3.292 \(\int \frac{(a \cos (e+f x))^m}{\sqrt{b \csc (e+f x)}} \, dx\)

Optimal. Leaf size=78 \[ -\frac{\sqrt [4]{\sin ^2(e+f x)} \sqrt{b \csc (e+f x)} (a \cos (e+f x))^{m+1} \, _2F_1\left (\frac{1}{4},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{a b f (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.097994, antiderivative size = 78, normalized size of antiderivative = 1., 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 = {2586, 2576} \[ -\frac{\sqrt [4]{\sin ^2(e+f x)} \sqrt{b \csc (e+f x)} (a \cos (e+f x))^{m+1} \, _2F_1\left (\frac{1}{4},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{a b f (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[e + f*x])^m/Sqrt[b*Csc[e + f*x]],x]

[Out]

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

Rule 2586

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(1*(b*Cos[e +
f*x])^(n + 1)*(b*Sec[e + f*x])^(n + 1))/b^2, 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] &&  !IntegerQ[n] && LtQ[n, 1]

Rule 2576

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

Rubi steps

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

Mathematica [C]  time = 1.71423, size = 225, normalized size = 2.88 \[ \frac{14 b (a \cos (e+f x))^m F_1\left (\frac{3}{4};-m,m+\frac{3}{2};\frac{7}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{3 f (b \csc (e+f x))^{3/2} \left (7 F_1\left (\frac{3}{4};-m,m+\frac{3}{2};\frac{7}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \left (2 m F_1\left (\frac{7}{4};1-m,m+\frac{3}{2};\frac{11}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+(2 m+3) F_1\left (\frac{7}{4};-m,m+\frac{5}{2};\frac{11}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a*Cos[e + f*x])^m/Sqrt[b*Csc[e + f*x]],x]

[Out]

(14*b*AppellF1[3/4, -m, 3/2 + m, 7/4, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(a*Cos[e + f*x])^m)/(3*f*(b*Csc
[e + f*x])^(3/2)*(7*AppellF1[3/4, -m, 3/2 + m, 7/4, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*(2*m*AppellF1
[7/4, 1 - m, 3/2 + m, 11/4, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3 + 2*m)*AppellF1[7/4, -m, 5/2 + m, 11
/4, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))

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Maple [F]  time = 0.365, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a\cos \left ( fx+e \right ) \right ) ^{m}{\frac{1}{\sqrt{b\csc \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \cos \left (f x + e\right )\right )^{m}}{\sqrt{b \csc \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cos(f*x + e))^m/sqrt(b*csc(f*x + e)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \csc \left (f x + e\right )} \left (a \cos \left (f x + e\right )\right )^{m}}{b \csc \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*csc(f*x + e))*(a*cos(f*x + e))^m/(b*csc(f*x + e)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \cos{\left (e + f x \right )}\right )^{m}}{\sqrt{b \csc{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(f*x+e))**m/(b*csc(f*x+e))**(1/2),x)

[Out]

Integral((a*cos(e + f*x))**m/sqrt(b*csc(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \cos \left (f x + e\right )\right )^{m}}{\sqrt{b \csc \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cos(f*x + e))^m/sqrt(b*csc(f*x + e)), x)