3.1.33 \(\int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx\) [33]

Optimal. Leaf size=129 \[ \frac {10 a e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 d \sqrt {e \sin (c+d x)}}-\frac {10 a e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e} \]

[Out]

-2/7*a*e*cos(d*x+c)*(e*sin(d*x+c))^(5/2)/d+2/9*b*(e*sin(d*x+c))^(9/2)/d/e-10/21*a*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)-10/21*a*e^3*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.06, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2748, 2715, 2721, 2720} \begin {gather*} \frac {10 a e^4 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{21 d \sqrt {e \sin (c+d x)}}-\frac {10 a e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*a*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*d*Sqrt[e*Sin[c + d*x]]) - (10*a*e^3*Cos[c +
d*x]*Sqrt[e*Sin[c + d*x]])/(21*d) - (2*a*e*Cos[c + d*x]*(e*Sin[c + d*x])^(5/2))/(7*d) + (2*b*(e*Sin[c + d*x])
^(9/2))/(9*d*e)

Rule 2715

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] && Integ
erQ[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 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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])

Rubi steps

\begin {align*} \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx &=\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}+a \int (e \sin (c+d x))^{7/2} \, dx\\ &=-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}+\frac {1}{7} \left (5 a e^2\right ) \int (e \sin (c+d x))^{3/2} \, dx\\ &=-\frac {10 a e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}+\frac {1}{21} \left (5 a e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx\\ &=-\frac {10 a e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}+\frac {\left (5 a e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{21 \sqrt {e \sin (c+d x)}}\\ &=\frac {10 a e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 d \sqrt {e \sin (c+d x)}}-\frac {10 a e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\\ \end {align*}

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Mathematica [A]
time = 0.91, size = 108, normalized size = 0.84 \begin {gather*} \frac {e^3 \left (-120 a F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+(21 b-138 a \cos (c+d x)-28 b \cos (2 (c+d x))+18 a \cos (3 (c+d x))+7 b \cos (4 (c+d x))) \sqrt {\sin (c+d x)}\right ) \sqrt {e \sin (c+d x)}}{252 d \sqrt {\sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*(-120*a*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + (21*b - 138*a*Cos[c + d*x] - 28*b*Cos[2*(c + d*x)] + 18*a*C
os[3*(c + d*x)] + 7*b*Cos[4*(c + d*x)])*Sqrt[Sin[c + d*x]])*Sqrt[e*Sin[c + d*x]])/(252*d*Sqrt[Sin[c + d*x]])

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Maple [A]
time = 0.14, size = 127, normalized size = 0.98

method	result	size
default	\(\frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9 e}-\frac {e^{4} a \left (-6 \left (\sin ^{5}\left (d x +c \right )\right )+5 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-4 \left (\sin ^{3}\left (d x +c \right )\right )+10 \sin \left (d x +c \right )\right )}{21 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2/9/e*b*(e*sin(d*x+c))^(9/2)-1/21*e^4*a*(-6*sin(d*x+c)^5+5*(-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))-4*sin(d*x+c)^3+10*sin(d*x+c))/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 130, normalized size = 1.01 \begin {gather*} \frac {15 \, \sqrt {2} \sqrt {-i} a e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, \sqrt {2} \sqrt {i} a e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (7 \, b \cos \left (d x + c\right )^{4} e^{\frac {7}{2}} + 9 \, a \cos \left (d x + c\right )^{3} e^{\frac {7}{2}} - 14 \, b \cos \left (d x + c\right )^{2} e^{\frac {7}{2}} - 24 \, a \cos \left (d x + c\right ) e^{\frac {7}{2}} + 7 \, b e^{\frac {7}{2}}\right )} \sqrt {\sin \left (d x + c\right )}}{63 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/63*(15*sqrt(2)*sqrt(-I)*a*e^(7/2)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*sqrt(2)*sqrt
(I)*a*e^(7/2)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(7*b*cos(d*x + c)^4*e^(7/2) + 9*a*c
os(d*x + c)^3*e^(7/2) - 14*b*cos(d*x + c)^2*e^(7/2) - 24*a*cos(d*x + c)*e^(7/2) + 7*b*e^(7/2))*sqrt(sin(d*x +
c)))/d

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4846 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+b*cos(d*x+c))*(e*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

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

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

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)),x)

[Out]

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

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