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

Optimal. Leaf size=89 \[ \frac {4 a^4 \sqrt {e \cos (c+d x)}}{3 d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {2 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2670, 2680, 2642, 2641} \[ \frac {4 a^4 \sqrt {e \cos (c+d x)}}{3 d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {2 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*e^2*Sqrt[e*Cos[c + d*x]]) + (4*a^4*Sqrt[e*Cos[c + d
*x]])/(3*d*e^3*(a^2 - a^2*Sin[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]

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 2670

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

Rule 2680

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx &=\frac {a^4 \int \frac {(e \cos (c+d x))^{3/2}}{(a-a \sin (c+d x))^2} \, dx}{e^4}\\ &=\frac {4 a^4 \sqrt {e \cos (c+d x)}}{3 d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {a^2 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{3 e^2}\\ &=\frac {4 a^4 \sqrt {e \cos (c+d x)}}{3 d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (a^2 \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^2 \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 {4 a^4 \sqrt {e \cos (c+d x)}}{3 d e^3 \left (a^2-a^2 \sin (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.07, size = 66, normalized size = 0.74 \[ \frac {4 \sqrt [4]{2} a^2 (\sin (c+d x)+1)^{3/4} \, _2F_1\left (-\frac {3}{4},-\frac {1}{4};\frac {1}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*2^(1/4)*a^2*Hypergeometric2F1[-3/4, -1/4, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4))/(3*d*e*(e*Co
s[c + d*x])^(3/2))

________________________________________________________________________________________

fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \sqrt {e \cos \left (d x + c\right )}}{e^{3} \cos \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 1.04, size = 193, normalized size = 2.17 \[ \frac {2 \left (2 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^2/(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)-4*sin(1/2*d*x+1/
2*c)^2*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c))*a^2/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________