3.47 \(\int \frac{(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=124 \[ \frac{2 \left (a^2-2 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{2 a b \sqrt{e \sin (c+d x)}}{3 d e^3}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.139355, antiderivative size = 124, 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, 2642, 2641} \[ \frac{2 \left (a^2-2 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{2 a b \sqrt{e \sin (c+d x)}}{3 d e^3}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2/(e*Sin[c + d*x])^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.26479, size = 76, normalized size = 0.61 \[ -\frac{2 \left (\left (a^2+b^2\right ) \cos (c+d x)+\left (a^2-2 b^2\right ) \sin ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )+2 a b\right )}{3 d e (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2/(e*Sin[c + d*x])^(5/2),x]

[Out]

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

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Maple [A]  time = 1.883, size = 190, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -{\frac{4\,ab}{3\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{3\,{e}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) } \left ( \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },{\frac{\sqrt{2}}{2}} \right ){a}^{2}-2\,{b}^{2}\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{5/2}{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \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))^(5/2),x)

[Out]

(-4/3*a*b/e/(e*sin(d*x+c))^(3/2)-1/3/e^2*((2*a^2+2*b^2)*sin(d*x+c)*cos(d*x+c)^2+(1-sin(d*x+c))^(1/2)*(2+2*sin(
d*x+c))^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-2*b^2*(1-sin(d*x+c))^(1/2)*(2+2
*sin(d*x+c))^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2)))/sin(d*x+c)^2/cos(d*x+c)/(e*si
n(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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^2/(e*sin(d*x + c))^(5/2), 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 )}}{{\left (e^{3} \cos \left (d x + c\right )^{2} - e^{3}\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))^(5/2),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^3*cos(d*x + c)^2 - e^3)*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))**(5/2),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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^2/(e*sin(d*x + c))^(5/2), x)