\(\int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx\) [269]

Optimal result

Integrand size = 25, antiderivative size = 154 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{77 a^4 d \sqrt {e \cos (c+d x)}}-\frac {4 e \sqrt {e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

-2/77*e^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)/a^4/d/(e*cos(d*x+c))^(1/2)-4/11*e*(e*cos(d*x+c))^(1/2)/a/d/(a+a*sin(d*x+c))^3+2/77*e*(e*cos(d*x+c))^(1/2)/d
/(a^2+a^2*sin(d*x+c))^2+2/77*e*(e*cos(d*x+c))^(1/2)/d/(a^4+a^4*sin(d*x+c))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2759, 2760, 2762, 2721, 2720} \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{77 a^4 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^4 \sin (c+d x)+a^4\right )}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {4 e \sqrt {e \cos (c+d x)}}{11 a d (a \sin (c+d x)+a)^3} \]

[In]

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

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*a^4*d*Sqrt[e*Cos[c + d*x]]) - (4*e*Sqrt[e*Cos[c + d*
x]])/(11*a*d*(a + a*Sin[c + d*x])^3) + (2*e*Sqrt[e*Cos[c + d*x]])/(77*d*(a^2 + a^2*Sin[c + d*x])^2) + (2*e*Sqr
t[e*Cos[c + d*x]])/(77*d*(a^4 + a^4*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 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]

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 {4 e \sqrt {e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}-\frac {e^2 \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx}{11 a^2} \\ & = -\frac {4 e \sqrt {e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {\left (3 e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))} \, dx}{77 a^3} \\ & = -\frac {4 e \sqrt {e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{77 a^4} \\ & = -\frac {4 e \sqrt {e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {\left (e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 a^4 \sqrt {e \cos (c+d x)}} \\ & = -\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{77 a^4 d \sqrt {e \cos (c+d x)}}-\frac {4 e \sqrt {e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e \sqrt {e \cos (c+d x)}}{77 d \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 = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {(e \cos (c+d x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {15}{4},\frac {9}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{10\ 2^{3/4} a^4 d e (1+\sin (c+d x))^{5/4}} \]

[In]

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

[Out]

-1/10*((e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[5/4, 15/4, 9/4, (1 - Sin[c + d*x])/2])/(2^(3/4)*a^4*d*e*(1 + S
in[c + d*x])^(5/4))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(582\) vs. \(2(162)=324\).

Time = 9.14 (sec) , antiderivative size = 583, normalized size of antiderivative = 3.79

method	result	size
default	\(\frac {2 \left (32 \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 ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-80 \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 ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 \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 ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+176 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-40 \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 ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-144 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+10 \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 )-176 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-78 \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 )+176 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{2}}{77 \left (32 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-80 \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{4} \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}\, d}\)	\(583\)

[In]

int((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

2/77/(32*sin(1/2*d*x+1/2*c)^10-80*sin(1/2*d*x+1/2*c)^8+80*sin(1/2*d*x+1/2*c)^6-40*sin(1/2*d*x+1/2*c)^4+10*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)*(32*(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)^10+32*sin(1/2*d*x+
1/2*c)^10*cos(1/2*d*x+1/2*c)-80*(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)^8-64*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+80*(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)^6+176*sin(1/
2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-40*(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)^4-144*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+10*(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)^2-176*
sin(1/2*d*x+1/2*c)^5-78*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)^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))+176*sin(1/2*d*x+1/2*c)^3+12*sin(1/2*d*x+1/2*c))*e^2/d

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.49 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {{\left (3 i \, \sqrt {2} e \cos \left (d x + c\right )^{2} + {\left (i \, \sqrt {2} e \cos \left (d x + c\right )^{2} - 4 i \, \sqrt {2} e\right )} \sin \left (d x + c\right ) - 4 i \, \sqrt {2} e\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (-3 i \, \sqrt {2} e \cos \left (d x + c\right )^{2} + {\left (-i \, \sqrt {2} e \cos \left (d x + c\right )^{2} + 4 i \, \sqrt {2} e\right )} \sin \left (d x + c\right ) + 4 i \, \sqrt {2} e\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (e \cos \left (d x + c\right )^{2} - 3 \, e \sin \left (d x + c\right ) + 11 \, e\right )} \sqrt {e \cos \left (d x + c\right )}}{77 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/77*((3*I*sqrt(2)*e*cos(d*x + c)^2 + (I*sqrt(2)*e*cos(d*x + c)^2 - 4*I*sqrt(2)*e)*sin(d*x + c) - 4*I*sqrt(2)*
e)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (-3*I*sqrt(2)*e*cos(d*x + c)^2 + (-I*sq
rt(2)*e*cos(d*x + c)^2 + 4*I*sqrt(2)*e)*sin(d*x + c) + 4*I*sqrt(2)*e)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d
*x + c) - I*sin(d*x + c)) + 2*(e*cos(d*x + c)^2 - 3*e*sin(d*x + c) + 11*e)*sqrt(e*cos(d*x + c)))/(3*a^4*d*cos(
d*x + c)^2 - 4*a^4*d + (a^4*d*cos(d*x + c)^2 - 4*a^4*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]

[In]

integrate((e*cos(d*x+c))**(3/2)/(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]

[In]

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

[Out]

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