\(\int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx\) [315]

Optimal result

Integrand size = 27, antiderivative size = 218 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {2 e^{5/2} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^3+a^3 \cos (c+d x)+a^3 \sin (c+d x)\right )}-\frac {2 e^{5/2} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^3+a^3 \cos (c+d x)+a^3 \sin (c+d x)\right )} \]

[Out]

-4/3*e*(e*cos(d*x+c))^(3/2)/a/d/(a+a*sin(d*x+c))^(3/2)-2*e^(5/2)*arcsinh((e*cos(d*x+c))^(1/2)/e^(1/2))*(1+cos(
d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d/(a^3+a^3*cos(d*x+c)+a^3*sin(d*x+c))-2*e^(5/2)*arctan(sin(d*x+c)*e^(1/2)
/(e*cos(d*x+c))^(1/2)/(1+cos(d*x+c))^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d/(a^3+a^3*cos(d*x+c)+
a^3*sin(d*x+c))

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2759, 2763, 2854, 209, 2912, 65, 221} \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 e^{5/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d \left (a^3 \sin (c+d x)+a^3 \cos (c+d x)+a^3\right )}-\frac {2 e^{5/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d \left (a^3 \sin (c+d x)+a^3 \cos (c+d x)+a^3\right )}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

(-4*e*(e*Cos[c + d*x])^(3/2))/(3*a*d*(a + a*Sin[c + d*x])^(3/2)) - (2*e^(5/2)*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqr
t[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a^3 + a^3*Cos[c + d*x] + a^3*Sin[c + d*x])) - (2*e^
(5/2)*ArcTan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt
[a + a*Sin[c + d*x]])/(d*(a^3 + a^3*Cos[c + d*x] + a^3*Sin[c + d*x]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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 2763

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

Rule 2854

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-2*(b/f), Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.37 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {\sqrt [4]{2} (e \cos (c+d x))^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {7}{4},\frac {11}{4},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {a (1+\sin (c+d x))}}{7 a^3 d e (1+\sin (c+d x))^{9/4}} \]

[In]

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

[Out]

-1/7*(2^(1/4)*(e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[7/4, 7/4, 11/4, (1 - Sin[c + d*x])/2]*Sqrt[a*(1 + Sin[c
+ d*x])])/(a^3*d*e*(1 + Sin[c + d*x])^(9/4))

Maple [A] (verified)

Time = 3.32 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.32

method	result	size
default	\(-\frac {2 \left (3 \sin \left (d x +c \right ) \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-3 \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+2 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-2 \sin \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-3 \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+2 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {e \cos \left (d x +c \right )}\, e^{2}}{3 d \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) a^{2} \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\sin \left (d x +c \right )\right )}}\)	\(287\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 1334, normalized size of antiderivative = 6.12 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/6*(3*(a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d - (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x + c))*(-
e^10/(a^10*d^4))^(1/4)*log((2*(e^7*sin(d*x + c) + (a^5*d^2*e^2*cos(d*x + c) + a^5*d^2*e^2)*sqrt(-e^10/(a^10*d^
4)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) + (2*a^8*d^3*cos(d*x + c)^2 + a^8*d^3*cos(d*x + c) - a^8*d^
3*sin(d*x + c) - a^8*d^3)*(-e^10/(a^10*d^4))^(3/4) + (a^3*d*e^5*cos(d*x + c) + a^3*d*e^5 + (2*a^3*d*e^5*cos(d*
x + c) + a^3*d*e^5)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - 3*(a^3*d*cos(
d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d - (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1
/4)*log((2*(e^7*sin(d*x + c) + (a^5*d^2*e^2*cos(d*x + c) + a^5*d^2*e^2)*sqrt(-e^10/(a^10*d^4)))*sqrt(e*cos(d*x
+ c))*sqrt(a*sin(d*x + c) + a) - (2*a^8*d^3*cos(d*x + c)^2 + a^8*d^3*cos(d*x + c) - a^8*d^3*sin(d*x + c) - a^
8*d^3)*(-e^10/(a^10*d^4))^(3/4) - (a^3*d*e^5*cos(d*x + c) + a^3*d*e^5 + (2*a^3*d*e^5*cos(d*x + c) + a^3*d*e^5)
*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - 3*(I*a^3*d*cos(d*x + c)^2 - I*a^
3*d*cos(d*x + c) - 2*I*a^3*d + (-I*a^3*d*cos(d*x + c) - 2*I*a^3*d)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4)*log(
(2*(e^7*sin(d*x + c) - (a^5*d^2*e^2*cos(d*x + c) + a^5*d^2*e^2)*sqrt(-e^10/(a^10*d^4)))*sqrt(e*cos(d*x + c))*s
qrt(a*sin(d*x + c) + a) + (2*I*a^8*d^3*cos(d*x + c)^2 + I*a^8*d^3*cos(d*x + c) - I*a^8*d^3*sin(d*x + c) - I*a^
8*d^3)*(-e^10/(a^10*d^4))^(3/4) + (-I*a^3*d*e^5*cos(d*x + c) - I*a^3*d*e^5 + (-2*I*a^3*d*e^5*cos(d*x + c) - I*
a^3*d*e^5)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - 3*(-I*a^3*d*cos(d*x +
c)^2 + I*a^3*d*cos(d*x + c) + 2*I*a^3*d + (I*a^3*d*cos(d*x + c) + 2*I*a^3*d)*sin(d*x + c))*(-e^10/(a^10*d^4))^
(1/4)*log((2*(e^7*sin(d*x + c) - (a^5*d^2*e^2*cos(d*x + c) + a^5*d^2*e^2)*sqrt(-e^10/(a^10*d^4)))*sqrt(e*cos(d
*x + c))*sqrt(a*sin(d*x + c) + a) + (-2*I*a^8*d^3*cos(d*x + c)^2 - I*a^8*d^3*cos(d*x + c) + I*a^8*d^3*sin(d*x
+ c) + I*a^8*d^3)*(-e^10/(a^10*d^4))^(3/4) + (I*a^3*d*e^5*cos(d*x + c) + I*a^3*d*e^5 + (2*I*a^3*d*e^5*cos(d*x
+ c) + I*a^3*d*e^5)*sin(d*x + c))*(-e^10/(a^10*d^4))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - 8*(e^2*cos(d*
x + c) - e^2*sin(d*x + c) + e^2)*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^2 - a^3*d*
cos(d*x + c) - 2*a^3*d - (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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