3.1.43 \(\int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx\) [43]

Optimal. Leaf size=154 \[ \frac {2 \left (7 a^2+2 b^2\right ) e^2 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 d \sqrt {e \sin (c+d x)}}-\frac {2 \left (7 a^2+2 b^2\right ) e \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}+\frac {18 a b (e \sin (c+d x))^{5/2}}{35 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{7 d e} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 154, 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 {\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 e \left (7 a^2+2 b^2\right ) \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}+\frac {18 a b (e \sin (c+d x))^{5/2}}{35 d e}+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.87, size = 117, normalized size = 0.76 \begin {gather*} \frac {\left (-\frac {1}{2} \left (5 \left (28 a^2+5 b^2\right ) \cos (c+d x)+3 b (-28 a+28 a \cos (2 (c+d x))+5 b \cos (3 (c+d x)))\right ) \csc (c+d x)-\frac {10 \left (7 a^2+2 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}}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*((5*(28*a^2 + 5*b^2)*Cos[c + d*x] + 3*b*(-28*a + 28*a*Cos[2*(c + d*x)] + 5*b*Cos[3*(c + d*x)]))*Csc[c +
d*x]) - (10*(7*a^2 + 2*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, 2])/Sin[c + d*x]^(3/2))*(e*Sin[c + d*x])^(3/2))/
(105*d)

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Maple [A]
time = 0.13, size = 229, normalized size = 1.49

method	result	size
default	\(-\frac {e^{2} \left (30 b^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+35 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+10 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}+84 a b \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+70 a^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-84 a b \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{105 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^2*(30*b^2*cos(d*x+c)^4*sin(d*x+c)+35*(-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+84*a*b*cos(d*x+c)^3*sin(d*x
+c)+70*a^2*cos(d*x+c)^2*sin(d*x+c)-10*b^2*cos(d*x+c)^2*sin(d*x+c)-84*a*b*cos(d*x+c)*sin(d*x+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((a+b*cos(d*x+c))^2*(e*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

e^(3/2)*integrate((b*cos(d*x + c) + a)^2*sin(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 = 151, normalized size = 0.98 \begin {gather*} \frac {5 \, \sqrt {2} \sqrt {-i} {\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 \, \sqrt {2} \sqrt {i} {\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 (15 \, b^{2} \cos \left (d x + c\right )^{3} e^{\frac {3}{2}} + 42 \, a b \cos \left (d x + c\right )^{2} e^{\frac {3}{2}} - 42 \, a b e^{\frac {3}{2}} + 5 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right ) e^{\frac {3}{2}}\right )} \sqrt {\sin \left (d x + c\right )}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(5*sqrt(2)*sqrt(-I)*(7*a^2 + 2*b^2)*e^(3/2)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5
*sqrt(2)*sqrt(I)*(7*a^2 + 2*b^2)*e^(3/2)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(15*b^2*
cos(d*x + c)^3*e^(3/2) + 42*a*b*cos(d*x + c)^2*e^(3/2) - 42*a*b*e^(3/2) + 5*(7*a^2 - b^2)*cos(d*x + c)*e^(3/2)
)*sqrt(sin(d*x + c)))/d

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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 )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((b*cos(d*x + c) + a)^2*e^(3/2)*sin(d*x + c)^(3/2), 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 )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,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))^2,x)

[Out]

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

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