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

Optimal. Leaf size=76 \[ -\frac {4 (e \cos (c+d x))^{3/2}}{21 a d e (a \sin (c+d x)+a)^{3/2}}-\frac {2 (e \cos (c+d x))^{3/2}}{7 d e (a \sin (c+d x)+a)^{5/2}} \]

[Out]

-2/7*(e*cos(d*x+c))^(3/2)/d/e/(a+a*sin(d*x+c))^(5/2)-4/21*(e*cos(d*x+c))^(3/2)/a/d/e/(a+a*sin(d*x+c))^(3/2)

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Rubi [A]  time = 0.13, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ -\frac {4 (e \cos (c+d x))^{3/2}}{21 a d e (a \sin (c+d x)+a)^{3/2}}-\frac {2 (e \cos (c+d x))^{3/2}}{7 d e (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(e*Cos[c + d*x])^(3/2))/(7*d*e*(a + a*Sin[c + d*x])^(5/2)) - (4*(e*Cos[c + d*x])^(3/2))/(21*a*d*e*(a + a*S
in[c + d*x])^(3/2))

Rule 2671

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac {2 (e \cos (c+d x))^{3/2}}{7 d e (a+a \sin (c+d x))^{5/2}}+\frac {2 \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx}{7 a}\\ &=-\frac {2 (e \cos (c+d x))^{3/2}}{7 d e (a+a \sin (c+d x))^{5/2}}-\frac {4 (e \cos (c+d x))^{3/2}}{21 a d e (a+a \sin (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 59, normalized size = 0.78 \[ -\frac {2 (2 \sin (c+d x)+5) \sqrt {a (\sin (c+d x)+1)} (e \cos (c+d x))^{3/2}}{21 a^3 d e (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(e*Cos[c + d*x])^(3/2)*Sqrt[a*(1 + Sin[c + d*x])]*(5 + 2*Sin[c + d*x]))/(21*a^3*d*e*(1 + Sin[c + d*x])^3)

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fricas [B]  time = 0.63, size = 148, normalized size = 1.95 \[ \frac {2 \, \sqrt {e \cos \left (d x + c\right )} {\left (2 \, \cos \left (d x + c\right )^{2} + {\left (2 \, \cos \left (d x + c\right ) - 3\right )} \sin \left (d x + c\right ) + 5 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{21 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/21*sqrt(e*cos(d*x + c))*(2*cos(d*x + c)^2 + (2*cos(d*x + c) - 3)*sin(d*x + c) + 5*cos(d*x + c) + 3)*sqrt(a*s
in(d*x + c) + a)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(
d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 44, normalized size = 0.58 \[ -\frac {2 \left (2 \sin \left (d x +c \right )+5\right ) \cos \left (d x +c \right ) \sqrt {e \cos \left (d x +c \right )}}{21 d \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-2/21/d*(2*sin(d*x+c)+5)*cos(d*x+c)*(e*cos(d*x+c))^(1/2)/(a*(1+sin(d*x+c)))^(5/2)

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maxima [B]  time = 0.70, size = 207, normalized size = 2.72 \[ -\frac {2 \, {\left (5 \, \sqrt {a} \sqrt {e} + \frac {4 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} \sqrt {-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{21 \, {\left (a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/21*(5*sqrt(a)*sqrt(e) + 4*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(a)*sqrt(e)*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 - 5*sqrt(a)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*sqrt(-sin(d*x + c)/(cos(d*x +
c) + 1) + 1)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((a^3 + 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a
^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2))

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mupad [B]  time = 7.05, size = 145, normalized size = 1.91 \[ -\frac {8\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\sqrt {-e\,\left (2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}\,\left (-58\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+18\,{\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2+26\,\sin \left (2\,c+2\,d\,x\right )-\sin \left (4\,c+4\,d\,x\right )+20\right )}{21\,a^3\,d\,\left (240\,{\sin \left (c+d\,x\right )}^2+210\,\sin \left (c+d\,x\right )-20\,{\sin \left (2\,c+2\,d\,x\right )}^2-45\,\sin \left (3\,c+3\,d\,x\right )+\sin \left (5\,c+5\,d\,x\right )+16\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(1/2)/(a + a*sin(c + d*x))^(5/2),x)

[Out]

-(8*(a*(sin(c + d*x) + 1))^(1/2)*(-e*(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(26*sin(2*c + 2*d*x) - sin(4*c + 4*d*
x) - 58*sin(c/2 + (d*x)/2)^2 + 18*sin((3*c)/2 + (3*d*x)/2)^2 + 20))/(21*a^3*d*(210*sin(c + d*x) - 45*sin(3*c +
 3*d*x) + sin(5*c + 5*d*x) - 20*sin(2*c + 2*d*x)^2 + 240*sin(c + d*x)^2 + 16))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \cos {\left (c + d x \right )}}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(1/2)/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Integral(sqrt(e*cos(c + d*x))/(a*(sin(c + d*x) + 1))**(5/2), x)

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