3.3.12 \(\int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{9/2}} \, dx\) [212]

Optimal. Leaf size=114 \[ \frac {2 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d e^4 \sqrt {e \cos (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{7 d e^3 (e \cos (c+d x))^{3/2}}+\frac {4 \left (a^2+a^2 \sin (c+d x)\right )}{7 d e (e \cos (c+d x))^{7/2}} \]

[Out]

2/7*a^2*sin(d*x+c)/d/e^3/(e*cos(d*x+c))^(3/2)+4/7*(a^2+a^2*sin(d*x+c))/d/e/(e*cos(d*x+c))^(7/2)+2/7*a^2*(cos(1
/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/e^4/(e*cos(
d*x+c))^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2755, 2716, 2721, 2720} \begin {gather*} \frac {2 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d e^4 \sqrt {e \cos (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{7 d e^3 (e \cos (c+d x))^{3/2}}+\frac {4 \left (a^2 \sin (c+d x)+a^2\right )}{7 d e (e \cos (c+d x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^2/(e*Cos[c + d*x])^(9/2),x]

[Out]

(2*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*d*e^4*Sqrt[e*Cos[c + d*x]]) + (2*a^2*Sin[c + d*x])/(7*
d*e^3*(e*Cos[c + d*x])^(3/2)) + (4*(a^2 + a^2*Sin[c + d*x]))/(7*d*e*(e*Cos[c + d*x])^(7/2))

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2755

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{9/2}} \, dx &=\frac {4 \left (a^2+a^2 \sin (c+d x)\right )}{7 d e (e \cos (c+d x))^{7/2}}+\frac {\left (3 a^2\right ) \int \frac {1}{(e \cos (c+d x))^{5/2}} \, dx}{7 e^2}\\ &=\frac {2 a^2 \sin (c+d x)}{7 d e^3 (e \cos (c+d x))^{3/2}}+\frac {4 \left (a^2+a^2 \sin (c+d x)\right )}{7 d e (e \cos (c+d x))^{7/2}}+\frac {a^2 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{7 e^4}\\ &=\frac {2 a^2 \sin (c+d x)}{7 d e^3 (e \cos (c+d x))^{3/2}}+\frac {4 \left (a^2+a^2 \sin (c+d x)\right )}{7 d e (e \cos (c+d x))^{7/2}}+\frac {\left (a^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 e^4 \sqrt {e \cos (c+d x)}}\\ &=\frac {2 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d e^4 \sqrt {e \cos (c+d x)}}+\frac {2 a^2 \sin (c+d x)}{7 d e^3 (e \cos (c+d x))^{3/2}}+\frac {4 \left (a^2+a^2 \sin (c+d x)\right )}{7 d e (e \cos (c+d x))^{7/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.06, size = 66, normalized size = 0.58 \begin {gather*} \frac {2 \sqrt [4]{2} a^2 \, _2F_1\left (-\frac {7}{4},\frac {3}{4};-\frac {3}{4};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{7/4}}{7 d e (e \cos (c+d x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^2/(e*Cos[c + d*x])^(9/2),x]

[Out]

(2*2^(1/4)*a^2*Hypergeometric2F1[-7/4, 3/4, -3/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(7/4))/(7*d*e*(e*Co
s[c + d*x])^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(374\) vs. \(2(126)=252\).
time = 4.53, size = 375, normalized size = 3.29

method	result	size
default	\(-\frac {2 \left (8 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{7 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{4} d}\)	\(375\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(9/2),x,method=_RETURNVERBOSE)

[Out]

-2/7/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*
d*x+1/2*c)^2*e+e)^(1/2)/e^4*(8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2
*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^6-12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+6*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2
-8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*sin(1/2*d*x+1/2*c))*a^2/d

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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((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

e^(-9/2)*integrate((a*sin(d*x + c) + a)^2/cos(d*x + c)^(9/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 176, normalized size = 1.54 \begin {gather*} \frac {{\left (-i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} a^{2} \sin \left (d x + c\right ) + 2 i \, \sqrt {2} a^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} a^{2} \sin \left (d x + c\right ) - 2 i \, \sqrt {2} a^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \sqrt {\cos \left (d x + c\right )}}{7 \, {\left (d \cos \left (d x + c\right )^{2} e^{\frac {9}{2}} + 2 \, d e^{\frac {9}{2}} \sin \left (d x + c\right ) - 2 \, d e^{\frac {9}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

1/7*((-I*sqrt(2)*a^2*cos(d*x + c)^2 - 2*I*sqrt(2)*a^2*sin(d*x + c) + 2*I*sqrt(2)*a^2)*weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c)) + (I*sqrt(2)*a^2*cos(d*x + c)^2 + 2*I*sqrt(2)*a^2*sin(d*x + c) - 2*I*sqrt(2)
*a^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(a^2*sin(d*x + c) - 2*a^2)*sqrt(cos(d*x +
c)))/(d*cos(d*x + c)^2*e^(9/2) + 2*d*e^(9/2)*sin(d*x + c) - 2*d*e^(9/2))

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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((a+a*sin(d*x+c))**2/(e*cos(d*x+c))**(9/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4847 deep

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Giac [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((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^2*e^(-9/2)/cos(d*x + c)^(9/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^2/(e*cos(c + d*x))^(9/2),x)

[Out]

int((a + a*sin(c + d*x))^2/(e*cos(c + d*x))^(9/2), x)

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