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

Optimal. Leaf size=158 \[ -\frac {e (e \cos (c+d x))^{p-1} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} F_1\left (1-p;\frac {1-p}{2},\frac {1-p}{2};2-p;\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right )}{b d (1-p)} \]

[Out]

-e*AppellF1(1-p,1/2-1/2*p,1/2-1/2*p,2-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/(1-p)

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Rubi [A]  time = 0.08, antiderivative size = 158, 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} \[ -\frac {e (e \cos (c+d x))^{p-1} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} F_1\left (1-p;\frac {1-p}{2},\frac {1-p}{2};2-p;\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right )}{b d (1-p)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*AppellF1[1 - p, (1 - p)/2, (1 - p)/2, 2 - 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*(1 - p)))

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)} \, dx &=-\frac {e F_1\left (1-p;\frac {1-p}{2},\frac {1-p}{2};2-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 (1-p)}\\ \end {align*}

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Mathematica [B]  time = 20.27, size = 3815, normalized size = 24.15 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((e*Cos[c + d*x])^p*Tan[c + d*x]*(a*Sqrt[Sec[c + d*x]^2] + b*Tan[c + d*x])*(-(b*AppellF1[1, (1 + p)/2, 1, 2, -
Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Tan[c + d*x]) - (6*a^5*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]
^2, ((-a^2 + b^2)*Tan[c + d*x]^2)/a^2])/((Sec[c + d*x]^2)^(p/2)*(a^2 + (a^2 - b^2)*Tan[c + d*x]^2)*(-3*a^2*App
ellF1[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))))/(2*a^2*d*Sqrt[Sec[c + d*x]^2]*(a + b*Sin[c + d*x])*(a + (
b*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])*((Sqrt[Sec[c + d*x]^2]*(a*Sqrt[Sec[c + d*x]^2] + b*Tan[c + d*x])*(-(b*Ap
pellF1[1, (1 + p)/2, 1, 2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Tan[c + d*x]) - (6*a^5*AppellF1[1/2
, p/2, 1, 3/2, -Tan[c + d*x]^2, ((-a^2 + b^2)*Tan[c + d*x]^2)/a^2])/((Sec[c + d*x]^2)^(p/2)*(a^2 + (a^2 - b^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))))/(2*a^2*(a + (b*Tan[c + d*x])/
Sqrt[Sec[c + d*x]^2])) - (Tan[c + d*x]^2*(a*Sqrt[Sec[c + d*x]^2] + b*Tan[c + d*x])*(-(b*AppellF1[1, (1 + p)/2,
1, 2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Tan[c + d*x]) - (6*a^5*AppellF1[1/2, p/2, 1, 3/2, -Tan[
c + d*x]^2, ((-a^2 + b^2)*Tan[c + d*x]^2)/a^2])/((Sec[c + d*x]^2)^(p/2)*(a^2 + (a^2 - b^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))))/(2*a^2*Sqrt[Sec[c + d*x]^2]*(a + (b*Tan[c + d*x])
/Sqrt[Sec[c + d*x]^2])) + (Tan[c + d*x]*(b*Sec[c + d*x]^2 + a*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(-(b*AppellF1
[1, (1 + p)/2, 1, 2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Tan[c + d*x]) - (6*a^5*AppellF1[1/2, p/2,
1, 3/2, -Tan[c + d*x]^2, ((-a^2 + b^2)*Tan[c + d*x]^2)/a^2])/((Sec[c + d*x]^2)^(p/2)*(a^2 + (a^2 - b^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))))/(2*a^2*Sqrt[Sec[c + d*x]^2]*(a + (b
*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])) - (Tan[c + d*x]*(a*Sqrt[Sec[c + d*x]^2] + b*Tan[c + d*x])*(b*Sqrt[Sec[c
+ d*x]^2] - (b*Tan[c + d*x]^2)/Sqrt[Sec[c + d*x]^2])*(-(b*AppellF1[1, (1 + p)/2, 1, 2, -Tan[c + d*x]^2, (-1 +
b^2/a^2)*Tan[c + d*x]^2]*Tan[c + d*x]) - (6*a^5*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, ((-a^2 + b^2)*Tan[
c + d*x]^2)/a^2])/((Sec[c + d*x]^2)^(p/2)*(a^2 + (a^2 - b^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))))/(2*a^2*Sqrt[Sec[c + d*x]^2]*(a + (b*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^2) + (T
an[c + d*x]*(a*Sqrt[Sec[c + d*x]^2] + b*Tan[c + d*x])*(-(b*AppellF1[1, (1 + p)/2, 1, 2, -Tan[c + d*x]^2, (-1 +
b^2/a^2)*Tan[c + d*x]^2]*Sec[c + d*x]^2) - b*Tan[c + d*x]*((-1 + b^2/a^2)*AppellF1[2, (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] - ((1 + p)*AppellF1[2, 1 + (1 + p)/2, 1,
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) + (12*a^5*(a^2 - b^2)*Appe
llF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, ((-a^2 + b^2)*Tan[c + d*x]^2)/a^2]*(Sec[c + d*x]^2)^(1 - p/2)*Tan[c +
d*x])/((a^2 + (a^2 - b^2)*Tan[c + d*x]^2)^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)) + (6
*a^5*p*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, ((-a^2 + b^2)*Tan[c + d*x]^2)/a^2]*Tan[c + d*x])/((Sec[c +
d*x]^2)^(p/2)*(a^2 + (a^2 - b^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))
- (6*a^5*(-1/3*(p*AppellF1[3/2, 1 + p/2, 1, 5/2, -Tan[c + d*x]^2, ((-a^2 + b^2)*Tan[c + d*x]^2)/a^2]*Sec[c +
d*x]^2*Tan[c + d*x]) + (2*(-a^2 + b^2)*AppellF1[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, ((-a^2 + b^2)*Tan[c + d*x]^
2)/a^2]*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2)))/((Sec[c + d*x]^2)^(p/2)*(a^2 + (a^2 - b^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)) + (6*a^5*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2,
((-a^2 + b^2)*Tan[c + d*x]^2)/a^2]*(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])*S
ec[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))))/((Sec[c + d*x]^2)^(p/2)*(a^2 + (a^2 - b^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)*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])*Tan[c + d*x]^2)^2)))/(2*a^2*Sqrt[Sec[c + d*x]^2]*(a + (b*T
an[c + d*x])/Sqrt[Sec[c + d*x]^2]))))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \cos \left (d x + c\right )\right )^{p}}{b \sin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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