3.627 \(\int \frac{(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=154 \[ -\frac{2 e (e \cos (c+d x))^{p-1} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}} F_1\left (-\frac{1}{2};\frac{1-p}{2},\frac{1-p}{2};\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{b d \sqrt{a+b \sin (c+d x)}} \]

[Out]

(-2*e*AppellF1[-1/2, (1 - p)/2, (1 - p)/2, 1/2, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*(e
*Cos[c + d*x])^(-1 + p)*(1 - (a + b*Sin[c + d*x])/(a - b))^((1 - p)/2)*(1 - (a + b*Sin[c + d*x])/(a + b))^((1
- p)/2))/(b*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.11651, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2704, 138} \[ -\frac{2 e (e \cos (c+d x))^{p-1} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}} F_1\left (-\frac{1}{2};\frac{1-p}{2},\frac{1-p}{2};\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{b d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*AppellF1[-1/2, (1 - p)/2, (1 - p)/2, 1/2, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*(e
*Cos[c + d*x])^(-1 + p)*(1 - (a + b*Sin[c + d*x])/(a - b))^((1 - p)/2)*(1 - (a + b*Sin[c + d*x])/(a + b))^((1
- p)/2))/(b*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2704

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(g*(g*
Cos[e + f*x])^(p - 1))/(f*(1 - (a + b*Sin[e + f*x])/(a - b))^((p - 1)/2)*(1 - (a + b*Sin[e + f*x])/(a + b))^((
p - 1)/2)), Subst[Int[(-(b/(a - b)) - (b*x)/(a - b))^((p - 1)/2)*(b/(a + b) - (b*x)/(a + b))^((p - 1)/2)*(a +
b*x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] &&  !IGtQ[m, 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx &=\frac{\left (e (e \cos (c+d x))^{-1+p} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{b}{a-b}-\frac{b x}{a-b}\right )^{\frac{1}{2} (-1+p)} \left (\frac{b}{a+b}-\frac{b x}{a+b}\right )^{\frac{1}{2} (-1+p)}}{(a+b x)^{3/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{2 e F_1\left (-\frac{1}{2};\frac{1-p}{2},\frac{1-p}{2};\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}}}{b d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.92009, size = 185, normalized size = 1.2 \[ -\frac{2 e (e \cos (c+d x))^{p-1} \left (\frac{\sqrt{b^2}-b \sin (c+d x)}{a+\sqrt{b^2}}\right )^{\frac{1-p}{2}} \left (\frac{\sqrt{b^2}+b \sin (c+d x)}{\sqrt{b^2}-a}\right )^{\frac{1-p}{2}} F_1\left (-\frac{1}{2};\frac{1-p}{2},\frac{1-p}{2};\frac{1}{2};\frac{a+b \sin (c+d x)}{a-\sqrt{b^2}},\frac{a+b \sin (c+d x)}{a+\sqrt{b^2}}\right )}{b d \sqrt{a+b \sin (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*e*AppellF1[-1/2, (1 - p)/2, (1 - p)/2, 1/2, (a + b*Sin[c + d*x])/(a - Sqrt[b^2]), (a + b*Sin[c + d*x])/(a
+ Sqrt[b^2])]*(e*Cos[c + d*x])^(-1 + p)*((Sqrt[b^2] - b*Sin[c + d*x])/(a + Sqrt[b^2]))^((1 - p)/2)*((Sqrt[b^2]
+ b*Sin[c + d*x])/(-a + Sqrt[b^2]))^((1 - p)/2))/(b*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [F]  time = 0.129, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*(e*cos(d*x + c))^p/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2), x
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**p/(a+b*sin(d*x+c))**(3/2),x)

[Out]

Integral((e*cos(c + d*x))**p/(a + b*sin(c + d*x))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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