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

Optimal. Leaf size=102 \[ \frac{2 a \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 b}{3 d e (e \sin (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0707494, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2669, 2636, 2642, 2641} \[ \frac{2 a \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 b}{3 d e (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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 \frac{a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx &=-\frac{2 b}{3 d e (e \sin (c+d x))^{3/2}}+a \int \frac{1}{(e \sin (c+d x))^{5/2}} \, dx\\ &=-\frac{2 b}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac{a \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{3 e^2}\\ &=-\frac{2 b}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac{\left (a \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}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac{2 a 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)}}\\ \end{align*}

Mathematica [A]  time = 0.151196, size = 59, normalized size = 0.58 \[ -\frac{2 \left (a \cos (c+d x)+a \sin ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )+b\right )}{3 d e (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b + a*Cos[c + d*x] + a*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 = 1.694, size = 124, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -{\frac{2\,b}{3\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{a}{3\,{e}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) } \left ( \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 ) -2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+2\,\sin \left ( dx+c \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))/(e*sin(d*x+c))^(5/2),x)

[Out]

(-2/3*b/e/(e*sin(d*x+c))^(3/2)-1/3*a/e^2*((1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(5/2)*Ellipti
cF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^3+2*sin(d*x+c))/sin(d*x+c)^2/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{b \cos \left (d x + c\right ) + a}{\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))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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