\(\int \frac {(a+c x^2)^{5/2}}{(d+e x)^2} \, dx\) [550]

Optimal result

Integrand size = 19, antiderivative size = 219 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {5 c \left (8 d \left (c d^2+a e^2\right )-e \left (4 c d^2+3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}-\frac {5 c (4 d-3 e x) \left (a+c x^2\right )^{3/2}}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)}+\frac {5 \sqrt {c} \left (8 c^2 d^4+12 a c d^2 e^2+3 a^2 e^4\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 e^6}+\frac {5 c d \left (c d^2+a e^2\right )^{3/2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^6} \]

[Out]

-5/12*c*(-3*e*x+4*d)*(c*x^2+a)^(3/2)/e^3-(c*x^2+a)^(5/2)/e/(e*x+d)+5*c*d*(a*e^2+c*d^2)^(3/2)*arctanh((-c*d*x+a
*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/e^6+5/8*(3*a^2*e^4+12*a*c*d^2*e^2+8*c^2*d^4)*arctanh(x*c^(1/2)/(c*x^2
+a)^(1/2))*c^(1/2)/e^6-5/8*c*(8*d*(a*e^2+c*d^2)-e*(3*a*e^2+4*c*d^2)*x)*(c*x^2+a)^(1/2)/e^5

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {747, 829, 858, 223, 212, 739} \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {5 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right )}{8 e^6}+\frac {5 c d \left (a e^2+c d^2\right )^{3/2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^6}-\frac {5 c \sqrt {a+c x^2} \left (8 d \left (a e^2+c d^2\right )-e x \left (3 a e^2+4 c d^2\right )\right )}{8 e^5}-\frac {5 c \left (a+c x^2\right )^{3/2} (4 d-3 e x)}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)} \]

[In]

Int[(a + c*x^2)^(5/2)/(d + e*x)^2,x]

[Out]

(-5*c*(8*d*(c*d^2 + a*e^2) - e*(4*c*d^2 + 3*a*e^2)*x)*Sqrt[a + c*x^2])/(8*e^5) - (5*c*(4*d - 3*e*x)*(a + c*x^2
)^(3/2))/(12*e^3) - (a + c*x^2)^(5/2)/(e*(d + e*x)) + (5*Sqrt[c]*(8*c^2*d^4 + 12*a*c*d^2*e^2 + 3*a^2*e^4)*ArcT
anh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*e^6) + (5*c*d*(c*d^2 + a*e^2)^(3/2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a
*e^2]*Sqrt[a + c*x^2])])/e^6

Rule 212

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

Rule 223

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

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 747

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 1))), x] - Dist[2*c*(p/(e*(m + 1))), Int[x*(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c,
d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 829

