3.2.73 \(\int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx\) [173]

Optimal. Leaf size=79 \[ -\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{4} (-1+2 m);\frac {1}{4} (3+2 m);\sin ^2(e+f x)\right ) (a \sin (e+f x))^m}{b f (1-2 m) \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \]

[Out]

-2*hypergeom([-1/4, -1/4+1/2*m],[3/4+1/2*m],sin(f*x+e)^2)*(a*sin(f*x+e))^m/b/f/(1-2*m)/(cos(f*x+e)^2)^(1/4)/(b
*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2682, 2657} \begin {gather*} -\frac {2 (a \sin (e+f x))^m \, _2F_1\left (-\frac {1}{4},\frac {1}{4} (2 m-1);\frac {1}{4} (2 m+3);\sin ^2(e+f x)\right )}{b f (1-2 m) \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2657

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b^(2*IntPart[
(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2682

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[a*Cos[e + f*
x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b*(a*Sin[e + f*x])^(n + 1))), 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]

Rubi steps

\begin {align*} \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx &=\frac {\left (a \sqrt {a \sin (e+f x)}\right ) \int \cos ^{\frac {3}{2}}(e+f x) (a \sin (e+f x))^{-\frac {3}{2}+m} \, dx}{b \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{4} (-1+2 m);\frac {1}{4} (3+2 m);\sin ^2(e+f x)\right ) (a \sin (e+f x))^m}{b f (1-2 m) \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(224\) vs. \(2(79)=158\).
time = 5.55, size = 224, normalized size = 2.84 \begin {gather*} \frac {\sec ^4(e+f x) \sec ^2(e+f x)^{\frac {1}{2} (-4+m)} (a \sin (e+f x))^m \left (\, _2F_1\left (\frac {m}{2},\frac {1}{4} (-1+2 m);\frac {1}{4} (3+2 m);-\tan ^2(e+f x)\right )+\frac {\cos (2 (e+f x)) \sec ^2(e+f x) \left (-\left ((3+2 m) \, _2F_1\left (\frac {m}{2},\frac {1}{4} (-1+2 m);\frac {1}{4} (3+2 m);-\tan ^2(e+f x)\right )\right )+2 (-1+2 m) \, _2F_1\left (\frac {2+m}{2},\frac {1}{4} (3+2 m);\frac {1}{4} (7+2 m);-\tan ^2(e+f x)\right ) \tan ^2(e+f x)\right )}{(3+2 m) \left (-1+\tan ^2(e+f x)\right )}\right )}{b f (-1+2 m) \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \frac {\left (a \sin \left (f x +e \right )\right )^{m}}{\left (b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

integral(sqrt(b*tan(f*x + e))*(a*sin(f*x + e))^m/(b^2*tan(f*x + e)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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