\(\int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx\) [210]

Optimal result

Integrand size = 25, antiderivative size = 89 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {4 a^4 \sqrt {e \cos (c+d x)}}{3 d e^3 \left (a^2-a^2 \sin (c+d x)\right )} \]

[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] (verified)

Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2749, 2759, 2721, 2720} \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx=\frac {4 a^4 \sqrt {e \cos (c+d x)}}{3 d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}} \]

[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 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 2749

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 2759

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*} \text {integral}& = \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)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.74 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx=\frac {4 \sqrt [4]{2} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4},\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{3/4}}{3 d e (e \cos (c+d x))^{3/2}} \]

[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))

Maple [A] (verified)

Time = 4.24 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.16

method	result	size
default	\(-\frac {2 \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\)	\(192\)
parts	\(-\frac {2 a^{2} \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\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}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )+\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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 e^{2} \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}-\frac {4 a^{2} \left (2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\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}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 e^{2} \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}+\frac {4 a^{2}}{3 \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} e d}\)	\(510\)

[In]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.46 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {4 \, \sqrt {e \cos \left (d x + c\right )} a^{2} - {\left (i \, \sqrt {2} a^{2} \sin \left (d x + c\right ) - i \, \sqrt {2} a^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (-i \, \sqrt {2} a^{2} \sin \left (d x + c\right ) + i \, \sqrt {2} a^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{3 \, {\left (d e^{3} \sin \left (d x + c\right ) - d e^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[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)