3.551 \(\int \frac{(a+c x^2)^{5/2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=213 \[ -\frac{5 c^{3/2} d \left (3 a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 e^6}+\frac{5 c \sqrt{a+c x^2} \left (a e^2+4 c d^2-2 c d e x\right )}{2 e^5}-\frac{5 c \sqrt{a e^2+c d^2} \left (a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 e^6}+\frac{5 c \left (a+c x^2\right )^{3/2} (4 d+e x)}{6 e^3 (d+e x)}-\frac{\left (a+c x^2\right )^{5/2}}{2 e (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.238605, antiderivative size = 213, 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, 813, 815, 844, 217, 206, 725} \[ -\frac{5 c^{3/2} d \left (3 a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 e^6}+\frac{5 c \sqrt{a+c x^2} \left (a e^2+4 c d^2-2 c d e x\right )}{2 e^5}-\frac{5 c \sqrt{a e^2+c d^2} \left (a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 e^6}+\frac{5 c \left (a+c x^2\right )^{3/2} (4 d+e x)}{6 e^3 (d+e x)}-\frac{\left (a+c x^2\right )^{5/2}}{2 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^(5/2)/(d + e*x)^3,x]

[Out]

(5*c*(4*c*d^2 + a*e^2 - 2*c*d*e*x)*Sqrt[a + c*x^2])/(2*e^5) + (5*c*(4*d + e*x)*(a + c*x^2)^(3/2))/(6*e^3*(d +
e*x)) - (a + c*x^2)^(5/2)/(2*e*(d + e*x)^2) - (5*c^(3/2)*d*(4*c*d^2 + 3*a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*
x^2]])/(2*e^6) - (5*c*Sqrt[c*d^2 + a*e^2]*(4*c*d^2 + a*e^2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a
+ c*x^2])])/(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 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 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)^3} \, dx &=-\frac{\left (a+c x^2\right )^{5/2}}{2 e (d+e x)^2}+\frac{(5 c) \int \frac{x \left (a+c x^2\right )^{3/2}}{(d+e x)^2} \, dx}{2 e}\\ &=\frac{5 c (4 d+e x) \left (a+c x^2\right )^{3/2}}{6 e^3 (d+e x)}-\frac{\left (a+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{(5 c) \int \frac{(-2 a e+8 c d x) \sqrt{a+c x^2}}{d+e x} \, dx}{4 e^3}\\ &=\frac{5 c \left (4 c d^2+a e^2-2 c d e x\right ) \sqrt{a+c x^2}}{2 e^5}+\frac{5 c (4 d+e x) \left (a+c x^2\right )^{3/2}}{6 e^3 (d+e x)}-\frac{\left (a+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{5 \int \frac{-4 a c e \left (2 c d^2+a e^2\right )+4 c^2 d \left (4 c d^2+3 a e^2\right ) x}{(d+e x) \sqrt{a+c x^2}} \, dx}{8 e^5}\\ &=\frac{5 c \left (4 c d^2+a e^2-2 c d e x\right ) \sqrt{a+c x^2}}{2 e^5}+\frac{5 c (4 d+e x) \left (a+c x^2\right )^{3/2}}{6 e^3 (d+e x)}-\frac{\left (a+c x^2\right )^{5/2}}{2 e (d+e x)^2}+\frac{\left (5 c \left (c d^2+a e^2\right ) \left (4 c d^2+a e^2\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{2 e^6}-\frac{\left (5 c^2 d \left (4 c d^2+3 a e^2\right )\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{2 e^6}\\ &=\frac{5 c \left (4 c d^2+a e^2-2 c d e x\right ) \sqrt{a+c x^2}}{2 e^5}+\frac{5 c (4 d+e x) \left (a+c x^2\right )^{3/2}}{6 e^3 (d+e x)}-\frac{\left (a+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{\left (5 c \left (c d^2+a e^2\right ) \left (4 c d^2+a e^2\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 )}{2 e^6}-\frac{\left (5 c^2 d \left (4 c d^2+3 a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{2 e^6}\\ &=\frac{5 c \left (4 c d^2+a e^2-2 c d e x\right ) \sqrt{a+c x^2}}{2 e^5}+\frac{5 c (4 d+e x) \left (a+c x^2\right )^{3/2}}{6 e^3 (d+e x)}-\frac{\left (a+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{5 c^{3/2} d \left (4 c d^2+3 a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 e^6}-\frac{5 c \sqrt{c d^2+a e^2} \left (4 c d^2+a e^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{2 e^6}\\ \end{align*}

