3.506 \(\int \frac{(b \sec (a+b x))^n}{(c \sin (a+b x))^{3/2}} \, dx\)

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

[Out]

-((Hypergeometric2F1[5/4, (1 - n)/2, (3 - n)/2, Cos[a + b*x]^2]*(b*Sec[a + b*x])^(-1 + n)*(Sin[a + b*x]^2)^(1/
4))/(c*(1 - n)*Sqrt[c*Sin[a + b*x]]))

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Rubi [A]  time = 0.116657, 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 = {2587, 2576} \[ -\frac{\sqrt [4]{\sin ^2(a+b x)} (b \sec (a+b x))^{n-1} \, _2F_1\left (\frac{5}{4},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(a+b x)\right )}{c (1-n) \sqrt{c \sin (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[a + b*x])^n/(c*Sin[a + b*x])^(3/2),x]

[Out]

-((Hypergeometric2F1[5/4, (1 - n)/2, (3 - n)/2, Cos[a + b*x]^2]*(b*Sec[a + b*x])^(-1 + n)*(Sin[a + b*x]^2)^(1/
4))/(c*(1 - n)*Sqrt[c*Sin[a + b*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]

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

Mathematica [A]  time = 0.148636, size = 73, normalized size = 0.94 \[ -\frac{\sin (2 (a+b x)) \cos ^2(a+b x)^{\frac{n-1}{2}} (b \sec (a+b x))^n \, _2F_1\left (-\frac{1}{4},\frac{n+1}{2};\frac{3}{4};\sin ^2(a+b x)\right )}{b (c \sin (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[a + b*x])^n/(c*Sin[a + b*x])^(3/2),x]

[Out]

-(((Cos[a + b*x]^2)^((-1 + n)/2)*Hypergeometric2F1[-1/4, (1 + n)/2, 3/4, Sin[a + b*x]^2]*(b*Sec[a + b*x])^n*Si
n[2*(a + b*x)])/(b*(c*Sin[a + b*x])^(3/2)))

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\sec \left ( bx+a \right ) \right ) ^{n} \left ( c\sin \left ( bx+a \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x)

[Out]

int((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \sec \left (b x + a\right )\right )^{n}}{\left (c \sin \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(b*x + a))^n/(c*sin(b*x + a))^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{c \sin \left (b x + a\right )} \left (b \sec \left (b x + a\right )\right )^{n}}{c^{2} \cos \left (b x + a\right )^{2} - c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(c*sin(b*x + a))*(b*sec(b*x + a))^n/(c^2*cos(b*x + a)^2 - c^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(b*x+a))**n/(c*sin(b*x+a))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \sec \left (b x + a\right )\right )^{n}}{\left (c \sin \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(b*x + a))^n/(c*sin(b*x + a))^(3/2), x)