3.384 \(\int (b \csc (e+f x))^m \sqrt {d \tan (e+f x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2618, 2602, 2577} \[ \frac {2 \cos ^2(e+f x)^{3/4} (d \tan (e+f x))^{3/2} (b \csc (e+f x))^m \, _2F_1\left (\frac {3}{4},\frac {1}{4} (3-2 m);\frac {1}{4} (7-2 m);\sin ^2(e+f x)\right )}{d f (3-2 m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2577

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)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2602

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]

Rule 2618

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^FracPart[m]*(Sin[e + f*x]/a)^FracPart[m], Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 3.20, size = 87, normalized size = 1.10 \[ -\frac {2 (d \tan (e+f x))^{3/2} \sec ^2(e+f x)^{-m/2} (b \csc (e+f x))^m \, _2F_1\left (\frac {1}{4} (3-2 m),1-\frac {m}{2};\frac {1}{4} (7-2 m);-\tan ^2(e+f x)\right )}{d f (2 m-3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*Csc[e + f*x])^m*Hypergeometric2F1[(3 - 2*m)/4, 1 - m/2, (7 - 2*m)/4, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^
(3/2))/(d*f*(-3 + 2*m)*(Sec[e + f*x]^2)^(m/2))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(f*x + e))*(b*csc(f*x + e))^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \tan \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*tan(f*x + e))*(b*csc(f*x + e))^m, x)

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maple [F]  time = 0.57, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x +e \right )\right )^{m} \sqrt {d \tan \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \tan \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^m*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(f*x + e))*(b*csc(f*x + e))^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))**m*(d*tan(f*x+e))**(1/2),x)

[Out]

Integral((b*csc(e + f*x))**m*sqrt(d*tan(e + f*x)), x)

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