### 3.553 $$\int \frac{(a+c x^2)^{5/2}}{(d+e x)^5} \, dx$$

Optimal. Leaf size=287 $-\frac{5 c^2 \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{8 e^6 \left (a e^2+c d^2\right )^{3/2}}+\frac{5 c^2 \sqrt{a+c x^2} \left (e x \left (3 a e^2+4 c d^2\right )+8 d \left (a e^2+c d^2\right )\right )}{8 e^5 (d+e x) \left (a e^2+c d^2\right )}-\frac{5 c^{5/2} d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{e^6}-\frac{5 c \left (a+c x^2\right )^{3/2} \left (3 e x \left (a e^2+2 c d^2\right )+d \left (a e^2+4 c d^2\right )\right )}{24 e^3 (d+e x)^3 \left (a e^2+c d^2\right )}-\frac{\left (a+c x^2\right )^{5/2}}{4 e (d+e x)^4}$

[Out]

(5*c^2*(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*(c*d^2 + a*e^2)*(d + e*x)) - (5
*c*(d*(4*c*d^2 + a*e^2) + 3*e*(2*c*d^2 + a*e^2)*x)*(a + c*x^2)^(3/2))/(24*e^3*(c*d^2 + a*e^2)*(d + e*x)^3) - (
a + c*x^2)^(5/2)/(4*e*(d + e*x)^4) - (5*c^(5/2)*d*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/e^6 - (5*c^2*(8*c^2*d^
4 + 12*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(8*e^6*(c*d^2 +
a*e^2)^(3/2))

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Rubi [A]  time = 0.302573, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.368, Rules used = {733, 811, 813, 844, 217, 206, 725} $-\frac{5 c^2 \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{8 e^6 \left (a e^2+c d^2\right )^{3/2}}+\frac{5 c^2 \sqrt{a+c x^2} \left (e x \left (3 a e^2+4 c d^2\right )+8 d \left (a e^2+c d^2\right )\right )}{8 e^5 (d+e x) \left (a e^2+c d^2\right )}-\frac{5 c^{5/2} d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{e^6}-\frac{5 c \left (a+c x^2\right )^{3/2} \left (3 e x \left (a e^2+2 c d^2\right )+d \left (a e^2+4 c d^2\right )\right )}{24 e^3 (d+e x)^3 \left (a e^2+c d^2\right )}-\frac{\left (a+c x^2\right )^{5/2}}{4 e (d+e x)^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + c*x^2)^(5/2)/(d + e*x)^5,x]

[Out]

(5*c^2*(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*(c*d^2 + a*e^2)*(d + e*x)) - (5
*c*(d*(4*c*d^2 + a*e^2) + 3*e*(2*c*d^2 + a*e^2)*x)*(a + c*x^2)^(3/2))/(24*e^3*(c*d^2 + a*e^2)*(d + e*x)^3) - (
a + c*x^2)^(5/2)/(4*e*(d + e*x)^4) - (5*c^(5/2)*d*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/e^6 - (5*c^2*(8*c^2*d^
4 + 12*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(8*e^6*(c*d^2 +
a*e^2)^(3/2))

