3.45 \(\int \frac{(a+b \cos (c+d x))^2}{\sqrt{e \sin (c+d x)}} \, dx\)

Optimal. Leaf size=114 \[ \frac{2 \left (3 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d \sqrt{e \sin (c+d x)}}+\frac{10 a b \sqrt{e \sin (c+d x)}}{3 d e}+\frac{2 b \sqrt{e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e} \]

[Out]

(2*(3*a^2 + 2*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) + (10*a*b*S
qrt[e*Sin[c + d*x]])/(3*d*e) + (2*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(3*d*e)

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Rubi [A]  time = 0.129492, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2692, 2669, 2642, 2641} \[ \frac{2 \left (3 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d \sqrt{e \sin (c+d x)}}+\frac{10 a b \sqrt{e \sin (c+d x)}}{3 d e}+\frac{2 b \sqrt{e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*a^2 + 2*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) + (10*a*b*S
qrt[e*Sin[c + d*x]])/(3*d*e) + (2*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(3*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 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 \frac{(a+b \cos (c+d x))^2}{\sqrt{e \sin (c+d x)}} \, dx &=\frac{2 b (a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{3 d e}+\frac{2}{3} \int \frac{\frac{3 a^2}{2}+b^2+\frac{5}{2} a b \cos (c+d x)}{\sqrt{e \sin (c+d x)}} \, dx\\ &=\frac{10 a b \sqrt{e \sin (c+d x)}}{3 d e}+\frac{2 b (a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{3 d e}+\frac{1}{3} \left (3 a^2+2 b^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx\\ &=\frac{10 a b \sqrt{e \sin (c+d x)}}{3 d e}+\frac{2 b (a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{3 d e}+\frac{\left (\left (3 a^2+2 b^2\right ) \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{3 \sqrt{e \sin (c+d x)}}\\ &=\frac{2 \left (3 a^2+2 b^2\right ) F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d \sqrt{e \sin (c+d x)}}+\frac{10 a b \sqrt{e \sin (c+d x)}}{3 d e}+\frac{2 b (a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{3 d e}\\ \end{align*}

Mathematica [A]  time = 0.379236, size = 79, normalized size = 0.69 \[ \frac{2 b \sin (c+d x) (6 a+b \cos (c+d x))-2 \left (3 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )}{3 d \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*a^2 + 2*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sqrt[Sin[c + d*x]] + 2*b*(6*a + b*Cos[c + d*x])*Sin[c
+ d*x])/(3*d*Sqrt[e*Sin[c + d*x]])

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Maple [A]  time = 1.77, size = 170, normalized size = 1.5 \begin{align*} -{\frac{1}{3\,d\cos \left ( dx+c \right ) } \left ( 3\,\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}+2\,\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}-2\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-12\,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))^(1/2),x)

[Out]

-1/3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(3*(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+2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*Elliptic
F((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-2*b^2*sin(d*x+c)*cos(d*x+c)^2-12*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 \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt{e \sin \left (d x + c\right )}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt{e \sin \left (d x + c\right )}}{e \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))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \cos{\left (c + d x \right )}\right )^{2}}{\sqrt{e \sin{\left (c + d x \right )}}}\, dx \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))**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt{e \sin \left (d x + c\right )}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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