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

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B]  time = 25.54, size = 4727, normalized size = 27.81 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((e*Cos[c + d*x])^p*Tan[c + d*x]*(b*(a^2 - b^2)*AppellF1[1, (-1 + p)/2, 2, 2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*
Tan[c + d*x]^2]*Tan[c + d*x] + (3*a^5*((-2*a^2*b^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*
Tan[c + d*x]^2])/((-3*a^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (4*(a^2
- b^2)*AppellF1[3/2, p/2, 3, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 +
p)/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)*(b^2*Tan[c + d*x]^2 - a^2*(1 +
Tan[c + d*x]^2))^2) + ((a^2 + b^2)*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])
/((-3*a^2*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (2*(a^2 - b^2)*AppellF1
[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 1, 5/2, -T
an[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)*(-(b^2*Tan[c + d*x]^2) + a^2*(1 + Tan[c + d*x]^
2)))))/(1 + Tan[c + d*x]^2)^(p/2)))/(a^3*(-a^2 + b^2)*d*(a + b*Sin[c + d*x])^2*((Sec[c + d*x]^2*(b*(a^2 - b^2)
*AppellF1[1, (-1 + p)/2, 2, 2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Tan[c + d*x] + (3*a^5*((-2*a^2*
b^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])/((-3*a^2*AppellF1[1/2, p/2, 2,
3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (4*(a^2 - b^2)*AppellF1[3/2, p/2, 3, 5/2, -Tan[c + d*x
]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*T
an[c + d*x]^2])*Tan[c + d*x]^2)*(b^2*Tan[c + d*x]^2 - a^2*(1 + Tan[c + d*x]^2))^2) + ((a^2 + b^2)*AppellF1[1/2
, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])/((-3*a^2*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d
*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (2*(a^2 - b^2)*AppellF1[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a
^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 1, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*
Tan[c + d*x]^2)*(-(b^2*Tan[c + d*x]^2) + a^2*(1 + Tan[c + d*x]^2)))))/(1 + Tan[c + d*x]^2)^(p/2)))/(a^3*(-a^2
+ b^2)) + (Tan[c + d*x]*(b*(a^2 - b^2)*AppellF1[1, (-1 + p)/2, 2, 2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d
*x]^2]*Sec[c + d*x]^2 + b*(a^2 - b^2)*Tan[c + d*x]*(-1/2*((-1 + p)*AppellF1[2, 1 + (-1 + p)/2, 2, 3, -Tan[c +
d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x]) + 2*(-1 + b^2/a^2)*AppellF1[2, (-1 + p)/2,
3, 3, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x]) - 3*a^5*p*Sec[c + d*x]^2*T
an[c + d*x]*(1 + Tan[c + d*x]^2)^(-1 - p/2)*((-2*a^2*b^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2
/a^2)*Tan[c + d*x]^2])/((-3*a^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (
4*(a^2 - b^2)*AppellF1[3/2, p/2, 3, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2,
(2 + p)/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)*(b^2*Tan[c + d*x]^2 - a^2
*(1 + Tan[c + d*x]^2))^2) + ((a^2 + b^2)*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*
x]^2])/((-3*a^2*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (2*(a^2 - b^2)*Ap
pellF1[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 1, 5
/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)*(-(b^2*Tan[c + d*x]^2) + a^2*(1 + Tan[c +
d*x]^2)))) + (3*a^5*((4*a^2*b^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*(-
2*a^2*Sec[c + d*x]^2*Tan[c + d*x] + 2*b^2*Sec[c + d*x]^2*Tan[c + d*x]))/((-3*a^2*AppellF1[1/2, p/2, 2, 3/2, -T
an[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (4*(a^2 - b^2)*AppellF1[3/2, p/2, 3, 5/2, -Tan[c + d*x]^2, (-1
+ b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d
*x]^2])*Tan[c + d*x]^2)*(b^2*Tan[c + d*x]^2 - a^2*(1 + Tan[c + d*x]^2))^3) - (2*a^2*b^2*(-1/3*(p*AppellF1[3/2,
1 + p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x]) + (4*(-1 + b^2/
a^2)*AppellF1[3/2, p/2, 3, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/3
))/((-3*a^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (4*(a^2 - b^2)*Appell
