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

Optimal. Leaf size=219 \[ \frac{5 \sqrt{c} \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 e^6}-\frac{5 c \sqrt{a+c x^2} \left (8 d \left (a e^2+c d^2\right )-e x \left (3 a e^2+4 c d^2\right )\right )}{8 e^5}+\frac{5 c d \left (a e^2+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^6}-\frac{5 c \left (a+c x^2\right )^{3/2} (4 d-3 e x)}{12 e^3}-\frac{\left (a+c x^2\right )^{5/2}}{e (d+e x)} \]

[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

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Rubi [A]  time = 0.243901, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {733, 815, 844, 217, 206, 725} \[ \frac{5 \sqrt{c} \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 e^6}-\frac{5 c \sqrt{a+c x^2} \left (8 d \left (a e^2+c d^2\right )-e x \left (3 a e^2+4 c d^2\right )\right )}{8 e^5}+\frac{5 c d \left (a e^2+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^6}-\frac{5 c \left (a+c x^2\right )^{3/2} (4 d-3 e x)}{12 e^3}-\frac{\left (a+c x^2\right )^{5/2}}{e (d+e x)} \]

Antiderivative was successfully verified.

[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 733

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 815

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 844

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]

Rule 217

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 206

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

Rule 725

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]

Rubi steps

\begin{align*} \int \frac{\left (a+c x^2\right )^{5/2}}{(d+e x)^2} \, dx &=-\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 ) \operatorname{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 ) \operatorname{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]  time = 0.24748, size = 239, normalized size = 1.09 \[ \frac{15 \sqrt{c} \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )+e \sqrt{a+c x^2} \left (9 c e x \left (3 a e^2+4 c d^2\right )-\frac{24 \left (a e^2+c d^2\right )^2}{d+e x}-16 c d \left (7 a e^2+6 c d^2\right )-16 c^2 d e^2 x^2+6 c^2 e^3 x^3\right )+120 c d \left (a e^2+c d^2\right )^{3/2} \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )-120 c d \left (a e^2+c d^2\right )^{3/2} \log (d+e x)}{24 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(-16*c*d*(6*c*d^2 + 7*a*e^2) + 9*c*e*(4*c*d^2 + 3*a*e^2)*x - 16*c^2*d*e^2*x^2 + 6*c^2*e^3*x
^3 - (24*(c*d^2 + a*e^2)^2)/(d + e*x)) - 120*c*d*(c*d^2 + a*e^2)^(3/2)*Log[d + e*x] + 15*Sqrt[c]*(8*c^2*d^4 +
12*a*c*d^2*e^2 + 3*a^2*e^4)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]] + 120*c*d*(c*d^2 + a*e^2)^(3/2)*Log[a*e - c*d*x
 + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/(24*e^6)

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Maple [B]  time = 0.194, size = 1796, normalized size = 8.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*e^2+c*d^2)/(d/e+x)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(7/2)-1/e*c*d/(a*e^2+c*d^2)*(c*(d/e+x
)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(5/2)+5/4/e^2*c^2*d^2/(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2
+c*d^2)/e^2)^(3/2)*x+35/8/e^2*c^2*d^2/(a*e^2+c*d^2)*a*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x+
75/8/e^2*c^(3/2)*d^2/(a*e^2+c*d^2)*ln((-c*d/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^
2)^(1/2))*a^2-5/3/e*c*d/(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*a-5/3/e^3*c^2*d^3/
(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)+5/2/e^4*c^3*d^4/(a*e^2+c*d^2)*(c*(d/e+x)^2
-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x+25/2/e^4*c^(5/2)*d^4/(a*e^2+c*d^2)*ln((-c*d/e+(d/e+x)*c)/c^(1/2)+(
c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))*a-5/e*c*d/(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*
e^2+c*d^2)/e^2)^(1/2)*a^2-10/e^3*c^2*d^3/(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*a
-5/e^5*c^3*d^5/(a*e^2+c*d^2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)+5/e^6*c^(7/2)*d^6/(a*e^2+c*
d^2)*ln((-c*d/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))+5/e*c*d/(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*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(d/e+x)^2-2
*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))*a^3+15/e^3*c^2*d^3/(a*e^2+c*d^2)/((a*e^2+c*d^2)/e^2)^(1/2)*l
n((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/
e^2)^(1/2))/(d/e+x))*a^2+15/e^5*c^3*d^5/(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*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))*a+5/e^7*
c^4*d^7/(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*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^
(1/2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))+1/(a*e^2+c*d^2)*c*(c*(d/e+x)^2-2*c*d/e*(
d/e+x)+(a*e^2+c*d^2)/e^2)^(5/2)*x+5/4/(a*e^2+c*d^2)*c*a*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*
x+15/8/(a*e^2+c*d^2)*c*a^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x+15/8/(a*e^2+c*d^2)*c^(1/2)*
a^3*ln((-c*d/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 89.5191, size = 2967, normalized size = 13.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + c x^{2}\right )^{\frac{5}{2}}}{\left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out