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

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

[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)

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Rubi [A]  time = 0.168032, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2692, 2669, 2635, 2642, 2641} \[ \frac{2 e^2 \left (7 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{21 d \sqrt{e \sin (c+d x)}}-\frac{2 e \left (7 a^2+2 b^2\right ) \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 d}+\frac{18 a b (e \sin (c+d x))^{5/2}}{35 d e}+\frac{2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e} \]

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 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 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))^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*}

Mathematica [A]  time = 0.811974, size = 117, normalized size = 0.76 \[ \frac{(e \sin (c+d x))^{3/2} \left (-\frac{1}{2} \csc (c+d x) \left (5 \left (28 a^2+5 b^2\right ) \cos (c+d x)+3 b (28 a \cos (2 (c+d x))-28 a+5 b \cos (3 (c+d x)))\right )-\frac{10 \left (7 a^2+2 b^2\right ) F\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )}{\sin ^{\frac{3}{2}}(c+d x)}\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-((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
])/2 - (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))/(1
05*d)

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Maple [A]  time = 1.75, size = 229, normalized size = 1.5 \begin{align*} -{\frac{{e}^{2}}{105\,d\cos \left ( dx+c \right ) } \left ( 30\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+35\,\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}+10\,\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}+84\,ab\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+70\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-10\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-84\,ab\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \end{align*}

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)

[Out]

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

[Out]

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

[Out]

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

[Out]

Timed out

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

[Out]

Timed out