3.1.51 \(\int (a+b \cos (c+d x))^3 (e \sin (c+d x))^{3/2} \, dx\) [51]

Optimal. Leaf size=202 \[ \frac {2 a \left (7 a^2+6 b^2\right ) e^2 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 d \sqrt {e \sin (c+d x)}}-\frac {2 a \left (7 a^2+6 b^2\right ) e \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}+\frac {2 b \left (89 a^2+28 b^2\right ) (e \sin (c+d x))^{5/2}}{315 d e}+\frac {26 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2}}{9 d e} \]

[Out]

2/315*b*(89*a^2+28*b^2)*(e*sin(d*x+c))^(5/2)/d/e+26/63*a*b*(a+b*cos(d*x+c))*(e*sin(d*x+c))^(5/2)/d/e+2/9*b*(a+
b*cos(d*x+c))^2*(e*sin(d*x+c))^(5/2)/d/e-2/21*a*(7*a^2+6*b^2)*e^2*(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)-2/21*a*
(7*a^2+6*b^2)*e*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2771, 2941, 2748, 2715, 2721, 2720} \begin {gather*} \frac {2 a e^2 \left (7 a^2+6 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 )}{21 d \sqrt {e \sin (c+d x)}}+\frac {2 b \left (89 a^2+28 b^2\right ) (e \sin (c+d x))^{5/2}}{315 d e}-\frac {2 a e \left (7 a^2+6 b^2\right ) \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))^2}{9 d e}+\frac {26 a b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{63 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(e*Sin[c + d*x])^(3/2),x]

[Out]

(2*a*(7*a^2 + 6*b^2)*e^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*d*Sqrt[e*Sin[c + d*x]]) - (2
*a*(7*a^2 + 6*b^2)*e*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(21*d) + (2*b*(89*a^2 + 28*b^2)*(e*Sin[c + d*x])^(5/2)
)/(315*d*e) + (26*a*b*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(5/2))/(63*d*e) + (2*b*(a + b*Cos[c + d*x])^2*(e*S
in[c + d*x])^(5/2))/(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])

Rule 2771

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])

Rule 2941

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

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^3 (e \sin (c+d x))^{3/2} \, dx &=\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2}}{9 d e}+\frac {2}{9} \int (a+b \cos (c+d x)) \left (\frac {9 a^2}{2}+2 b^2+\frac {13}{2} a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2} \, dx\\ &=\frac {26 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2}}{9 d e}+\frac {4}{63} \int \left (\frac {9}{4} a \left (7 a^2+6 b^2\right )+\frac {1}{4} b \left (89 a^2+28 b^2\right ) \cos (c+d x)\right ) (e \sin (c+d x))^{3/2} \, dx\\ &=\frac {2 b \left (89 a^2+28 b^2\right ) (e \sin (c+d x))^{5/2}}{315 d e}+\frac {26 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2}}{9 d e}+\frac {1}{7} \left (a \left (7 a^2+6 b^2\right )\right ) \int (e \sin (c+d x))^{3/2} \, dx\\ &=-\frac {2 a \left (7 a^2+6 b^2\right ) e \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}+\frac {2 b \left (89 a^2+28 b^2\right ) (e \sin (c+d x))^{5/2}}{315 d e}+\frac {26 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2}}{9 d e}+\frac {1}{21} \left (a \left (7 a^2+6 b^2\right ) e^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx\\ &=-\frac {2 a \left (7 a^2+6 b^2\right ) e \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}+\frac {2 b \left (89 a^2+28 b^2\right ) (e \sin (c+d x))^{5/2}}{315 d e}+\frac {26 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2}}{9 d e}+\frac {\left (a \left (7 a^2+6 b^2\right ) e^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{21 \sqrt {e \sin (c+d x)}}\\ &=\frac {2 a \left (7 a^2+6 b^2\right ) e^2 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 {2 a \left (7 a^2+6 b^2\right ) e \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}+\frac {2 b \left (89 a^2+28 b^2\right ) (e \sin (c+d x))^{5/2}}{315 d e}+\frac {26 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2}}{9 d e}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.30, size = 147, normalized size = 0.73 \begin {gather*} \frac {\left (-20 a \left (28 a^2+15 b^2\right ) \cot (c+d x)-\frac {2}{3} b \left (-756 a^2-147 b^2+28 \left (27 a^2+4 b^2\right ) \cos (2 (c+d x))+270 a b \cos (3 (c+d x))+35 b^2 \cos (4 (c+d x))\right ) \csc (c+d x)-\frac {80 a \left (7 a^2+6 b^2\right ) F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )}{\sin ^{\frac {3}{2}}(c+d x)}\right ) (e \sin (c+d x))^{3/2}}{840 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(e*Sin[c + d*x])^(3/2),x]