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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 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)^5} \, dx &=-\frac{\left (a+c x^2\right )^{5/2}}{4 e (d+e x)^4}+\frac{(5 c) \int \frac{x \left (a+c x^2\right )^{3/2}}{(d+e x)^4} \, dx}{4 e}\\ &=-\frac{5 c \left (d \left (4 c d^2+a e^2\right )+3 e \left (2 c d^2+a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{24 e^3 \left (c d^2+a e^2\right ) (d+e x)^3}-\frac{\left (a+c x^2\right )^{5/2}}{4 e (d+e x)^4}-\frac{(5 c) \int \frac{\left (4 a c d e-2 c \left (4 c d^2+3 a e^2\right ) x\right ) \sqrt{a+c x^2}}{(d+e x)^2} \, dx}{16 e^3 \left (c d^2+a e^2\right )}\\ &=\frac{5 c^2 \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 \left (c d^2+a e^2\right ) (d+e x)}-\frac{5 c \left (d \left (4 c d^2+a e^2\right )+3 e \left (2 c d^2+a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{24 e^3 \left (c d^2+a e^2\right ) (d+e x)^3}-\frac{\left (a+c x^2\right )^{5/2}}{4 e (d+e x)^4}+\frac{(5 c) \int \frac{4 a c e \left (4 c d^2+3 a e^2\right )-32 c^2 d \left (c d^2+a e^2\right ) x}{(d+e x) \sqrt{a+c x^2}} \, dx}{32 e^5 \left (c d^2+a e^2\right )}\\ &=\frac{5 c^2 \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 \left (c d^2+a e^2\right ) (d+e x)}-\frac{5 c \left (d \left (4 c d^2+a e^2\right )+3 e \left (2 c d^2+a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{24 e^3 \left (c d^2+a e^2\right ) (d+e x)^3}-\frac{\left (a+c x^2\right )^{5/2}}{4 e (d+e x)^4}-\frac{\left (5 c^3 d\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{e^6}+\frac{\left (5 c^2 \left (8 c^2 d^4+12 a c d^2 e^2+3 a^2 e^4\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{8 e^6 \left (c d^2+a e^2\right )}\\ &=\frac{5 c^2 \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 \left (c d^2+a e^2\right ) (d+e x)}-\frac{5 c \left (d \left (4 c d^2+a e^2\right )+3 e \left (2 c d^2+a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{24 e^3 \left (c d^2+a e^2\right ) (d+e x)^3}-\frac{\left (a+c x^2\right )^{5/2}}{4 e (d+e x)^4}-\frac{\left (5 c^3 d\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{e^6}-\frac{\left (5 c^2 \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}{c d^2+a e^2-x^2} \, dx,x,\frac{a e-c d x}{\sqrt{a+c x^2}}\right )}{8 e^6 \left (c d^2+a e^2\right )}\\ &=\frac{5 c^2 \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 \left (c d^2+a e^2\right ) (d+e x)}-\frac{5 c \left (d \left (4 c d^2+a e^2\right )+3 e \left (2 c d^2+a e^2\right ) x\right ) \left (a+c x^2\right )^{3/2}}{24 e^3 \left (c d^2+a e^2\right ) (d+e x)^3}-\frac{\left (a+c x^2\right )^{5/2}}{4 e (d+e x)^4}-\frac{5 c^{5/2} d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{e^6}-\frac{5 c^2 \left (8 c^2 d^4+12 a c d^2 e^2+3 a^2 e^4\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{8 e^6 \left (c d^2+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.545202, size = 299, normalized size = 1.04 $\frac{-\frac{15 c^2 \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )}{\left (a e^2+c d^2\right )^{3/2}}+\frac{15 c^2 \left (3 a^2 e^4+12 a c d^2 e^2+8 c^2 d^4\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}}+e \sqrt{a+c x^2} \left (\frac{c^2 d \left (139 a e^2+154 c d^2\right )}{(d+e x) \left (a e^2+c d^2\right )}-\frac{c \left (27 a e^2+86 c d^2\right )}{(d+e x)^2}+\frac{34 c d \left (a e^2+c d^2\right )}{(d+e x)^3}-\frac{6 \left (a e^2+c d^2\right )^2}{(d+e x)^4}+24 c^2\right )-120 c^{5/2} d \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{24 e^6}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(24*c^2 - (6*(c*d^2 + a*e^2)^2)/(d + e*x)^4 + (34*c*d*(c*d^2 + a*e^2))/(d + e*x)^3 - (c*(86
*c*d^2 + 27*a*e^2))/(d + e*x)^2 + (c^2*d*(154*c*d^2 + 139*a*e^2))/((c*d^2 + a*e^2)*(d + e*x))) + (15*c^2*(8*c^
2*d^4 + 12*a*c*d^2*e^2 + 3*a^2*e^4)*Log[d + e*x])/(c*d^2 + a*e^2)^(3/2) - 120*c^(5/2)*d*Log[c*x + Sqrt[c]*Sqrt
[a + c*x^2]] - (15*c^2*(8*c^2*d^4 + 12*a*c*d^2*e^2 + 3*a^2*e^4)*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a +
c*x^2]])/(c*d^2 + a*e^2)^(3/2))/(24*e^6)

