3.204 \(\int (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=168 \[ \frac {130 a^2 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d \sqrt {e \cos (c+d x)}}+\frac {130 a^2 e^3 \sin (c+d x) \sqrt {e \cos (c+d x)}}{231 d}-\frac {26 a^2 (e \cos (c+d x))^{9/2}}{99 d e}-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{9/2}}{11 d e}+\frac {26 a^2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{77 d} \]

[Out]

-26/99*a^2*(e*cos(d*x+c))^(9/2)/d/e+26/77*a^2*e*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/d-2/11*(e*cos(d*x+c))^(9/2)*(a
^2+a^2*sin(d*x+c))/d/e+130/231*a^2*e^4*(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*cos(d*x+c))^(1/2)+130/231*a^2*e^3*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.14, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2678, 2669, 2635, 2642, 2641} \[ \frac {130 a^2 e^3 \sin (c+d x) \sqrt {e \cos (c+d x)}}{231 d}+\frac {130 a^2 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d \sqrt {e \cos (c+d x)}}-\frac {26 a^2 (e \cos (c+d x))^{9/2}}{99 d e}-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{9/2}}{11 d e}+\frac {26 a^2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{77 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(7/2)*(a + a*Sin[c + d*x])^2,x]

[Out]

(-26*a^2*(e*Cos[c + d*x])^(9/2))/(99*d*e) + (130*a^2*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(231*d*
Sqrt[e*Cos[c + d*x]]) + (130*a^2*e^3*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (26*a^2*e*(e*Cos[c + d*x])^(
5/2)*Sin[c + d*x])/(77*d) - (2*(e*Cos[c + d*x])^(9/2)*(a^2 + a^2*Sin[c + d*x]))/(11*d*e)

Rule 2635

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

Rule 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2669

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

Rule 2678

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

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2 \, dx &=-\frac {2 (e \cos (c+d x))^{9/2} \left (a^2+a^2 \sin (c+d x)\right )}{11 d e}+\frac {1}{11} (13 a) \int (e \cos (c+d x))^{7/2} (a+a \sin (c+d x)) \, dx\\ &=-\frac {26 a^2 (e \cos (c+d x))^{9/2}}{99 d e}-\frac {2 (e \cos (c+d x))^{9/2} \left (a^2+a^2 \sin (c+d x)\right )}{11 d e}+\frac {1}{11} \left (13 a^2\right ) \int (e \cos (c+d x))^{7/2} \, dx\\ &=-\frac {26 a^2 (e \cos (c+d x))^{9/2}}{99 d e}+\frac {26 a^2 e (e \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}-\frac {2 (e \cos (c+d x))^{9/2} \left (a^2+a^2 \sin (c+d x)\right )}{11 d e}+\frac {1}{77} \left (65 a^2 e^2\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac {26 a^2 (e \cos (c+d x))^{9/2}}{99 d e}+\frac {130 a^2 e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac {26 a^2 e (e \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}-\frac {2 (e \cos (c+d x))^{9/2} \left (a^2+a^2 \sin (c+d x)\right )}{11 d e}+\frac {1}{231} \left (65 a^2 e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {26 a^2 (e \cos (c+d x))^{9/2}}{99 d e}+\frac {130 a^2 e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac {26 a^2 e (e \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}-\frac {2 (e \cos (c+d x))^{9/2} \left (a^2+a^2 \sin (c+d x)\right )}{11 d e}+\frac {\left (65 a^2 e^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{231 \sqrt {e \cos (c+d x)}}\\ &=-\frac {26 a^2 (e \cos (c+d x))^{9/2}}{99 d e}+\frac {130 a^2 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d \sqrt {e \cos (c+d x)}}+\frac {130 a^2 e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac {26 a^2 e (e \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}-\frac {2 (e \cos (c+d x))^{9/2} \left (a^2+a^2 \sin (c+d x)\right )}{11 d e}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 66, normalized size = 0.39 \[ -\frac {32 \sqrt [4]{2} a^2 (e \cos (c+d x))^{9/2} \, _2F_1\left (-\frac {13}{4},\frac {9}{4};\frac {13}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{9 d e (\sin (c+d x)+1)^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(7/2)*(a + a*Sin[c + d*x])^2,x]

[Out]

(-32*2^(1/4)*a^2*(e*Cos[c + d*x])^(9/2)*Hypergeometric2F1[-13/4, 9/4, 13/4, (1 - Sin[c + d*x])/2])/(9*d*e*(1 +
Sin[c + d*x])^(9/4))

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} e^{3} \cos \left (d x + c\right )^{5} - 2 \, a^{2} e^{3} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} e^{3} \cos \left (d x + c\right )^{3}\right )} \sqrt {e \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-(a^2*e^3*cos(d*x + c)^5 - 2*a^2*e^3*cos(d*x + c)^3*sin(d*x + c) - 2*a^2*e^3*cos(d*x + c)^3)*sqrt(e*c
os(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(7/2)*(a*sin(d*x + c) + a)^2, x)

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maple [A]  time = 0.75, size = 295, normalized size = 1.76 \[ -\frac {2 a^{2} e^{4} \left (-4032 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10080 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4928 \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8208 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12320 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2232 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12320 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+924 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6160 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+195 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-498 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-1540 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+154 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{693 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(7/2)*(a+a*sin(d*x+c))^2,x)

[Out]

-2/693/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^2*e^4*(-4032*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+10080*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-4928*sin(1/2*d*x+1/2*c)^11-8208*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^8+12320*sin(1/2*d*x+1/2*c)^9+2232*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12320*sin(1/2*d*x+1/2
*c)^7+924*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6160*sin(1/2*d*x+1/2*c)^5+195*(sin(1/2*d*x+1/2*c)^2)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-498*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2
*c)-1540*sin(1/2*d*x+1/2*c)^3+154*sin(1/2*d*x+1/2*c))/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(7/2)*(a*sin(d*x + c) + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(7/2)*(a + a*sin(c + d*x))^2,x)

[Out]

int((e*cos(c + d*x))^(7/2)*(a + a*sin(c + d*x))^2, 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))**(7/2)*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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