\(\int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx\) [292]

Optimal result

Integrand size = 27, antiderivative size = 239 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}-\frac {5 a^2 \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (1+\cos (c+d x)+\sin (c+d x))}-\frac {5 a^2 \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (1+\cos (c+d x)+\sin (c+d x))}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2755, 2757, 2763, 2854, 209, 2912, 65, 221} \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a \sin (c+d x)+a}}-\frac {5 a^2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {5 a^2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d e^{3/2} (\sin (c+d x)+\cos (c+d x)+1)}+\frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}} \]

[In]

Int[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(3/2),x]

[Out]

(5*a^3*(e*Cos[c + d*x])^(3/2))/(d*e^3*Sqrt[a + a*Sin[c + d*x]]) - (5*a^2*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]
*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*e^(3/2)*(1 + Cos[c + d*x] + Sin[c + d*x])) - (5*a^2*ArcTa
n[(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*e^(3/2)*(1 + Cos[c + d*x] + Sin[c + d*x])) + (4*a*(a + a*Sin[c + d*x])^(3/2))/(d*e*Sqrt[e*Cos[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 2755

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

Rule 2757

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), 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] && GtQ[m, 0] && NeQ[m + p, 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 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^2\right ) \int \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)} \, dx}{e^2} \\ & = \frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^3\right ) \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 e^2} \\ & = \frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^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}{2 e (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (5 a^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}{2 e (a+a \cos (c+d x)+a \sin (c+d x))} \\ & = \frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^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 )}{2 d e (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (5 a^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 )}{d e (a+a \cos (c+d x)+a \sin (c+d x))} \\ & = \frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^3 \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 e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (5 a^3 \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 )}{d e^2 (a+a \cos (c+d x)+a \sin (c+d x))} \\ & = \frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^3 \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 e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {5 a^3 \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 e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.31 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {8 \sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {1}{4},\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^{5/2}}{d e \sqrt {e \cos (c+d x)} (1+\sin (c+d x))^{9/4}} \]

[In]

Integrate[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(3/2),x]

[Out]

(8*2^(1/4)*Hypergeometric2F1[-5/4, -1/4, 3/4, (1 - Sin[c + d*x])/2]*(a*(1 + Sin[c + d*x]))^(5/2))/(d*e*Sqrt[e*
Cos[c + d*x]]*(1 + Sin[c + d*x])^(9/4))

Maple [A] (verified)

Time = 6.72 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.28

method	result	size
default	\(\frac {\left (5 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )-5 \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+\cos ^{2}\left (d x +c \right )-\cos \left (d x +c \right ) \sin \left (d x +c \right )+5 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-5 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+9 \cos \left (d x +c \right )+8 \sin \left (d x +c \right )+8\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, a^{2}}{d \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) e \sqrt {e \cos \left (d x +c \right )}}\)	\(307\)

[In]

int((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(5*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-5*arctanh(sin(
d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+cos(d*
x+c)^2-cos(d*x+c)*sin(d*x+c)+5*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-5
*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+9*co
s(d*x+c)+8*sin(d*x+c)+8)*(a*(1+sin(d*x+c)))^(1/2)*a^2/(1+cos(d*x+c)+sin(d*x+c))/e/(e*cos(d*x+c))^(1/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 1093, normalized size of antiderivative = 4.57 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(5*d*e^2*(-a^10/(d^4*e^6))^(1/4)*cos(d*x + c)*log(125/2*(2*(a^7*sin(d*x + c) + (a^2*d^2*e^3*cos(d*x + c)
+ a^2*d^2*e^3)*sqrt(-a^10/(d^4*e^6)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) + (2*d^3*e^5*cos(d*x + c)^
2 + d^3*e^5*cos(d*x + c) - d^3*e^5*sin(d*x + c) - d^3*e^5)*(-a^10/(d^4*e^6))^(3/4) + (a^5*d*e^2*cos(d*x + c) +
a^5*d*e^2 + (2*a^5*d*e^2*cos(d*x + c) + a^5*d*e^2)*sin(d*x + c))*(-a^10/(d^4*e^6))^(1/4))/(cos(d*x + c) + sin
(d*x + c) + 1)) - 5*d*e^2*(-a^10/(d^4*e^6))^(1/4)*cos(d*x + c)*log(125/2*(2*(a^7*sin(d*x + c) + (a^2*d^2*e^3*c
os(d*x + c) + a^2*d^2*e^3)*sqrt(-a^10/(d^4*e^6)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) - (2*d^3*e^5*c
os(d*x + c)^2 + d^3*e^5*cos(d*x + c) - d^3*e^5*sin(d*x + c) - d^3*e^5)*(-a^10/(d^4*e^6))^(3/4) - (a^5*d*e^2*co
s(d*x + c) + a^5*d*e^2 + (2*a^5*d*e^2*cos(d*x + c) + a^5*d*e^2)*sin(d*x + c))*(-a^10/(d^4*e^6))^(1/4))/(cos(d*
x + c) + sin(d*x + c) + 1)) + 5*I*d*e^2*(-a^10/(d^4*e^6))^(1/4)*cos(d*x + c)*log(125/2*(2*(a^7*sin(d*x + c) -
(a^2*d^2*e^3*cos(d*x + c) + a^2*d^2*e^3)*sqrt(-a^10/(d^4*e^6)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)
- (2*I*d^3*e^5*cos(d*x + c)^2 + I*d^3*e^5*cos(d*x + c) - I*d^3*e^5*sin(d*x + c) - I*d^3*e^5)*(-a^10/(d^4*e^6))
^(3/4) - (-I*a^5*d*e^2*cos(d*x + c) - I*a^5*d*e^2 + (-2*I*a^5*d*e^2*cos(d*x + c) - I*a^5*d*e^2)*sin(d*x + c))*
(-a^10/(d^4*e^6))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - 5*I*d*e^2*(-a^10/(d^4*e^6))^(1/4)*cos(d*x + c)*l
og(125/2*(2*(a^7*sin(d*x + c) - (a^2*d^2*e^3*cos(d*x + c) + a^2*d^2*e^3)*sqrt(-a^10/(d^4*e^6)))*sqrt(e*cos(d*x
+ c))*sqrt(a*sin(d*x + c) + a) - (-2*I*d^3*e^5*cos(d*x + c)^2 - I*d^3*e^5*cos(d*x + c) + I*d^3*e^5*sin(d*x +
c) + I*d^3*e^5)*(-a^10/(d^4*e^6))^(3/4) - (I*a^5*d*e^2*cos(d*x + c) + I*a^5*d*e^2 + (2*I*a^5*d*e^2*cos(d*x + c
) + I*a^5*d*e^2)*sin(d*x + c))*(-a^10/(d^4*e^6))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) + 4*(a^2*sin(d*x +
c) - 9*a^2)*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*e^2*cos(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))**(5/2)/(e*cos(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)/(e*cos(d*x + c))^(3/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int((a + a*sin(c + d*x))^(5/2)/(e*cos(c + d*x))^(3/2),x)

[Out]

int((a + a*sin(c + d*x))^(5/2)/(e*cos(c + d*x))^(3/2), x)