + 5 \, a^{2} b\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}\right )}}{5 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

2/5*(6*sqrt(2)*sqrt(-a*b)*a^2*b*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e
))) + 6*sqrt(2)*sqrt(-a*b)*a^2*b*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x +
e))) + (a^2*b*cos(f*x + e)^2 + 5*a^2*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e)))/f

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Sympy [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((a*sin(f*x+e))**(5/2)*(b*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming sageVARa near 0Simplification assuming sageVARf near 0Simplification assuming sageV
ARx near 0S

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(e + f*x))^(5/2)*(b*tan(e + f*x))^(3/2),x)

[Out]

int((a*sin(e + f*x))^(5/2)*(b*tan(e + f*x))^(3/2), x)

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