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Maple [B]  time = 0.201, size = 5406, normalized size = 18.8 \begin{align*} \text{output 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)^5,x)

[Out]

result too large to display

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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)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 103.694, size = 7640, normalized size = 26.62 \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)^5,x, algorithm="fricas")

[Out]

[1/48*(120*(c^4*d^9 + 2*a*c^3*d^7*e^2 + a^2*c^2*d^5*e^4 + (c^4*d^5*e^4 + 2*a*c^3*d^3*e^6 + a^2*c^2*d*e^8)*x^4
+ 4*(c^4*d^6*e^3 + 2*a*c^3*d^4*e^5 + a^2*c^2*d^2*e^7)*x^3 + 6*(c^4*d^7*e^2 + 2*a*c^3*d^5*e^4 + a^2*c^2*d^3*e^6
)*x^2 + 4*(c^4*d^8*e + 2*a*c^3*d^6*e^3 + a^2*c^2*d^4*e^5)*x)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*
x - a) + 15*(8*c^4*d^8 + 12*a*c^3*d^6*e^2 + 3*a^2*c^2*d^4*e^4 + (8*c^4*d^4*e^4 + 12*a*c^3*d^2*e^6 + 3*a^2*c^2*
e^8)*x^4 + 4*(8*c^4*d^5*e^3 + 12*a*c^3*d^3*e^5 + 3*a^2*c^2*d*e^7)*x^3 + 6*(8*c^4*d^6*e^2 + 12*a*c^3*d^4*e^4 +
3*a^2*c^2*d^2*e^6)*x^2 + 4*(8*c^4*d^7*e + 12*a*c^3*d^5*e^3 + 3*a^2*c^2*d^3*e^5)*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*(120*c^4*d^8*e + 220*a*c^3*d^6*e^3 + 89*a^2*c^2*d^4*e^5 - 17*a^3*c*d^2*e^7
- 6*a^4*e^9 + 24*(c^4*d^4*e^5 + 2*a*c^3*d^2*e^7 + a^2*c^2*e^9)*x^4 + 5*(50*c^4*d^5*e^4 + 97*a*c^3*d^3*e^6 + 4
7*a^2*c^2*d*e^8)*x^3 + (520*c^4*d^6*e^3 + 968*a*c^3*d^4*e^5 + 421*a^2*c^2*d^2*e^7 - 27*a^3*c*e^9)*x^2 + 5*(84*
c^4*d^7*e^2 + 155*a*c^3*d^5*e^4 + 67*a^2*c^2*d^3*e^6 - 4*a^3*c*d*e^8)*x)*sqrt(c*x^2 + a))/(c^2*d^8*e^6 + 2*a*c
*d^6*e^8 + a^2*d^4*e^10 + (c^2*d^4*e^10 + 2*a*c*d^2*e^12 + a^2*e^14)*x^4 + 4*(c^2*d^5*e^9 + 2*a*c*d^3*e^11 + a
^2*d*e^13)*x^3 + 6*(c^2*d^6*e^8 + 2*a*c*d^4*e^10 + a^2*d^2*e^12)*x^2 + 4*(c^2*d^7*e^7 + 2*a*c*d^5*e^9 + a^2*d^
3*e^11)*x), 1/48*(240*(c^4*d^9 + 2*a*c^3*d^7*e^2 + a^2*c^2*d^5*e^4 + (c^4*d^5*e^4 + 2*a*c^3*d^3*e^6 + a^2*c^2*
