3.282 \(\int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=78 \[ \frac {\sqrt [4]{\cos ^2(a+b x)} \csc ^{p-1}(a+b x) \, _2F_1\left (\frac {5}{4},\frac {1-p}{2};\frac {3-p}{2};\sin ^2(a+b x)\right )}{b d (1-p) \sqrt {d \cos (a+b x)}} \]

[Out]

(cos(b*x+a)^2)^(1/4)*csc(b*x+a)^(-1+p)*hypergeom([5/4, 1/2-1/2*p],[3/2-1/2*p],sin(b*x+a)^2)/b/d/(1-p)/(d*cos(b
*x+a))^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2587, 2577} \[ \frac {\sqrt [4]{\cos ^2(a+b x)} \csc ^{p-1}(a+b x) \, _2F_1\left (\frac {5}{4},\frac {1-p}{2};\frac {3-p}{2};\sin ^2(a+b x)\right )}{b d (1-p) \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^p/(d*Cos[a + b*x])^(3/2),x]

[Out]

((Cos[a + b*x]^2)^(1/4)*Csc[a + b*x]^(-1 + p)*Hypergeometric2F1[5/4, (1 - p)/2, (3 - p)/2, Sin[a + b*x]^2])/(b
*d*(1 - p)*Sqrt[d*Cos[a + b*x]])

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 2587

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

Rubi steps

\begin {align*} \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx &=\left (\csc ^p(a+b x) \sin ^p(a+b x)\right ) \int \frac {\sin ^{-p}(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\\ &=\frac {\sqrt [4]{\cos ^2(a+b x)} \csc ^{-1+p}(a+b x) \, _2F_1\left (\frac {5}{4},\frac {1-p}{2};\frac {3-p}{2};\sin ^2(a+b x)\right )}{b d (1-p) \sqrt {d \cos (a+b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 68, normalized size = 0.87 \[ \frac {2 \sin ^2(a+b x)^{\frac {p-1}{2}} \csc ^{p-1}(a+b x) \, _2F_1\left (-\frac {1}{4},\frac {p+1}{2};\frac {3}{4};\cos ^2(a+b x)\right )}{b d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^p/(d*Cos[a + b*x])^(3/2),x]

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \cos \left (b x + a\right )} \csc \left (b x + a\right )^{p}}{d^{2} \cos \left (b x + a\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*cos(b*x + a))*csc(b*x + a)^p/(d^2*cos(b*x + a)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )^{p}}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^p/(d*cos(b*x + a))^(3/2), x)

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{p}\left (b x +a \right )}{\left (d \cos \left (b x +a \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^p/(d*cos(b*x+a))^(3/2),x)

[Out]

int(csc(b*x+a)^p/(d*cos(b*x+a))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )^{p}}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^p/(d*cos(b*x + a))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^p}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/sin(a + b*x))^p/(d*cos(a + b*x))^(3/2),x)

[Out]

int((1/sin(a + b*x))^p/(d*cos(a + b*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{p}{\left (a + b x \right )}}{\left (d \cos {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**p/(d*cos(b*x+a))**(3/2),x)

[Out]

Integral(csc(a + b*x)**p/(d*cos(a + b*x))**(3/2), x)

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