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

Optimal. Leaf size=97 \[ \frac {2 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 a}{3 d e (e \cos (c+d x))^{3/2}}+\frac {2 a \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}} \]

[Out]

2/3*a/d/e/(e*cos(d*x+c))^(3/2)+2/3*a*sin(d*x+c)/d/e/(e*cos(d*x+c))^(3/2)+2/3*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co
s(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/e^2/(e*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 97, 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} \[ \frac {2 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 a}{3 d e (e \cos (c+d x))^{3/2}}+\frac {2 a \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.46, size = 86, normalized size = 0.89 \[ \frac {2 a \left (\cos (c+d x)-(\sin (c+d x)-1) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{3 d e^2 \sqrt {e \cos (c+d x)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(Cos[c + d*x] - Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(-1 + Sin[c + d*x])))/(3*d*e^2*Sqrt[e*Cos[c
+ d*x]]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}}{e^{3} \cos \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)/(e^3*cos(d*x + c)^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.88, size = 189, normalized size = 1.95 \[ -\frac {2 \left (2 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}+2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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