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

Optimal. Leaf size=170 \[ -\frac{e (e \cos (c+d x))^{p-1} \left (-\frac{b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac{1-p}{2}} \left (\frac{b (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac{1-p}{2}} F_1\left (3-p;\frac{1-p}{2},\frac{1-p}{2};4-p;\frac{a+b}{a+b \sin (c+d x)},\frac{a-b}{a+b \sin (c+d x)}\right )}{b d (3-p) (a+b \sin (c+d x))^2} \]

[Out]

-((e*AppellF1[3 - p, (1 - p)/2, (1 - p)/2, 4 - p, (a + b)/(a + b*Sin[c + d*x]), (a - b)/(a + b*Sin[c + d*x])]*
(e*Cos[c + d*x])^(-1 + p)*(-((b*(1 - Sin[c + d*x]))/(a + b*Sin[c + d*x])))^((1 - p)/2)*((b*(1 + Sin[c + d*x]))
/(a + b*Sin[c + d*x]))^((1 - p)/2))/(b*d*(3 - p)*(a + b*Sin[c + d*x])^2))

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Rubi [A]  time = 0.0689775, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {2703} \[ -\frac{e (e \cos (c+d x))^{p-1} \left (-\frac{b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac{1-p}{2}} \left (\frac{b (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac{1-p}{2}} F_1\left (3-p;\frac{1-p}{2},\frac{1-p}{2};4-p;\frac{a+b}{a+b \sin (c+d x)},\frac{a-b}{a+b \sin (c+d x)}\right )}{b d (3-p) (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*AppellF1[3 - p, (1 - p)/2, (1 - p)/2, 4 - p, (a + b)/(a + b*Sin[c + d*x]), (a - b)/(a + b*Sin[c + d*x])]*
(e*Cos[c + d*x])^(-1 + p)*(-((b*(1 - Sin[c + d*x]))/(a + b*Sin[c + d*x])))^((1 - p)/2)*((b*(1 + Sin[c + d*x]))
/(a + b*Sin[c + d*x]))^((1 - p)/2))/(b*d*(3 - p)*(a + b*Sin[c + d*x])^2))

Rule 2703

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*AppellF1[-p - m, (1 - p)/2, (1 - p)/2, 1 - p - m, (a + b)/(
a + b*Sin[e + f*x]), (a - b)/(a + b*Sin[e + f*x])])/(b*f*(m + p)*(-((b*(1 - Sin[e + f*x]))/(a + b*Sin[e + f*x]
)))^((p - 1)/2)*((b*(1 + Sin[e + f*x]))/(a + b*Sin[e + f*x]))^((p - 1)/2)), x] /; FreeQ[{a, b, e, f, g, p}, x]
&& NeQ[a^2 - b^2, 0] && ILtQ[m, 0] &&  !IGtQ[m + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx &=-\frac{e F_1\left (3-p;\frac{1-p}{2},\frac{1-p}{2};4-p;\frac{a+b}{a+b \sin (c+d x)},\frac{a-b}{a+b \sin (c+d x)}\right ) (e \cos (c+d x))^{-1+p} \left (-\frac{b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac{1-p}{2}} \left (\frac{b (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac{1-p}{2}}}{b d (3-p) (a+b \sin (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 28.951, size = 7904, normalized size = 46.49 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.744, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\cos \left ( dx+c \right ) \right ) ^{p}}{ \left ( a+b\sin \left ( dx+c \right ) \right ) ^{3}}}\, 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,x)

[Out]

int((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^3,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 )}^{3}}\,{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,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\left (e \cos \left (d x + c\right )\right )^{p}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} +{\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, 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,x, algorithm="fricas")

[Out]

integral(-(e*cos(d*x + c))^p/(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 + (b^3*cos(d*x + c)^2 - 3*a^2*b - b^3)*si
n(d*x + c)), 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((e*cos(d*x+c))**p/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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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 )}^{3}}\,{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,x, algorithm="giac")

[Out]

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