3.3.2 \(\int \frac {a+a \sin (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx\) [202]

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

[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 = 0.05, 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 = {2748, 2716, 2721, 2720} \begin {gather*} \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}} \end {gather*}

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 2716

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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 = 0.29, size = 86, normalized size = 0.89 \begin {gather*} \frac {2 a \left (\cos (c+d x)-\sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (-1+\sin (c+d x))\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} \end {gather*}

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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Maple [A]
time = 2.64, size = 189, normalized size = 1.95

method	result	size
default	\(-\frac {2 \left (2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\)	\(189\)

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,method=_RETURNVERBOSE)

[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*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 108, normalized size = 1.11 \begin {gather*} \frac {{\left (-i \, \sqrt {2} a \sin \left (d x + c\right ) + i \, \sqrt {2} a\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (i \, \sqrt {2} a \sin \left (d x + c\right ) - i \, \sqrt {2} a\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, a \sqrt {\cos \left (d x + c\right )}}{3 \, {\left (d e^{\frac {5}{2}} \sin \left (d x + c\right ) - d e^{\frac {5}{2}}\right )}} \end {gather*}

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]

1/3*((-I*sqrt(2)*a*sin(d*x + c) + I*sqrt(2)*a)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (I*
sqrt(2)*a*sin(d*x + c) - I*sqrt(2)*a)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*a*sqrt(cos
(d*x + c)))/(d*e^(5/2)*sin(d*x + c) - d*e^(5/2))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^(-5/2)/cos(d*x + c)^(5/2), x)

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

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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