\(\int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx\) [270]

Optimal result

Integrand size = 25, antiderivative size = 191 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 a^4 d \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{13 d e (a+a \sin (c+d x))^4}-\frac {10 (e \cos (c+d x))^{3/2}}{117 a d e (a+a \sin (c+d x))^3}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

-2/13*(e*cos(d*x+c))^(3/2)/d/e/(a+a*sin(d*x+c))^4-10/117*(e*cos(d*x+c))^(3/2)/a/d/e/(a+a*sin(d*x+c))^3-2/39*(e
*cos(d*x+c))^(3/2)/d/e/(a^2+a^2*sin(d*x+c))^2-2/39*(e*cos(d*x+c))^(3/2)/d/e/(a^4+a^4*sin(d*x+c))-2/39*(cos(1/2
*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/a^4/d/cos(d
*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2760, 2762, 2721, 2719} \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^4 \sin (c+d x)+a^4\right )}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{39 a^4 d \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {10 (e \cos (c+d x))^{3/2}}{117 a d e (a \sin (c+d x)+a)^3}-\frac {2 (e \cos (c+d x))^{3/2}}{13 d e (a \sin (c+d x)+a)^4} \]

[In]

Int[Sqrt[e*Cos[c + d*x]]/(a + a*Sin[c + d*x])^4,x]

[Out]

(-2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(39*a^4*d*Sqrt[Cos[c + d*x]]) - (2*(e*Cos[c + d*x])^(3/2))
/(13*d*e*(a + a*Sin[c + d*x])^4) - (10*(e*Cos[c + d*x])^(3/2))/(117*a*d*e*(a + a*Sin[c + d*x])^3) - (2*(e*Cos[
c + d*x])^(3/2))/(39*d*e*(a^2 + a^2*Sin[c + d*x])^2) - (2*(e*Cos[c + d*x])^(3/2))/(39*d*e*(a^4 + a^4*Sin[c + d
*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1))), x] + Dist[(m + p + 1)/(a*(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] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2762

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (e \cos (c+d x))^{3/2}}{13 d e (a+a \sin (c+d x))^4}+\frac {5 \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^3} \, dx}{13 a} \\ & = -\frac {2 (e \cos (c+d x))^{3/2}}{13 d e (a+a \sin (c+d x))^4}-\frac {10 (e \cos (c+d x))^{3/2}}{117 a d e (a+a \sin (c+d x))^3}+\frac {5 \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx}{39 a^2} \\ & = -\frac {2 (e \cos (c+d x))^{3/2}}{13 d e (a+a \sin (c+d x))^4}-\frac {10 (e \cos (c+d x))^{3/2}}{117 a d e (a+a \sin (c+d x))^3}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {\int \frac {\sqrt {e \cos (c+d x)}}{a+a \sin (c+d x)} \, dx}{39 a^3} \\ & = -\frac {2 (e \cos (c+d x))^{3/2}}{13 d e (a+a \sin (c+d x))^4}-\frac {10 (e \cos (c+d x))^{3/2}}{117 a d e (a+a \sin (c+d x))^3}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^4+a^4 \sin (c+d x)\right )}-\frac {\int \sqrt {e \cos (c+d x)} \, dx}{39 a^4} \\ & = -\frac {2 (e \cos (c+d x))^{3/2}}{13 d e (a+a \sin (c+d x))^4}-\frac {10 (e \cos (c+d x))^{3/2}}{117 a d e (a+a \sin (c+d x))^3}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^4+a^4 \sin (c+d x)\right )}-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)} \, dx}{39 a^4 \sqrt {\cos (c+d x)}} \\ & = -\frac {2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 a^4 d \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{13 d e (a+a \sin (c+d x))^4}-\frac {10 (e \cos (c+d x))^{3/2}}{117 a d e (a+a \sin (c+d x))^3}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {2 (e \cos (c+d x))^{3/2}}{39 d e \left (a^4+a^4 \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 = 0.35 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx=-\frac {(e \cos (c+d x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {17}{4},\frac {7}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{12 \sqrt [4]{2} a^4 d e (1+\sin (c+d x))^{3/4}} \]

[In]

Integrate[Sqrt[e*Cos[c + d*x]]/(a + a*Sin[c + d*x])^4,x]

[Out]

-1/12*((e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[3/4, 17/4, 7/4, (1 - Sin[c + d*x])/2])/(2^(1/4)*a^4*d*e*(1 + S
in[c + d*x])^(3/4))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(693\) vs. \(2(195)=390\).

