3.4.41 \(\int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx\) [341]

Optimal. Leaf size=105 \[ -\frac {2^{-2+\frac {p}{2}} (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {6-p}{2},\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{1-\frac {p}{2}}}{a d e (1+p) (a+a \sin (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2768, 72, 71} \begin {gather*} -\frac {2^{\frac {p}{2}-2} (\sin (c+d x)+1)^{1-\frac {p}{2}} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {6-p}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{a d e (p+1) (a \sin (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^p/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-((2^(-2 + p/2)*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(6 - p)/2, (1 + p)/2, (3 + p)/2, (1 - Sin[c + d*x])
/2]*(1 + Sin[c + d*x])^(1 - p/2))/(a*d*e*(1 + p)*(a + a*Sin[c + d*x])^(3/2)))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, 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))^{5/2}} \, dx &=\frac {\left (a^2 (e \cos (c+d x))^{1+p} (a-a \sin (c+d x))^{\frac {1}{2} (-1-p)} (a+a \sin (c+d x))^{\frac {1}{2} (-1-p)}\right ) \text {Subst}\left (\int (a-a x)^{\frac {1}{2} (-1+p)} (a+a x)^{-\frac {5}{2}+\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac {\left (2^{-3+\frac {p}{2}} (e \cos (c+d x))^{1+p} (a-a \sin (c+d x))^{\frac {1}{2} (-1-p)} (a+a \sin (c+d x))^{-1+\frac {1}{2} (-1-p)+\frac {p}{2}} \left (\frac {a+a \sin (c+d x)}{a}\right )^{1-\frac {p}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {5}{2}+\frac {1}{2} (-1+p)} (a-a x)^{\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac {2^{-2+\frac {p}{2}} (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {6-p}{2},\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{1-\frac {p}{2}}}{a d e (1+p) (a+a \sin (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 102, normalized size = 0.97 \begin {gather*} -\frac {2^{-2+\frac {p}{2}} \cos (c+d x) (e \cos (c+d x))^p \, _2F_1\left (3-\frac {p}{2},\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-p/2}}{a^2 d (1+p) \sqrt {a (1+\sin (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^p/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

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

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x)

[Out]

int((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((cos(d*x + c)*e)^p/(a*sin(d*x + c) + a)^(5/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(d*x + c) + a)*(cos(d*x + c)*e)^p/(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*
a^3)*sin(d*x + c)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**p/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Integral((e*cos(c + d*x))**p/(a*(sin(c + d*x) + 1))**(5/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^p/(a + a*sin(c + d*x))^(5/2),x)

[Out]

int((e*cos(c + d*x))^p/(a + a*sin(c + d*x))^(5/2), x)

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