\(\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx\) [620]
Optimal result
Integrand size = 23, antiderivative size = 170 \[
\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=-\frac {e \operatorname {AppellF1}\left (2-p,\frac {1-p}{2},\frac {1-p}{2},3-p,\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right ) (e \cos (c+d x))^{-1+p} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}}}{b d (2-p) (a+b \sin (c+d x))}
\]
[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))
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}
\[
\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=-\frac {e (e \cos (c+d x))^{p-1} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (2-p,\frac {1-p}{2},\frac {1-p}{2},3-p,\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right )}{b d (2-p) (a+b \sin (c+d x))}
\]
[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 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 (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*}
Mathematica [B] (warning: unable to verify)
Leaf count is larger than twice the leaf count of optimal. \(4727\) vs. \(2(170)=340\).
Time = 25.71 (sec) , antiderivative size = 4727, normalized size of antiderivative = 27.81
\[
\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=\text {Result too large to show}
\]
[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))))
Maple [F]
\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +b \sin \left (d x +c \right )\right )^{2}}d x\]
[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)
Fricas [F]
\[
\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x }
\]
[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)
Sympy [F(-1)]
Timed out. \[
\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out}
\]
[In]
integrate((e*cos(d*x+c))**p/(a+b*sin(d*x+c))**2,x)
[Out]
Timed out
Maxima [F]
\[
\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x }
\]
[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)
Giac [F]
\[
\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x }
\]
[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)
Mupad [F(-1)]
Timed out. \[
\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x
\]
[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)