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

Optimal. Leaf size=193 \[ \frac {10 e^4 \left (11 a^2+2 b^2\right ) \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{231 d \sqrt {e \sin (c+d x)}}-\frac {10 e^3 \left (11 a^2+2 b^2\right ) \cos (c+d x) \sqrt {e \sin (c+d x)}}{231 d}-\frac {2 e \left (11 a^2+2 b^2\right ) \cos (c+d x) (e \sin (c+d x))^{5/2}}{77 d}+\frac {26 a b (e \sin (c+d x))^{9/2}}{99 d e}+\frac {2 b (e \sin (c+d x))^{9/2} (a+b \cos (c+d x))}{11 d e} \]

[Out]

-2/77*(11*a^2+2*b^2)*e*cos(d*x+c)*(e*sin(d*x+c))^(5/2)/d+26/99*a*b*(e*sin(d*x+c))^(9/2)/d/e+2/11*b*(a+b*cos(d*
x+c))*(e*sin(d*x+c))^(9/2)/d/e-10/231*(11*a^2+2*b^2)*e^4*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+
1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/(e*sin(d*x+c))^(1/2)-10/231*(11*a^2+2
*b^2)*e^3*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.19, antiderivative size = 193, 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 = {2692, 2669, 2635, 2642, 2641} \[ -\frac {10 e^3 \left (11 a^2+2 b^2\right ) \cos (c+d x) \sqrt {e \sin (c+d x)}}{231 d}+\frac {10 e^4 \left (11 a^2+2 b^2\right ) \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{231 d \sqrt {e \sin (c+d x)}}-\frac {2 e \left (11 a^2+2 b^2\right ) \cos (c+d x) (e \sin (c+d x))^{5/2}}{77 d}+\frac {26 a b (e \sin (c+d x))^{9/2}}{99 d e}+\frac {2 b (e \sin (c+d x))^{9/2} (a+b \cos (c+d x))}{11 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(11*a^2 + 2*b^2)*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(231*d*Sqrt[e*Sin[c + d*x]]) - (
10*(11*a^2 + 2*b^2)*e^3*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(231*d) - (2*(11*a^2 + 2*b^2)*e*Cos[c + d*x]*(e*Sin
[c + d*x])^(5/2))/(77*d) + (26*a*b*(e*Sin[c + d*x])^(9/2))/(99*d*e) + (2*b*(a + b*Cos[c + d*x])*(e*Sin[c + d*x
])^(9/2))/(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 2692

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[1/(m + p), Int[(g*Cos[e + f*x])^
p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ[{
a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m
])

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2} \, dx &=\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{9/2}}{11 d e}+\frac {2}{11} \int \left (\frac {11 a^2}{2}+b^2+\frac {13}{2} a b \cos (c+d x)\right ) (e \sin (c+d x))^{7/2} \, dx\\ &=\frac {26 a b (e \sin (c+d x))^{9/2}}{99 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{9/2}}{11 d e}+\frac {1}{11} \left (11 a^2+2 b^2\right ) \int (e \sin (c+d x))^{7/2} \, dx\\ &=-\frac {2 \left (11 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{5/2}}{77 d}+\frac {26 a b (e \sin (c+d x))^{9/2}}{99 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{9/2}}{11 d e}+\frac {1}{77} \left (5 \left (11 a^2+2 b^2\right ) e^2\right ) \int (e \sin (c+d x))^{3/2} \, dx\\ &=-\frac {10 \left (11 a^2+2 b^2\right ) e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{231 d}-\frac {2 \left (11 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{5/2}}{77 d}+\frac {26 a b (e \sin (c+d x))^{9/2}}{99 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{9/2}}{11 d e}+\frac {1}{231} \left (5 \left (11 a^2+2 b^2\right ) e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx\\ &=-\frac {10 \left (11 a^2+2 b^2\right ) e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{231 d}-\frac {2 \left (11 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{5/2}}{77 d}+\frac {26 a b (e \sin (c+d x))^{9/2}}{99 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{9/2}}{11 d e}+\frac {\left (5 \left (11 a^2+2 b^2\right ) e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{231 \sqrt {e \sin (c+d x)}}\\ &=\frac {10 \left (11 a^2+2 b^2\right ) e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{231 d \sqrt {e \sin (c+d x)}}-\frac {10 \left (11 a^2+2 b^2\right ) e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{231 d}-\frac {2 \left (11 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{5/2}}{77 d}+\frac {26 a b (e \sin (c+d x))^{9/2}}{99 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{9/2}}{11 d e}\\ \end {align*}

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Mathematica [A]  time = 1.67, size = 157, normalized size = 0.81 \[ \frac {(e \sin (c+d x))^{7/2} \left (\frac {1}{6} \csc ^3(c+d x) \left (-6 \left (506 a^2+71 b^2\right ) \cos (c+d x)+396 a^2 \cos (3 (c+d x))-1232 a b \cos (2 (c+d x))+308 a b \cos (4 (c+d x))+924 a b-117 b^2 \cos (3 (c+d x))+63 b^2 \cos (5 (c+d x))\right )-\frac {40 \left (11 a^2+2 b^2\right ) F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )}{\sin ^{\frac {7}{2}}(c+d x)}\right )}{924 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((((924*a*b - 6*(506*a^2 + 71*b^2)*Cos[c + d*x] - 1232*a*b*Cos[2*(c + d*x)] + 396*a^2*Cos[3*(c + d*x)] - 117*b
^2*Cos[3*(c + d*x)] + 308*a*b*Cos[4*(c + d*x)] + 63*b^2*Cos[5*(c + d*x)])*Csc[c + d*x]^3)/6 - (40*(11*a^2 + 2*
b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, 2])/Sin[c + d*x]^(7/2))*(e*Sin[c + d*x])^(7/2))/(924*d)

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fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b^{2} e^{3} \cos \left (d x + c\right )^{4} + 2 \, a b e^{3} \cos \left (d x + c\right )^{3} - 2 \, a b e^{3} \cos \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} e^{3} \cos \left (d x + c\right )^{2} - a^{2} e^{3}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 228, normalized size = 1.18 \[ \frac {\frac {4 a b \left (e \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9 e}-\frac {e^{4} \left (-42 b^{2} \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )+\left (-66 a^{2}+72 b^{2}\right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\left (176 a^{2}-10 b^{2}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+55 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+10 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}\right )}{231 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4/9/e*a*b*(e*sin(d*x+c))^(9/2)-1/231*e^4*(-42*b^2*sin(d*x+c)*cos(d*x+c)^6+(-66*a^2+72*b^2)*cos(d*x+c)^4*sin(d
*x+c)+(176*a^2-10*b^2)*cos(d*x+c)^2*sin(d*x+c)+55*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2
)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2+10*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^
(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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