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]

-2/3*b/d/e/(e*sin(d*x+c))^(3/2)-2/3*a*cos(d*x+c)/d/e/(e*sin(d*x+c))^(3/2)-2/3*a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^
(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/e^2/(e*sin(d*x
+c))^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 102, normalized size of antiderivative = 1.00, 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 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 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]

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 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])

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*}

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Mathematica [A]  time = 0.15, 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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fricas [F]  time = 1.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b \cos \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}}{{\left (e^{3} \cos \left (d x + c\right )^{2} - e^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \cos \left (d x + c\right ) + a}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.24, size = 124, normalized size = 1.22 \[ \frac {-\frac {2 b}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

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*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*Ellipt
icF((-sin(d*x+c)+1)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \cos \left (d x + c\right ) + a}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\cos \left (c+d\,x\right )}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \cos {\left (c + d x \right )}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

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]

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

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