3.4.36 \(\int (e \cos (c+d x))^p (a+a \sin (c+d x))^{5/2} \, dx\) [336]

Optimal. Leaf size=103 \[ -\frac {2^{3+\frac {p}{2}} a^3 (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {1}{2} (-4-p),\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-p/2}}{d e (1+p) \sqrt {a+a \sin (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 103, 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 {a^3 2^{\frac {p}{2}+3} (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {1}{2} (-p-4),\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (p+1) \sqrt {a \sin (c+d x)+a}} \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^(3 + p/2)*a^3*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(-4 - p)/2, (1 + p)/2, (3 + p)/2, (1 - Sin[c + d
*x])/2])/(d*e*(1 + p)*(1 + Sin[c + d*x])^(p/2)*Sqrt[a + a*Sin[c + d*x]]))

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 (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^{2+\frac {p}{2}} a^4 (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)+\frac {p}{2}} \left (\frac {a+a \sin (c+d x)}{a}\right )^{-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^{3+\frac {p}{2}} a^3 (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {1}{2} (-4-p),\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-p/2}}{d e (1+p) \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 102, normalized size = 0.99 \begin {gather*} -\frac {2^{3+\frac {p}{2}} a^3 \cos (c+d x) (e \cos (c+d x))^p \, _2F_1\left (-2-\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}}{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^(3 + p/2)*a^3*Cos[c + d*x]*(e*Cos[c + d*x])^p*Hypergeometric2F1[-2 - p/2, (1 + p)/2, (3 + p)/2, (1 - Sin[
c + d*x])/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 = 0.13, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}\, dx\]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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]

Exception raised: SystemError >> excessive stack use: stack is 6189 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\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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