3.6.49 \(\int (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2 \, dx\) [549]

Optimal. Leaf size=149 \[ -\frac {18 a b (e \cos (c+d x))^{5/2}}{35 d e}+\frac {2 \left (7 a^2+2 b^2\right ) e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {e \cos (c+d x)}}+\frac {2 \left (7 a^2+2 b^2\right ) e \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e} \]

[Out]

-18/35*a*b*(e*cos(d*x+c))^(5/2)/d/e-2/7*b*(e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))/d/e+2/21*(7*a^2+2*b^2)*e^2*(co
s(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)+2/21*(7*a^2+2*b^2)*e*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.11, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2771, 2748, 2715, 2721, 2720} \begin {gather*} \frac {2 e^2 \left (7 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {e \cos (c+d x)}}+\frac {2 e \left (7 a^2+2 b^2\right ) \sin (c+d x) \sqrt {e \cos (c+d x)}}{21 d}-\frac {18 a b (e \cos (c+d x))^{5/2}}{35 d e}-\frac {2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-18*a*b*(e*Cos[c + d*x])^(5/2))/(35*d*e) + (2*(7*a^2 + 2*b^2)*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2
])/(21*d*Sqrt[e*Cos[c + d*x]]) + (2*(7*a^2 + 2*b^2)*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(21*d) - (2*b*(e*Cos[
c + d*x])^(5/2)*(a + b*Sin[c + d*x]))/(7*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])

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2 \, dx &=-\frac {2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}+\frac {2}{7} \int (e \cos (c+d x))^{3/2} \left (\frac {7 a^2}{2}+b^2+\frac {9}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac {18 a b (e \cos (c+d x))^{5/2}}{35 d e}-\frac {2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}+\frac {1}{7} \left (7 a^2+2 b^2\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac {18 a b (e \cos (c+d x))^{5/2}}{35 d e}+\frac {2 \left (7 a^2+2 b^2\right ) e \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}+\frac {1}{21} \left (\left (7 a^2+2 b^2\right ) e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {18 a b (e \cos (c+d x))^{5/2}}{35 d e}+\frac {2 \left (7 a^2+2 b^2\right ) e \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}+\frac {\left (\left (7 a^2+2 b^2\right ) e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {e \cos (c+d x)}}\\ &=-\frac {18 a b (e \cos (c+d x))^{5/2}}{35 d e}+\frac {2 \left (7 a^2+2 b^2\right ) e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {e \cos (c+d x)}}+\frac {2 \left (7 a^2+2 b^2\right ) e \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}\\ \end {align*}

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Mathematica [A]
time = 1.16, size = 115, normalized size = 0.77 \begin {gather*} \frac {(e \cos (c+d x))^{3/2} \left (20 \left (7 a^2+2 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} \left (-84 a b \cos (2 (c+d x))+5 \left (28 a^2+5 b^2\right ) \sin (c+d x)-3 b (28 a+5 b \sin (3 (c+d x)))\right )\right )}{210 d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(3/2)*(20*(7*a^2 + 2*b^2)*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(-84*a*b*Cos[2*(c +
d*x)] + 5*(28*a^2 + 5*b^2)*Sin[c + d*x] - 3*b*(28*a + 5*b*Sin[3*(c + d*x)]))))/(210*d*Cos[c + d*x]^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(342\) vs. \(2(157)=314\).
time = 5.21, size = 343, normalized size = 2.30

method	result	size
default	\(-\frac {2 e^{2} \left (-240 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-336 a b \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+360 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+140 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+504 a b \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-140 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}-70 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-252 a b \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+42 a b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(343\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^2*(-240*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^8-336*a*b*sin(1/2*d*x+1/2*c)^7+360*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+140*a^2*cos(1/2*d*x+1/2*c)*
sin(1/2*d*x+1/2*c)^4+504*a*b*sin(1/2*d*x+1/2*c)^5-140*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+35*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+10*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-70*a^2*cos(1/2*d*x
+1/2*c)*sin(1/2*d*x+1/2*c)^2-252*a*b*sin(1/2*d*x+1/2*c)^3+10*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+42*a*
b*sin(1/2*d*x+1/2*c))/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((e*cos(d*x+c))^(3/2)*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 142, normalized size = 0.95 \begin {gather*} \frac {-5 i \, \sqrt {2} {\left (7 \, a^{2} + 2 \, 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 ) + 5 i \, \sqrt {2} {\left (7 \, a^{2} + 2 \, 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 (42 \, a b \cos \left (d x + c\right )^{2} e^{\frac {3}{2}} + 5 \, {\left (3 \, b^{2} \cos \left (d x + c\right )^{2} e^{\frac {3}{2}} - {\left (7 \, a^{2} + 2 \, b^{2}\right )} e^{\frac {3}{2}}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(-5*I*sqrt(2)*(7*a^2 + 2*b^2)*e^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sq
rt(2)*(7*a^2 + 2*b^2)*e^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(42*a*b*cos(d*x +
c)^2*e^(3/2) + 5*(3*b^2*cos(d*x + c)^2*e^(3/2) - (7*a^2 + 2*b^2)*e^(3/2))*sin(d*x + c))*sqrt(cos(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((e*cos(d*x+c))**(3/2)*(a+b*sin(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3064 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((e*cos(d*x+c))^(3/2)*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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