Mathematica [A]  time = 0.302595, size = 281, normalized size = 1.32 \[ \frac{\frac{e \sqrt{a+c x^2} \left (-3 a^2 e^4+a c e^2 \left (35 d^2+55 d e x+14 e^2 x^2\right )+c^2 \left (20 d^2 e^2 x^2+90 d^3 e x+60 d^4-5 d e^3 x^3+2 e^4 x^4\right )\right )}{(d+e x)^2}-\frac{15 c \left (a^2 e^4+5 a c d^2 e^2+4 c^2 d^4\right ) \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )}{\sqrt{a e^2+c d^2}}+\frac{15 c \left (a^2 e^4+5 a c d^2 e^2+4 c^2 d^4\right ) \log (d+e x)}{\sqrt{a e^2+c d^2}}-15 c^{3/2} d \left (3 a e^2+4 c d^2\right ) \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{6 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Sqrt[a + c*x^2]*(-3*a^2*e^4 + a*c*e^2*(35*d^2 + 55*d*e*x + 14*e^2*x^2) + c^2*(60*d^4 + 90*d^3*e*x + 20*d^2
*e^2*x^2 - 5*d*e^3*x^3 + 2*e^4*x^4)))/(d + e*x)^2 + (15*c*(4*c^2*d^4 + 5*a*c*d^2*e^2 + a^2*e^4)*Log[d + e*x])/
Sqrt[c*d^2 + a*e^2] - 15*c^(3/2)*d*(4*c*d^2 + 3*a*e^2)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]] - (15*c*(4*c^2*d^4 +
5*a*c*d^2*e^2 + a^2*e^4)*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/Sqrt[c*d^2 + a*e^2])/(6*e^6)

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Maple [B]  time = 0.195, size = 3342, normalized size = 15.7 \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)^3,x)

[Out]

5/6/e^3/(a*e^2+c*d^2)*c^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*d^2+3/2/e*c^2*d^2/(a*e^2+c*d^2
)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(5/2)+5/2/e^3*c^3*d^4/(a*e^2+c*d^2)^2*(c*(d/e+x)^2-2*c*d/e
*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)+15/2/e^5*c^4*d^6/(a*e^2+c*d^2)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/
e^2)^(1/2)-3/2*c^2*d/(a*e^2+c*d^2)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(5/2)*x-45/16*c^(3/2)*d/(
a*e^2+c*d^2)^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))+5/2/e/
(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)^(1/2)*a^2+3/2*c*d/(a*e^2+c*d^2)^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)+5/2/e^5/(a*e^2+c*d^2)*c^3*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^
2+c*d^2)/e^2)^(1/2)*d^4-5/2/e^6/(a*e^2+c*d^2)*c^(7/2)*d^5*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))-15/2/e^6*c^(9/2)*d^7/(a*e^2+c*d^2)^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))-25/4/e^4/(a*e^2+c*d^2)*c^(5/2)*d^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))*a-45/16*c^2*d/(a*e^2+c*d^2)^2*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+15/2/e*c^2*d^2/(a*e^2+c*d^2)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e
^2)^(1/2)*a^2-15/4/e^4*c^4*d^5/(a*e^2+c*d^2)^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*c^
2*d/(a*e^2+c*d^2)^2*a*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*x-225/16/e^2*c^(5/2)*d^3/(a*e^2+c*
d^2)^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^2-75/4/e^4*c^(7/
2)*d^5/(a*e^2+c*d^2)^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-
15/8/e^2*c^3*d^3/(a*e^2+c*d^2)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*x+5/2/e*c^2*d^2/(a*e^2+
c*d^2)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*a-75/16/e^2/(a*e^2+c*d^2)*c^(3/2)*d*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+15/e^3*c^3*d^4/(a*e^2+c*d^2)^2*(
c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*a-15/2/e^7*c^5*d^8/(a*e^2+c*d^2)^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))+1/2/e/(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)-1/2/e/(a
*e^2+c*d^2)/(d/e+x)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(7/2)+5/6/e/(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)^(3/2)*a+5/e^3/(a*e^2+c*d^2)*c^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)
/e^2)^(1/2)*a*d^2-5/2/e/(a*e^2+c*d^2)*c/((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^3-5/2/e^7/(a*e^2+c*d
^2)*c^4/((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))*d^6-5/8/e^2/(a*e^2+c*d^2)*c^2*d*(c*(d/e+x)^2-2*c*d/e*(
d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*x-5/4/e^4/(a*e^2+c*d^2)*c^3*d^3*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2
)^(1/2)*x-105/16/e^2*c^3*d^3/(a*e^2+c*d^2)^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-15/2/e*
c^2*d^2/(a*e^2+c*d^2)^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^3-45/2/e^3*c^3*d^4/(a*e^2+c*d^2)^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^2-45/2/e^5*c^4*d^6/(a*e^2+c*d^2)^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-35/16/e^2/(a*e^2+c*d^2)*c^2*d*a*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*
x-15/2/e^3/(a*e^2+c*d^2)*c^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^2*d^2-15/2/e^5/(a*e^2+c*d^2)*c^
3/((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*d^4