d*e^8)*x^4 + 4*(c^4*d^6*e^3 + 2*a*c^3*d^4*e^5 + a^2*c^2*d^2*e^7)*x^3 + 6*(c^4*d^7*e^2 + 2*a*c^3*d^5*e^4 + a^2*
c^2*d^3*e^6)*x^2 + 4*(c^4*d^8*e + 2*a*c^3*d^6*e^3 + a^2*c^2*d^4*e^5)*x)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2
+ a)) + 15*(8*c^4*d^8 + 12*a*c^3*d^6*e^2 + 3*a^2*c^2*d^4*e^4 + (8*c^4*d^4*e^4 + 12*a*c^3*d^2*e^6 + 3*a^2*c^2*e
^8)*x^4 + 4*(8*c^4*d^5*e^3 + 12*a*c^3*d^3*e^5 + 3*a^2*c^2*d*e^7)*x^3 + 6*(8*c^4*d^6*e^2 + 12*a*c^3*d^4*e^4 + 3
*a^2*c^2*d^2*e^6)*x^2 + 4*(8*c^4*d^7*e + 12*a*c^3*d^5*e^3 + 3*a^2*c^2*d^3*e^5)*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*(120*c^4*d^8*e + 220*a*c^3*d^6*e^3 + 89*a^2*c^2*d^4*e^5 - 17*a^3*c*d^2*e^7
- 6*a^4*e^9 + 24*(c^4*d^4*e^5 + 2*a*c^3*d^2*e^7 + a^2*c^2*e^9)*x^4 + 5*(50*c^4*d^5*e^4 + 97*a*c^3*d^3*e^6 + 47
*a^2*c^2*d*e^8)*x^3 + (520*c^4*d^6*e^3 + 968*a*c^3*d^4*e^5 + 421*a^2*c^2*d^2*e^7 - 27*a^3*c*e^9)*x^2 + 5*(84*c
^4*d^7*e^2 + 155*a*c^3*d^5*e^4 + 67*a^2*c^2*d^3*e^6 - 4*a^3*c*d*e^8)*x)*sqrt(c*x^2 + a))/(c^2*d^8*e^6 + 2*a*c*
d^6*e^8 + a^2*d^4*e^10 + (c^2*d^4*e^10 + 2*a*c*d^2*e^12 + a^2*e^14)*x^4 + 4*(c^2*d^5*e^9 + 2*a*c*d^3*e^11 + a^
2*d*e^13)*x^3 + 6*(c^2*d^6*e^8 + 2*a*c*d^4*e^10 + a^2*d^2*e^12)*x^2 + 4*(c^2*d^7*e^7 + 2*a*c*d^5*e^9 + a^2*d^3
*e^11)*x), -1/24*(15*(8*c^4*d^8 + 12*a*c^3*d^6*e^2 + 3*a^2*c^2*d^4*e^4 + (8*c^4*d^4*e^4 + 12*a*c^3*d^2*e^6 + 3
*a^2*c^2*e^8)*x^4 + 4*(8*c^4*d^5*e^3 + 12*a*c^3*d^3*e^5 + 3*a^2*c^2*d*e^7)*x^3 + 6*(8*c^4*d^6*e^2 + 12*a*c^3*d
^4*e^4 + 3*a^2*c^2*d^2*e^6)*x^2 + 4*(8*c^4*d^7*e + 12*a*c^3*d^5*e^3 + 3*a^2*c^2*d^3*e^5)*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)) -
60*(c^4*d^9 + 2*a*c^3*d^7*e^2 + a^2*c^2*d^5*e^4 + (c^4*d^5*e^4 + 2*a*c^3*d^3*e^6 + a^2*c^2*d*e^8)*x^4 + 4*(c^4
*d^6*e^3 + 2*a*c^3*d^4*e^5 + a^2*c^2*d^2*e^7)*x^3 + 6*(c^4*d^7*e^2 + 2*a*c^3*d^5*e^4 + a^2*c^2*d^3*e^6)*x^2 +
4*(c^4*d^8*e + 2*a*c^3*d^6*e^3 + a^2*c^2*d^4*e^5)*x)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) -
(120*c^4*d^8*e + 220*a*c^3*d^6*e^3 + 89*a^2*c^2*d^4*e^5 - 17*a^3*c*d^2*e^7 - 6*a^4*e^9 + 24*(c^4*d^4*e^5 + 2*
