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

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

[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)

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Rubi [A]  time = 0.297824, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2692, 2862, 2669, 2635, 2642, 2641} \[ \frac{2 a e^2 \left (7 a^2+6 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{21 d \sqrt{e \sin (c+d x)}}+\frac{2 b \left (89 a^2+28 b^2\right ) (e \sin (c+d x))^{5/2}}{315 d e}-\frac{2 a e \left (7 a^2+6 b^2\right ) \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 d}+\frac{2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))^2}{9 d e}+\frac{26 a b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{63 d e} \]

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 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
])

Rule 2862

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] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*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 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 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 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]

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 = 1.30724, size = 147, normalized size = 0.73 \[ \frac{(e \sin (c+d x))^{3/2} \left (-20 a \left (28 a^2+15 b^2\right ) \cot (c+d x)-\frac{2}{3} b \csc (c+d x) \left (28 \left (27 a^2+4 b^2\right ) \cos (2 (c+d x))-756 a^2+270 a b \cos (3 (c+d x))+35 b^2 \cos (4 (c+d x))-147 b^2\right )-\frac{80 a \left (7 a^2+6 b^2\right ) F\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )}{\sin ^{\frac{3}{2}}(c+d x)}\right )}{840 d} \]

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)

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Maple [A]  time = 2.476, size = 226, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{2\,b \left ( 5\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+27\,{a}^{2}+4\,{b}^{2} \right ) }{45\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{2}a}{21\,\cos \left ( dx+c \right ) } \left ( 18\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+ \left ( 14\,{a}^{2}-6\,{b}^{2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +7\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){a}^{2}+6\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){b}^{2} \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \right ) } \end{align*}

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)

[Out]

(2/45/e*b*(e*sin(d*x+c))^(5/2)*(5*b^2*cos(d*x+c)^2+27*a^2+4*b^2)-1/21*e^2*a*(18*b^2*sin(d*x+c)*cos(d*x+c)^4+(1
4*a^2-6*b^2)*cos(d*x+c)^2*sin(d*x+c)+7*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF(
(1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+6*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF
((1-sin(d*x+c))^(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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

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]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} e \cos \left (d x + c\right )^{3} + 3 \, a b^{2} e \cos \left (d x + c\right )^{2} + 3 \, a^{2} b e \cos \left (d x + c\right ) + a^{3} e\right )} \sqrt{e \sin \left (d x + c\right )} \sin \left (d x + c\right ), x\right ) \end{align*}

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]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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

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*sin(d*x + c))^(3/2), x)