Time = 11.76 (sec) , antiderivative size = 694, normalized size of antiderivative = 3.63

method	result	size
default	\(\text {Expression too large to display}\)	\(694\)

[In]

int((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

2/117/(64*sin(1/2*d*x+1/2*c)^12-192*sin(1/2*d*x+1/2*c)^10+240*sin(1/2*d*x+1/2*c)^8-160*sin(1/2*d*x+1/2*c)^6+60
*sin(1/2*d*x+1/2*c)^4-12*sin(1/2*d*x+1/2*c)^2+1)/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(3
84*sin(1/2*d*x+1/2*c)^14*cos(1/2*d*x+1/2*c)-192*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),
2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^12-1152*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^12+576*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*
x+1/2*c)^10+1472*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-720*EllipticE(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)^8-1024*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+480*EllipticE(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)^6+280*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-180*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+40*sin(1/2*d*x+1/2*c)^4*cos(
1/2*d*x+1/2*c)+36*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*sin(1/2*d*x+1/2*c)^2+208*sin(1/2*d*x+1/2*c)^5+120*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-208*sin(1/2*d*x+1/2*
c)^3-20*sin(1/2*d*x+1/2*c))*e/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 = 472, normalized size of antiderivative = 2.47 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx=\frac {3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{4} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 8 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 4 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 4 i \, \sqrt {2} \cos \left (d x + c\right ) - 8 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) - 4 i \, \sqrt {2} \cos \left (d x + c\right ) - 8 i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{4} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 8 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 4 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 4 i \, \sqrt {2} \cos \left (d x + c\right ) + 8 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) + 4 i \, \sqrt {2} \cos \left (d x + c\right ) + 8 i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, \cos \left (d x + c\right )^{4} + 12 \, \cos \left (d x + c\right )^{3} - 14 \, \cos \left (d x + c\right )^{2} + {\left (3 \, \cos \left (d x + c\right )^{3} - 9 \, \cos \left (d x + c\right )^{2} - 23 \, \cos \left (d x + c\right ) + 9\right )} \sin \left (d x + c\right ) - 32 \, \cos \left (d x + c\right ) - 9\right )} \sqrt {e \cos \left (d x + c\right )}}{117 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 3 \, a^{4} d \cos \left (d x + c\right )^{3} - 8 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + 8 \, a^{4} d - {\left (a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d \cos \left (d x + c\right ) - 8 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/117*(3*(-I*sqrt(2)*cos(d*x + c)^4 + 3*I*sqrt(2)*cos(d*x + c)^3 + 8*I*sqrt(2)*cos(d*x + c)^2 + (I*sqrt(2)*cos
(d*x + c)^3 + 4*I*sqrt(2)*cos(d*x + c)^2 - 4*I*sqrt(2)*cos(d*x + c) - 8*I*sqrt(2))*sin(d*x + c) - 4*I*sqrt(2)*
cos(d*x + c) - 8*I*sqrt(2))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x
+ c))) + 3*(I*sqrt(2)*cos(d*x + c)^4 - 3*I*sqrt(2)*cos(d*x + c)^3 - 8*I*sqrt(2)*cos(d*x + c)^2 + (-I*sqrt(2)*
cos(d*x + c)^3 - 4*I*sqrt(2)*cos(d*x + c)^2 + 4*I*sqrt(2)*cos(d*x + c) + 8*I*sqrt(2))*sin(d*x + c) + 4*I*sqrt(
2)*cos(d*x + c) + 8*I*sqrt(2))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(
d*x + c))) + 2*(3*cos(d*x + c)^4 + 12*cos(d*x + c)^3 - 14*cos(d*x + c)^2 + (3*cos(d*x + c)^3 - 9*cos(d*x + c)^
2 - 23*cos(d*x + c) + 9)*sin(d*x + c) - 32*cos(d*x + c) - 9)*sqrt(e*cos(d*x + c)))/(a^4*d*cos(d*x + c)^4 - 3*a
^4*d*cos(d*x + c)^3 - 8*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + 8*a^4*d - (a^4*d*cos(d*x + c)^3 + 4*a^4*
d*cos(d*x + c)^2 - 4*a^4*d*cos(d*x + c) - 8*a^4*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]

[In]

integrate((e*cos(d*x+c))**(1/2)/(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sqrt {e \cos \left (d x + c\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate(sqrt(e*cos(d*x + c))/(a*sin(d*x + c) + a)^4, x)

Giac [F]

\[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sqrt {e \cos \left (d x + c\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(sqrt(e*cos(d*x + c))/(a*sin(d*x + c) + a)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {\sqrt {e\,\cos \left (c+d\,x\right )}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]

[In]

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

[Out]

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