3.622 \(\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^8} \, dx\)

Optimal. Leaf size=170 \[ -\frac {e (e \cos (c+d x))^{p-1} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} F_1\left (8-p;\frac {1-p}{2},\frac {1-p}{2};9-p;\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right )}{b d (8-p) (a+b \sin (c+d x))^7} \]

[Out]

-e*AppellF1(8-p,1/2-1/2*p,1/2-1/2*p,9-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/(8-p)/(a+b
*sin(d*x+c))^7

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Rubi [A]  time = 0.07, antiderivative size = 170, normalized size of antiderivative = 1.00, 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} \[ -\frac {e (e \cos (c+d x))^{p-1} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} F_1\left (8-p;\frac {1-p}{2},\frac {1-p}{2};9-p;\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right )}{b d (8-p) (a+b \sin (c+d x))^7} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^p/(a + b*Sin[c + d*x])^8,x]

[Out]

-((e*AppellF1[8 - p, (1 - p)/2, (1 - p)/2, 9 - 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*(8 - p)*(a + b*Sin[c + d*x])^7))

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))^8} \, dx &=-\frac {e F_1\left (8-p;\frac {1-p}{2},\frac {1-p}{2};9-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 (8-p) (a+b \sin (c+d x))^7}\\ \end {align*}

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Mathematica [F]  time = 67.68, size = 0, normalized size = 0.00 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^8} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(e*Cos[c + d*x])^p/(a + b*Sin[c + d*x])^8,x]

[Out]

Integrate[(e*Cos[c + d*x])^p/(a + b*Sin[c + d*x])^8, x]

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \cos \left (d x + c\right )\right )^{p}}{b^{8} \cos \left (d x + c\right )^{8} + a^{8} + 28 \, a^{6} b^{2} + 70 \, a^{4} b^{4} + 28 \, a^{2} b^{6} + b^{8} - 4 \, {\left (7 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{6} + 2 \, {\left (35 \, a^{4} b^{4} + 42 \, a^{2} b^{6} + 3 \, b^{8}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, a^{6} b^{2} + 35 \, a^{4} b^{4} + 21 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{2} - 8 \, {\left (a b^{7} \cos \left (d x + c\right )^{6} - a^{7} b - 7 \, a^{5} b^{3} - 7 \, a^{3} b^{5} - a b^{7} - {\left (7 \, a^{3} b^{5} + 3 \, a b^{7}\right )} \cos \left (d x + c\right )^{4} + {\left (7 \, a^{5} b^{3} + 14 \, a^{3} b^{5} + 3 \, a b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^p/(b^8*cos(d*x + c)^8 + a^8 + 28*a^6*b^2 + 70*a^4*b^4 + 28*a^2*b^6 + b^8 - 4*(7*a^2*
b^6 + b^8)*cos(d*x + c)^6 + 2*(35*a^4*b^4 + 42*a^2*b^6 + 3*b^8)*cos(d*x + c)^4 - 4*(7*a^6*b^2 + 35*a^4*b^4 + 2
1*a^2*b^6 + b^8)*cos(d*x + c)^2 - 8*(a*b^7*cos(d*x + c)^6 - a^7*b - 7*a^5*b^3 - 7*a^3*b^5 - a*b^7 - (7*a^3*b^5
+ 3*a*b^7)*cos(d*x + c)^4 + (7*a^5*b^3 + 14*a^3*b^5 + 3*a*b^7)*cos(d*x + c)^2)*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^8,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^p/(b*sin(d*x + c) + a)^8, x)

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maple [F]  time = 8.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +b \sin \left (d x +c \right )\right )^{8}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^8,x)

[Out]

int((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^8,x)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^p/(a + b*sin(c + d*x))^8,x)

[Out]

int((e*cos(c + d*x))^p/(a + b*sin(c + d*x))^8, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**p/(a+b*sin(d*x+c))**8,x)

[Out]

Timed out

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