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m +
2*p + 2))), x] + Dist[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)}+\frac {(5 c) \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x} \, dx}{e} \\ & = -\frac {5 c (4 d-3 e x) \left (a+c x^2\right )^{3/2}}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)}+\frac {5 \int \frac {\left (-a c d e+c \left (4 c d^2+3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{d+e x} \, dx}{4 e^3} \\ & = -\frac {5 c \left (8 d \left (c d^2+a e^2\right )-e \left (4 c d^2+3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}-\frac {5 c (4 d-3 e x) \left (a+c x^2\right )^{3/2}}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)}+\frac {5 \int \frac {-a c^2 d e \left (4 c d^2+5 a e^2\right )+c^2 \left (8 c^2 d^4+12 a c d^2 e^2+3 a^2 e^4\right ) x}{(d+e x) \sqrt {a+c x^2}} \, dx}{8 c e^5} \\ & = -\frac {5 c \left (8 d \left (c d^2+a e^2\right )-e \left (4 c d^2+3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}-\frac {5 c (4 d-3 e x) \left (a+c x^2\right )^{3/2}}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)}-\frac {\left (5 c d \left (c d^2+a e^2\right )^2\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{e^6}+\frac {\left (5 c \left (8 c^2 d^4+12 a c d^2 e^2+3 a^2 e^4\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 e^6} \\ & = -\frac {5 c \left (8 d \left (c d^2+a e^2\right )-e \left (4 c d^2+3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}-\frac {5 c (4 d-3 e x) \left (a+c x^2\right )^{3/2}}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)}+\frac {\left (5 c d \left (c d^2+a e^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{e^6}+\frac {\left (5 c \left (8 c^2 d^4+12 a c d^2 e^2+3 a^2 e^4\right )\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{8 e^6} \\ & = -\frac {5 c \left (8 d \left (c d^2+a e^2\right )-e \left (4 c d^2+3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 e^5}-\frac {5 c (4 d-3 e x) \left (a+c x^2\right )^{3/2}}{12 e^3}-\frac {\left (a+c x^2\right )^{5/2}}{e (d+e x)}+\frac {5 \sqrt {c} \left (8 c^2 d^4+12 a c d^2 e^2+3 a^2 e^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 e^6}+\frac {5 c d \left (c d^2+a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.14 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {\frac {e \sqrt {a+c x^2} \left (24 a^2 e^4+a c e^2 \left (160 d^2+85 d e x-27 e^2 x^2\right )+2 c^2 \left (60 d^4+30 d^3 e x-10 d^2 e^2 x^2+5 d e^3 x^3-3 e^4 x^4\right )\right )}{d+e x}-240 c d \left (-c d^2-a e^2\right )^{3/2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+15 \sqrt {c} \left (8 c^2 d^4+12 a c d^2 e^2+3 a^2 e^4\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{24 e^6} \]

[In]

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

[Out]

-1/24*((e*Sqrt[a + c*x^2]*(24*a^2*e^4 + a*c*e^2*(160*d^2 + 85*d*e*x - 27*e^2*x^2) + 2*c^2*(60*d^4 + 30*d^3*e*x
- 10*d^2*e^2*x^2 + 5*d*e^3*x^3 - 3*e^4*x^4)))/(d + e*x) - 240*c*d*(-(c*d^2) - a*e^2)^(3/2)*ArcTan[(Sqrt[c]*(d
+ e*x) - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]] + 15*Sqrt[c]*(8*c^2*d^4 + 12*a*c*d^2*e^2 + 3*a^2*e^4)*Log
[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/e^6

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(529\) vs. \(2(195)=390\).

Time = 2.01 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.42

method	result	size
risch	\(-\frac {c \left (-6 c \,x^{3} e^{3}+16 c d \,x^{2} e^{2}-27 a \,e^{3} x -36 c \,d^{2} e x +112 a d \,e^{2}+96 d^{3} c \right ) \sqrt {c \,x^{2}+a}}{24 e^{5}}+\frac {\frac {5 \sqrt {c}\, \left (3 a^{2} e^{4}+12 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{e}+\frac {48 c d \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}+\frac {\left (8 e^{6} a^{3}+24 d^{2} e^{4} a^{2} c +24 d^{4} e^{2} c^{2} a +8 c^{3} d^{6}\right ) \left (-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{\left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {c d e \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{e^{3}}}{8 e^{5}}\)	\(530\)
default	\(\text {Expression too large to display}\)	\(1309\)

[In]

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

[Out]

-1/24*c*(-6*c*e^3*x^3+16*c*d*e^2*x^2-27*a*e^3*x-36*c*d^2*e*x+112*a*d*e^2+96*c*d^3)*(c*x^2+a)^(1/2)/e^5+1/8/e^5
*(5*c^(1/2)*(3*a^2*e^4+12*a*c*d^2*e^2+8*c^2*d^4)/e*ln(c^(1/2)*x+(c*x^2+a)^(1/2))+48*c*d*(a^2*e^4+2*a*c*d^2*e^2
+c^2*d^4)/e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c
*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))+1/e^3*(8*a^3*e^6+24*a^2*c*d^2*e^4+24*a*c^2*d^4*e
^2+8*c^3*d^6)*(-1/(a*e^2+c*d^2)*e^2/(x+d/e)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)-c*d*e/(a*e^2
+c*d^2)/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(x+d/
e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))))

