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

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

[Out]

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

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Rubi [A]  time = 0.109772, antiderivative size = 76, 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} \[ -\frac{c \sqrt{c \sin (a+b x)} (b \sec (a+b x))^{n-1} \, _2F_1\left (-\frac{1}{4},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(a+b x)\right )}{(1-n) \sqrt [4]{\sin ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.757648, size = 104, normalized size = 1.37 \[ \frac{2 (c \sin (a+b x))^{5/2} \cos ^2(a+b x)^{\frac{n-1}{2}} (b \sec (a+b x))^{n-1} \left (5 \sin ^2(a+b x) \, _2F_1\left (\frac{9}{4},\frac{n+1}{2};\frac{13}{4};\sin ^2(a+b x)\right )+9 \, _2F_1\left (\frac{5}{4},\frac{n-1}{2};\frac{9}{4};\sin ^2(a+b x)\right )\right )}{45 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[a + b*x]^2)^((-1 + n)/2)*(b*Sec[a + b*x])^(-1 + n)*(c*Sin[a + b*x])^(5/2)*(9*Hypergeometric2F1[5/4, (-
1 + n)/2, 9/4, Sin[a + b*x]^2] + 5*Hypergeometric2F1[9/4, (1 + n)/2, 13/4, Sin[a + b*x]^2]*Sin[a + b*x]^2))/(4
5*c)

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Maple [F]  time = 0.111, 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 \left (c \sin \left (b x + a\right )\right )^{\frac{3}{2}} \left (b \sec \left (b x + a\right )\right )^{n}\,{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((c*sin(b*x + a))^(3/2)*(b*sec(b*x + a))^n, x)

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