3.505 \(\int \frac {(b \sec (a+b x))^n}{\sqrt {c \sin (a+b x)}} \, dx\)

Optimal. Leaf size=76 \[ -\frac {c \sin ^2(a+b x)^{3/4} (b \sec (a+b x))^{n-1} \, _2F_1\left (\frac {3}{4},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(a+b x)\right )}{(1-n) (c \sin (a+b x))^{3/2}} \]

[Out]

-c*hypergeom([3/4, 1/2-1/2*n],[3/2-1/2*n],cos(b*x+a)^2)*(b*sec(b*x+a))^(-1+n)*(sin(b*x+a)^2)^(3/4)/(1-n)/(c*si
n(b*x+a))^(3/2)

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Rubi [A]  time = 0.10, antiderivative size = 76, 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 = {2587, 2576} \[ -\frac {c \sin ^2(a+b x)^{3/4} (b \sec (a+b x))^{n-1} \, _2F_1\left (\frac {3}{4},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(a+b x)\right )}{(1-n) (c \sin (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

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Mathematica [A]  time = 0.14, size = 72, normalized size = 0.95 \[ \frac {\sin (2 (a+b x)) \cos ^2(a+b x)^{\frac {n-1}{2}} (b \sec (a+b x))^n \, _2F_1\left (\frac {1}{4},\frac {n+1}{2};\frac {5}{4};\sin ^2(a+b x)\right )}{b \sqrt {c \sin (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c \sin \left (b x + a\right )} \left (b \sec \left (b x + a\right )\right )^{n}}{c \sin \left (b x + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (b x + a\right )\right )^{n}}{\sqrt {c \sin \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (b x +a \right )\right )^{n}}{\sqrt {c \sin \left (b x +a \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(b*x+a))^n/(c*sin(b*x+a))^(1/2),x)

[Out]

int((b*sec(b*x+a))^n/(c*sin(b*x+a))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (b x + a\right )\right )^{n}}{\sqrt {c \sin \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(a + b*x))^n/(c*sin(a + b*x))^(1/2),x)

[Out]

int((b/cos(a + b*x))^n/(c*sin(a + b*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec {\left (a + b x \right )}\right )^{n}}{\sqrt {c \sin {\left (a + b x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(b*x+a))**n/(c*sin(b*x+a))**(1/2),x)

[Out]

Integral((b*sec(a + b*x))**n/sqrt(c*sin(a + b*x)), x)

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