Fricas [A] (verification not implemented)

none

Time = 8.79 (sec) , antiderivative size = 1372, normalized size of antiderivative = 6.26 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/48*(15*(8*c^2*d^5 + 12*a*c*d^3*e^2 + 3*a^2*d*e^4 + (8*c^2*d^4*e + 12*a*c*d^2*e^3 + 3*a^2*e^5)*x)*sqrt(c)*lo
g(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 120*(c^2*d^4 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*sqrt(c
*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*
x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(6*c^2*e^5*x^4 - 10*c^2*d*e^4*x^3 - 120*c^2*d^4*e - 1
60*a*c*d^2*e^3 - 24*a^2*e^5 + (20*c^2*d^2*e^3 + 27*a*c*e^5)*x^2 - 5*(12*c^2*d^3*e^2 + 17*a*c*d*e^4)*x)*sqrt(c*
x^2 + a))/(e^7*x + d*e^6), 1/48*(240*(c^2*d^4 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*sqrt(-c*d^2 - a*e^2)*
arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) + 15*
(8*c^2*d^5 + 12*a*c*d^3*e^2 + 3*a^2*d*e^4 + (8*c^2*d^4*e + 12*a*c*d^2*e^3 + 3*a^2*e^5)*x)*sqrt(c)*log(-2*c*x^2
- 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(6*c^2*e^5*x^4 - 10*c^2*d*e^4*x^3 - 120*c^2*d^4*e - 160*a*c*d^2*e^3 -
24*a^2*e^5 + (20*c^2*d^2*e^3 + 27*a*c*e^5)*x^2 - 5*(12*c^2*d^3*e^2 + 17*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(e^7*x
+ d*e^6), -1/24*(15*(8*c^2*d^5 + 12*a*c*d^3*e^2 + 3*a^2*d*e^4 + (8*c^2*d^4*e + 12*a*c*d^2*e^3 + 3*a^2*e^5)*x)*
sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 60*(c^2*d^4 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*sqrt(c*d^
2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x -
a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) - (6*c^2*e^5*x^4 - 10*c^2*d*e^4*x^3 - 120*c^2*d^4*e - 160*a*
c*d^2*e^3 - 24*a^2*e^5 + (20*c^2*d^2*e^3 + 27*a*c*e^5)*x^2 - 5*(12*c^2*d^3*e^2 + 17*a*c*d*e^4)*x)*sqrt(c*x^2 +
a))/(e^7*x + d*e^6), 1/24*(120*(c^2*d^4 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*sqrt(-c*d^2 - a*e^2)*arcta
n(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - 15*(8*c^
2*d^5 + 12*a*c*d^3*e^2 + 3*a^2*d*e^4 + (8*c^2*d^4*e + 12*a*c*d^2*e^3 + 3*a^2*e^5)*x)*sqrt(-c)*arctan(sqrt(-c)*
x/sqrt(c*x^2 + a)) + (6*c^2*e^5*x^4 - 10*c^2*d*e^4*x^3 - 120*c^2*d^4*e - 160*a*c*d^2*e^3 - 24*a^2*e^5 + (20*c^
2*d^2*e^3 + 27*a*c*e^5)*x^2 - 5*(12*c^2*d^3*e^2 + 17*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(e^7*x + d*e^6)]

Sympy [F]

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

[In]

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

[Out]

Integral((a + c*x**2)**(5/2)/(d + e*x)**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

int((a + c*x^2)^(5/2)/(d + e*x)^2,x)

[Out]

int((a + c*x^2)^(5/2)/(d + e*x)^2, x)