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 15.4451, size = 3295, normalized size = 15.47 \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)^3,x, algorithm="fricas")

[Out]

[1/12*(15*(4*c^2*d^5 + 3*a*c*d^3*e^2 + (4*c^2*d^3*e^2 + 3*a*c*d*e^4)*x^2 + 2*(4*c^2*d^4*e + 3*a*c*d^2*e^3)*x)*
sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 15*(4*c^2*d^4 + a*c*d^2*e^2 + (4*c^2*d^2*e^2 + a*c*e
^4)*x^2 + 2*(4*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*(2*c^2
*e^5*x^4 - 5*c^2*d*e^4*x^3 + 60*c^2*d^4*e + 35*a*c*d^2*e^3 - 3*a^2*e^5 + 2*(10*c^2*d^2*e^3 + 7*a*c*e^5)*x^2 +
5*(18*c^2*d^3*e^2 + 11*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(e^8*x^2 + 2*d*e^7*x + d^2*e^6), 1/12*(30*(4*c^2*d^5 + 3
*a*c*d^3*e^2 + (4*c^2*d^3*e^2 + 3*a*c*d*e^4)*x^2 + 2*(4*c^2*d^4*e + 3*a*c*d^2*e^3)*x)*sqrt(-c)*arctan(sqrt(-c)
*x/sqrt(c*x^2 + a)) + 15*(4*c^2*d^4 + a*c*d^2*e^2 + (4*c^2*d^2*e^2 + a*c*e^4)*x^2 + 2*(4*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*(2*c^2*e^5*x^4 - 5*c^2*d*e^4*x^3 + 60*c^2*
d^4*e + 35*a*c*d^2*e^3 - 3*a^2*e^5 + 2*(10*c^2*d^2*e^3 + 7*a*c*e^5)*x^2 + 5*(18*c^2*d^3*e^2 + 11*a*c*d*e^4)*x)
*sqrt(c*x^2 + a))/(e^8*x^2 + 2*d*e^7*x + d^2*e^6), -1/12*(30*(4*c^2*d^4 + a*c*d^2*e^2 + (4*c^2*d^2*e^2 + a*c*e
^4)*x^2 + 2*(4*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*(4*c^2*d^5 + 3*a*c*d^3*e^2 + (4*c^2*d^3*e^2 + 3*
a*c*d*e^4)*x^2 + 2*(4*c^2*d^4*e + 3*a*c*d^2*e^3)*x)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) -
2*(2*c^2*e^5*x^4 - 5*c^2*d*e^4*x^3 + 60*c^2*d^4*e + 35*a*c*d^2*e^3 - 3*a^2*e^5 + 2*(10*c^2*d^2*e^3 + 7*a*c*e^5
)*x^2 + 5*(18*c^2*d^3*e^2 + 11*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(e^8*x^2 + 2*d*e^7*x + d^2*e^6), -1/6*(15*(4*c^2
*d^4 + a*c*d^2*e^2 + (4*c^2*d^2*e^2 + a*c*e^4)*x^2 + 2*(4*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*(4*c^
2*d^5 + 3*a*c*d^3*e^2 + (4*c^2*d^3*e^2 + 3*a*c*d*e^4)*x^2 + 2*(4*c^2*d^4*e + 3*a*c*d^2*e^3)*x)*sqrt(-c)*arctan
(sqrt(-c)*x/sqrt(c*x^2 + a)) - (2*c^2*e^5*x^4 - 5*c^2*d*e^4*x^3 + 60*c^2*d^4*e + 35*a*c*d^2*e^3 - 3*a^2*e^5 +
2*(10*c^2*d^2*e^3 + 7*a*c*e^5)*x^2 + 5*(18*c^2*d^3*e^2 + 11*a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(e^8*x^2 + 2*d*e^7*
x + d^2*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 )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + c*x**2)**(5/2)/(d + e*x)**3, x)