a*c^3*d^2*e^7 + a^2*c^2*e^9)*x^4 + 5*(50*c^4*d^5*e^4 + 97*a*c^3*d^3*e^6 + 47*a^2*c^2*d*e^8)*x^3 + (520*c^4*d^6
*e^3 + 968*a*c^3*d^4*e^5 + 421*a^2*c^2*d^2*e^7 - 27*a^3*c*e^9)*x^2 + 5*(84*c^4*d^7*e^2 + 155*a*c^3*d^5*e^4 + 6
7*a^2*c^2*d^3*e^6 - 4*a^3*c*d*e^8)*x)*sqrt(c*x^2 + a))/(c^2*d^8*e^6 + 2*a*c*d^6*e^8 + a^2*d^4*e^10 + (c^2*d^4*
e^10 + 2*a*c*d^2*e^12 + a^2*e^14)*x^4 + 4*(c^2*d^5*e^9 + 2*a*c*d^3*e^11 + a^2*d*e^13)*x^3 + 6*(c^2*d^6*e^8 + 2
*a*c*d^4*e^10 + a^2*d^2*e^12)*x^2 + 4*(c^2*d^7*e^7 + 2*a*c*d^5*e^9 + a^2*d^3*e^11)*x), -1/24*(15*(8*c^4*d^8 +
12*a*c^3*d^6*e^2 + 3*a^2*c^2*d^4*e^4 + (8*c^4*d^4*e^4 + 12*a*c^3*d^2*e^6 + 3*a^2*c^2*e^8)*x^4 + 4*(8*c^4*d^5*e
^3 + 12*a*c^3*d^3*e^5 + 3*a^2*c^2*d*e^7)*x^3 + 6*(8*c^4*d^6*e^2 + 12*a*c^3*d^4*e^4 + 3*a^2*c^2*d^2*e^6)*x^2 +
4*(8*c^4*d^7*e + 12*a*c^3*d^5*e^3 + 3*a^2*c^2*d^3*e^5)*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)) - 120*(c^4*d^9 + 2*a*c^3*d^7*e^2 + a
^2*c^2*d^5*e^4 + (c^4*d^5*e^4 + 2*a*c^3*d^3*e^6 + a^2*c^2*d*e^8)*x^4 + 4*(c^4*d^6*e^3 + 2*a*c^3*d^4*e^5 + a^2*
c^2*d^2*e^7)*x^3 + 6*(c^4*d^7*e^2 + 2*a*c^3*d^5*e^4 + a^2*c^2*d^3*e^6)*x^2 + 4*(c^4*d^8*e + 2*a*c^3*d^6*e^3 +
a^2*c^2*d^4*e^5)*x)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (120*c^4*d^8*e + 220*a*c^3*d^6*e^3 + 89*a^2*
c^2*d^4*e^5 - 17*a^3*c*d^2*e^7 - 6*a^4*e^9 + 24*(c^4*d^4*e^5 + 2*a*c^3*d^2*e^7 + a^2*c^2*e^9)*x^4 + 5*(50*c^4*
d^5*e^4 + 97*a*c^3*d^3*e^6 + 47*a^2*c^2*d*e^8)*x^3 + (520*c^4*d^6*e^3 + 968*a*c^3*d^4*e^5 + 421*a^2*c^2*d^2*e^
7 - 27*a^3*c*e^9)*x^2 + 5*(84*c^4*d^7*e^2 + 155*a*c^3*d^5*e^4 + 67*a^2*c^2*d^3*e^6 - 4*a^3*c*d*e^8)*x)*sqrt(c*
x^2 + a))/(c^2*d^8*e^6 + 2*a*c*d^6*e^8 + a^2*d^4*e^10 + (c^2*d^4*e^10 + 2*a*c*d^2*e^12 + a^2*e^14)*x^4 + 4*(c^
2*d^5*e^9 + 2*a*c*d^3*e^11 + a^2*d*e^13)*x^3 + 6*(c^2*d^6*e^8 + 2*a*c*d^4*e^10 + a^2*d^2*e^12)*x^2 + 4*(c^2*d^
7*e^7 + 2*a*c*d^5*e^9 + a^2*d^3*e^11)*x)]

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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 )^{5}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError