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

Optimal. Leaf size=77 \[ \frac{(c \sin (a+b x))^{m+1} \, _2F_1\left (-\frac{1}{4},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c d (m+1) \sqrt [4]{\cos ^2(a+b x)} \sqrt{d \sec (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.107111, antiderivative size = 77, 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, 2577} \[ \frac{(c \sin (a+b x))^{m+1} \, _2F_1\left (-\frac{1}{4},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c d (m+1) \sqrt [4]{\cos ^2(a+b x)} \sqrt{d \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

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

Mathematica [A]  time = 3.82065, size = 116, normalized size = 1.51 \[ \frac{2 c \cos (2 (a+b x)) \left (-\tan ^2(a+b x)\right )^{\frac{1-m}{2}} (c \sin (a+b x))^{m-1} \, _2F_1\left (\frac{1}{4} (-2 m-3),\frac{1-m}{2};\frac{1}{4} (1-2 m);\sec ^2(a+b x)\right )}{b d (2 m+3) \left (\sec ^2(a+b x)-2\right ) \sqrt{d \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.095, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c\sin \left ( bx+a \right ) \right ) ^{m} \left ( d\sec \left ( bx+a \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((c*sin(b*x + a))^m/(d*sec(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{d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m}}{d^{2} \sec \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*sec(b*x + a))*(c*sin(b*x + a))^m/(d^2*sec(b*x + a)^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((c*sin(b*x+a))**m/(d*sec(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 (c \sin \left (b x + a\right )\right )^{m}}{\left (d \sec \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((c*sin(b*x+a))^m/(d*sec(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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