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

Optimal. Leaf size=118 \[ -\frac{2 \left (a^2+2 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d e^2 \sqrt{\sin (c+d x)}}-\frac{2 a b (e \sin (c+d x))^{3/2}}{d e^3}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{d e \sqrt{e \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.138044, antiderivative size = 118, 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 = {2691, 2669, 2640, 2639} \[ -\frac{2 \left (a^2+2 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d e^2 \sqrt{\sin (c+d x)}}-\frac{2 a b (e \sin (c+d x))^{3/2}}{d e^3}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2691

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[((g*C
os[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1)*(b + a*Sin[e + f*x]))/(f*g*(p + 1)), x] + Dist[1/(g^2*(p + 1
)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin
[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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}{(e \sin (c+d x))^{3/2}} \, dx &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{d e \sqrt{e \sin (c+d x)}}-\frac{2 \int \left (\frac{a^2}{2}+b^2+\frac{3}{2} a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)} \, dx}{e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{d e \sqrt{e \sin (c+d x)}}-\frac{2 a b (e \sin (c+d x))^{3/2}}{d e^3}-\frac{\left (a^2+2 b^2\right ) \int \sqrt{e \sin (c+d x)} \, dx}{e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{d e \sqrt{e \sin (c+d x)}}-\frac{2 a b (e \sin (c+d x))^{3/2}}{d e^3}-\frac{\left (\left (a^2+2 b^2\right ) \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{e^2 \sqrt{\sin (c+d x)}}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{d e \sqrt{e \sin (c+d x)}}-\frac{2 \left (a^2+2 b^2\right ) E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d e^2 \sqrt{\sin (c+d x)}}-\frac{2 a b (e \sin (c+d x))^{3/2}}{d e^3}\\ \end{align*}

Mathematica [A]  time = 0.23882, size = 75, normalized size = 0.64 \[ \frac{-2 \left (a^2+b^2\right ) \cos (c+d x)+2 \left (a^2+2 b^2\right ) \sqrt{\sin (c+d x)} E\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )-4 a b}{d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 1.618, size = 277, normalized size = 2.4 \begin{align*}{\frac{1}{ed\cos \left ( dx+c \right ) } \left ( \left ( -2\,{a}^{2}-2\,{b}^{2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-4\,ab\cos \left ( dx+c \right ) +2\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){a}^{2}+4\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){b}^{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 ) },{\frac{\sqrt{2}}{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} \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]

((-2*a^2-2*b^2)*cos(d*x+c)^2-4*a*b*cos(d*x+c)+2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*E
llipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+4*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*
EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*E
llipticF((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)*
EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2)/e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/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}}{\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 (-\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^{2} \cos \left (d x + c\right )^{2} - e^{2}}, 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*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(e*sin(d*x + c))/(e^2*cos(d*x + c)^2 - e^2), 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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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="giac")

[Out]

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