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

Optimal. Leaf size=109 \[ \frac{2 \left (3 a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{e \cos (c+d x)}}-\frac{10 a b \sqrt{e \cos (c+d x)}}{3 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e} \]

[Out]

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

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Rubi [A]  time = 0.128876, antiderivative size = 109, 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} \[ \frac{2 \left (3 a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{e \cos (c+d x)}}-\frac{10 a b \sqrt{e \cos (c+d x)}}{3 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.392046, size = 75, normalized size = 0.69 \[ \frac{2 \left (3 a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-2 b \cos (c+d x) (6 a+b \sin (c+d x))}{3 d \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.777, size = 210, normalized size = 1.9 \begin{align*} -{\frac{2}{3\,d} \left ( -4\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+3\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{2}+2\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){b}^{2}-12\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+2\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+6\,\sin \left ( 1/2\,dx+c/2 \right ) ab \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x)

[Out]

-2/3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(-4*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*
(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+2*(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^2-12*a*b*sin
(1/2*d*x+1/2*c)^3+2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+6*sin(1/2*d*x+1/2*c)*a*b)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(-(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)*sqrt(e*cos(d*x + c))/(e*cos(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*sin(d*x+c))**2/(e*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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