\(\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx\) [621]

Optimal result

Integrand size = 23, antiderivative size = 170 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx=-\frac {e \operatorname {AppellF1}\left (3-p,\frac {1-p}{2},\frac {1-p}{2},4-p,\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right ) (e \cos (c+d x))^{-1+p} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}}}{b d (3-p) (a+b \sin (c+d x))^2} \]

[Out]

-e*AppellF1(3-p,1/2-1/2*p,1/2-1/2*p,4-p,(a-b)/(a+b*sin(d*x+c)),(a+b)/(a+b*sin(d*x+c)))*(e*cos(d*x+c))^(-1+p)*(
-b*(1-sin(d*x+c))/(a+b*sin(d*x+c)))^(1/2-1/2*p)*(b*(1+sin(d*x+c))/(a+b*sin(d*x+c)))^(1/2-1/2*p)/b/d/(3-p)/(a+b
*sin(d*x+c))^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2782} \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx=-\frac {e (e \cos (c+d x))^{p-1} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (3-p,\frac {1-p}{2},\frac {1-p}{2},4-p,\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right )}{b d (3-p) (a+b \sin (c+d x))^2} \]

[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 2782

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(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)))*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])], 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*} \text {integral}& = -\frac {e \operatorname {AppellF1}\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] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(7781\) vs. \(2(170)=340\).

Time = 28.25 (sec) , antiderivative size = 7781, normalized size of antiderivative = 45.77 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F]

\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +b \sin \left (d x +c \right )\right )^{3}}d x\]

[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)

Fricas [F]

\[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^3} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]

[In]

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

[Out]

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