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

Optimal. Leaf size=93 \[ -\frac {2^{\frac {p-3}{2}} (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {5-p}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{a^2 d e (p+1)} \]

[Out]

-2^(-3/2+1/2*p)*(e*cos(d*x+c))^(1+p)*hypergeom([1/2+1/2*p, 5/2-1/2*p],[3/2+1/2*p],1/2-1/2*sin(d*x+c))*(1+sin(d
*x+c))^(-1/2-1/2*p)/a^2/d/e/(1+p)

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Rubi [A]  time = 0.09, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2688, 69} \[ -\frac {2^{\frac {p-3}{2}} (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {5-p}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{a^2 d e (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^((-3 + p)/2)*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(5 - p)/2, (1 + p)/2, (3 + p)/2, (1 - Sin[c + d*x
])/2]*(1 + Sin[c + d*x])^((-1 - p)/2))/(a^2*d*e*(1 + p)))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 2688

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^m*
(g*Cos[e + f*x])^(p + 1))/(f*g*(1 + Sin[e + f*x])^((p + 1)/2)*(1 - Sin[e + f*x])^((p + 1)/2)), Subst[Int[(1 +
(b*x)/a)^(m + (p - 1)/2)*(1 - (b*x)/a)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, p}, x] &
& EqQ[a^2 - b^2, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^2} \, dx &=\frac {\left ((e \cos (c+d x))^{1+p} (1-\sin (c+d x))^{\frac {1}{2} (-1-p)} (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}\right ) \operatorname {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+p)} (1+x)^{-2+\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{a^2 d e}\\ &=-\frac {2^{\frac {1}{2} (-3+p)} (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {5-p}{2},\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}}{a^2 d e (1+p)}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 94, normalized size = 1.01 \[ -\frac {2^{\frac {p-3}{2}} \cos (c+d x) (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^p \, _2F_1\left (\frac {5-p}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{a^2 d (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^((-3 + p)/2)*Cos[c + d*x]*(e*Cos[c + d*x])^p*Hypergeometric2F1[(5 - p)/2, (1 + p)/2, (3 + p)/2, (1 - Sin[
c + d*x])/2]*(1 + Sin[c + d*x])^((-1 - p)/2))/(a^2*d*(1 + p)))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (e \cos \left (d x + c\right )\right )^{p}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +a \sin \left (d x +c \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*cos(c + d*x))**p/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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