[Out]

((-20*a*(28*a^2 + 15*b^2)*Cot[c + d*x] - (2*b*(-756*a^2 - 147*b^2 + 28*(27*a^2 + 4*b^2)*Cos[2*(c + d*x)] + 270
*a*b*Cos[3*(c + d*x)] + 35*b^2*Cos[4*(c + d*x)])*Csc[c + d*x])/3 - (80*a*(7*a^2 + 6*b^2)*EllipticF[(-2*c + Pi
- 2*d*x)/4, 2])/Sin[c + d*x]^(3/2))*(e*Sin[c + d*x])^(3/2))/(840*d)

________________________________________________________________________________________

Maple [A]
time = 0.17, size = 291, normalized size = 1.44

method	result	size
default	\(-\frac {e^{2} \left (70 b^{3} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+270 a \,b^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+105 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{3}+90 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a \,b^{2}+378 a^{2} b \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-14 b^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+210 a^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-90 a \,b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-378 a^{2} b \cos \left (d x +c \right ) \sin \left (d x +c \right )-56 b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{315 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(291\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(e*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/315/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^2*(70*b^3*cos(d*x+c)^5*sin(d*x+c)+270*a*b^2*cos(d*x+c)^4*sin(d*x+c)+1
05*(-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^3+90*(-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*b^2+378*a^2*b*cos(d*x+c)^3*sin(d*x+c)-14*b^3*cos(d*x+c)^3*sin(d*x+c)+210*a^3*cos(d*x+c)^2*sin(d*x+c)-90*
a*b^2*cos(d*x+c)^2*sin(d*x+c)-378*a^2*b*cos(d*x+c)*sin(d*x+c)-56*b^3*cos(d*x+c)*sin(d*x+c))/d

________________________________________________________________________________________

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))^3*(e*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

e^(3/2)*integrate((b*cos(d*x + c) + a)^3*sin(d*x + c)^(3/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 190, normalized size = 0.94 \begin {gather*} \frac {15 \, \sqrt {2} \sqrt {-i} {\left (7 \, a^{3} + 6 \, a b^{2}\right )} e^{\frac {3}{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} {\left (7 \, a^{3} + 6 \, a b^{2}\right )} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (35 \, b^{3} \cos \left (d x + c\right )^{4} e^{\frac {3}{2}} + 135 \, a b^{2} \cos \left (d x + c\right )^{3} e^{\frac {3}{2}} + 7 \, {\left (27 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} e^{\frac {3}{2}} + 15 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) e^{\frac {3}{2}} - 7 \, {\left (27 \, a^{2} b + 4 \, b^{3}\right )} e^{\frac {3}{2}}\right )} \sqrt {\sin \left (d x + c\right )}}{315 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(e*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/315*(15*sqrt(2)*sqrt(-I)*(7*a^3 + 6*a*b^2)*e^(3/2)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))
+ 15*sqrt(2)*sqrt(I)*(7*a^3 + 6*a*b^2)*e^(3/2)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(3
5*b^3*cos(d*x + c)^4*e^(3/2) + 135*a*b^2*cos(d*x + c)^3*e^(3/2) + 7*(27*a^2*b - b^3)*cos(d*x + c)^2*e^(3/2) +
15*(7*a^3 - 3*a*b^2)*cos(d*x + c)*e^(3/2) - 7*(27*a^2*b + 4*b^3)*e^(3/2))*sqrt(sin(d*x + c)))/d

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (a + b \cos {\left (c + d x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3*(e*sin(d*x+c))**(3/2),x)

[Out]

Integral((e*sin(c + d*x))**(3/2)*(a + b*cos(c + d*x))**3, x)

________________________________________________________________________________________

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))^3*(e*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3*e^(3/2)*sin(d*x + c)^(3/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^(3/2)*(a + b*cos(c + d*x))^3,x)

[Out]

int((e*sin(c + d*x))^(3/2)*(a + b*cos(c + d*x))^3, x)

________________________________________________________________________________________