F1[3/2, p/2, 3, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 2, 5/2,
-Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)*(b^2*Tan[c + d*x]^2 - a^2*(1 + Tan[c + d*x]^2
))^2) - ((a^2 + b^2)*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*(2*a^2*Sec[c +
d*x]^2*Tan[c + d*x] - 2*b^2*Sec[c + d*x]^2*Tan[c + d*x]))/((-3*a^2*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2
, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (2*(a^2 - b^2)*AppellF1[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*T
an[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 1, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c
+ d*x]^2)*(-(b^2*Tan[c + d*x]^2) + a^2*(1 + Tan[c + d*x]^2))^2) + ((a^2 + b^2)*(-1/3*(p*AppellF1[3/2, 1 + p/2
, 1, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x]) + (2*(-1 + b^2/a^2)*App
ellF1[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/3))/((-3*
a^2*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (2*(a^2 - b^2)*AppellF1[3/2,
p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 1, 5/2, -Tan[c +
d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)*(-(b^2*Tan[c + d*x]^2) + a^2*(1 + Tan[c + d*x]^2))) -
((a^2 + b^2)*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*(2*(2*(a^2 - b^2)*App
ellF1[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 1, 5/
2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Sec[c + d*x]^2*Tan[c + d*x] - 3*a^2*(-1/3*(p*AppellF1[3/2,
1 + p/2, 1, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x]) + (2*(-1 + b^2/
a^2)*AppellF1[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/3
) + Tan[c + d*x]^2*(2*(a^2 - b^2)*((-3*p*AppellF1[5/2, 1 + p/2, 2, 7/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c
+ d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/5 + (12*(-1 + b^2/a^2)*AppellF1[5/2, p/2, 3, 7/2, -Tan[c + d*x]^2, (-1
+ b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/5) + a^2*p*((6*(-1 + b^2/a^2)*AppellF1[5/2, (2 + p)/2,
2, 7/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/5 - (3*(2 + p)*AppellF1[
5/2, 1 + (2 + p)/2, 1, 7/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/5))))
/((-3*a^2*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (2*(a^2 - b^2)*AppellF1
[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 1, 5/2, -T
an[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)^2*(-(b^2*Tan[c + d*x]^2) + a^2*(1 + Tan[c + d*x
]^2))) + (2*a^2*b^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*(2*(4*(a^2 - b^
2)*AppellF1[3/2, p/2, 3, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2,
2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Sec[c + d*x]^2*Tan[c + d*x] - 3*a^2*(-1/3*(p*AppellF
1[3/2, 1 + p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x]) + (4*(-1
+ b^2/a^2)*AppellF1[3/2, p/2, 3, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d
*x])/3) + Tan[c + d*x]^2*(4*(a^2 - b^2)*((-3*p*AppellF1[5/2, 1 + p/2, 3, 7/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*
Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/5 + (18*(-1 + b^2/a^2)*AppellF1[5/2, p/2, 4, 7/2, -Tan[c + d*x]^2
, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/5) + a^2*p*((12*(-1 + b^2/a^2)*AppellF1[5/2, (2
+ p)/2, 3, 7/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x])/5 - (3*(2 + p)*Ap
pellF1[5/2, 1 + (2 + p)/2, 2, 7/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2*Tan[c + d*x]
)/5))))/((-3*a^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (4*(a^2 - b^2)*A
ppellF1[3/2, p/2, 3, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 2,
5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)^2*(b^2*Tan[c + d*x]^2 - a^2*(1 + Tan[c +
d*x]^2))^2)))/(1 + Tan[c + d*x]^2)^(p/2)))/(a^3*(-a^2 + b^2))))

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (e \cos \left (d x + c\right )\right )^{p}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +b \sin \left (d x +c \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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