3.2.71 \(\int (a \sin (e+f x))^m \sqrt {b \tan (e+f x)} \, dx\) [171]

Optimal. Leaf size=79 \[ \frac {2 \cos ^2(e+f x)^{3/4} \, _2F_1\left (\frac {3}{4},\frac {1}{4} (3+2 m);\frac {1}{4} (7+2 m);\sin ^2(e+f x)\right ) (a \sin (e+f x))^m (b \tan (e+f x))^{3/2}}{b f (3+2 m)} \]

[Out]

2*(cos(f*x+e)^2)^(3/4)*hypergeom([3/4, 3/4+1/2*m],[7/4+1/2*m],sin(f*x+e)^2)*(a*sin(f*x+e))^m*(b*tan(f*x+e))^(3
/2)/b/f/(3+2*m)

________________________________________________________________________________________

Rubi [A]
time = 0.07, 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 \cos ^2(e+f x)^{3/4} (b \tan (e+f x))^{3/2} (a \sin (e+f x))^m \, _2F_1\left (\frac {3}{4},\frac {1}{4} (2 m+3);\frac {1}{4} (2 m+7);\sin ^2(e+f x)\right )}{b f (2 m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 (a \sin (e+f x))^m \sqrt {b \tan (e+f x)} \, dx &=\frac {\left (a \cos ^{\frac {3}{2}}(e+f x) (b \tan (e+f x))^{3/2}\right ) \int \frac {(a \sin (e+f x))^{\frac {1}{2}+m}}{\sqrt {\cos (e+f x)}} \, dx}{b (a \sin (e+f x))^{3/2}}\\ &=\frac {2 \cos ^2(e+f x)^{3/4} \, _2F_1\left (\frac {3}{4},\frac {1}{4} (3+2 m);\frac {1}{4} (7+2 m);\sin ^2(e+f x)\right ) (a \sin (e+f x))^m (b \tan (e+f x))^{3/2}}{b f (3+2 m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 3.44, size = 87, normalized size = 1.10 \begin {gather*} \frac {2 \, _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 ) \sec ^2(e+f x)^{m/2} (a \sin (e+f x))^m (b \tan (e+f x))^{3/2}}{b f (3+2 m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \left (a \sin \left (f x +e \right )\right )^{m} \sqrt {b \tan \left (f x +e \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e))*(a*sin(f*x + e))^m, x)

________________________________________________________________________________________

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))^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \sin {\left (e + f x \right )}\right )^{m} \sqrt {b \tan {\left (e + f x \right )}}\, 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))**(1/2),x)

[Out]

Integral((a*sin(e + f*x))**m*sqrt(b*tan(e + f*x)), x)

________________________________________________________________________________________

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))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*tan(f*x + e))*(a*sin(f*x + e))^m, x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________