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Giac [B]  time = 2.05407, size = 703, normalized size = 3.3 \begin{align*} \frac{5}{2} \,{\left (4 \, c^{\frac{5}{2}} d^{3} + 3 \, a c^{\frac{3}{2}} d e^{2}\right )} e^{\left (-6\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right ) + \frac{5 \,{\left (4 \, c^{3} d^{4} + 5 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} \arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} - a e^{2}}}\right ) e^{\left (-6\right )}}{\sqrt{-c d^{2} - a e^{2}}} + \frac{1}{6} \, \sqrt{c x^{2} + a}{\left ({\left (2 \, c^{2} x e^{\left (-3\right )} - 9 \, c^{2} d e^{\left (-4\right )}\right )} x + \frac{2 \,{\left (18 \, c^{3} d^{2} e^{13} + 7 \, a c^{2} e^{15}\right )} e^{\left (-18\right )}}{c}\right )} + \frac{{\left (10 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} c^{3} d^{4} e + 18 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} c^{\frac{7}{2}} d^{5} - 26 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a c^{3} d^{4} e + 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a c^{\frac{5}{2}} d^{3} e^{2} + 11 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} a c^{2} d^{2} e^{3} + 9 \, a^{2} c^{\frac{5}{2}} d^{3} e^{2} - 25 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a^{2} c^{2} d^{2} e^{3} - 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a^{2} c^{\frac{3}{2}} d e^{4} +{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} a^{2} c e^{5} + 9 \, a^{3} c^{\frac{3}{2}} d e^{4} +{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a^{3} c e^{5}\right )} e^{\left (-6\right )}}{{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} e + 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} \sqrt{c} d - a e\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*(4*c^(5/2)*d^3 + 3*a*c^(3/2)*d*e^2)*e^(-6)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a))) + 5*(4*c^3*d^4 + 5*a*c^2
*d^2*e^2 + a^2*c*e^4)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-6)/sqrt(
-c*d^2 - a*e^2) + 1/6*sqrt(c*x^2 + a)*((2*c^2*x*e^(-3) - 9*c^2*d*e^(-4))*x + 2*(18*c^3*d^2*e^13 + 7*a*c^2*e^15
)*e^(-18)/c) + (10*(sqrt(c)*x - sqrt(c*x^2 + a))^3*c^3*d^4*e + 18*(sqrt(c)*x - sqrt(c*x^2 + a))^2*c^(7/2)*d^5
- 26*(sqrt(c)*x - sqrt(c*x^2 + a))*a*c^3*d^4*e + 9*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a*c^(5/2)*d^3*e^2 + 11*(sqr
t(c)*x - sqrt(c*x^2 + a))^3*a*c^2*d^2*e^3 + 9*a^2*c^(5/2)*d^3*e^2 - 25*(sqrt(c)*x - sqrt(c*x^2 + a))*a^2*c^2*d
^2*e^3 - 9*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a^2*c^(3/2)*d*e^4 + (sqrt(c)*x - sqrt(c*x^2 + a))^3*a^2*c*e^5 + 9*a
^3*c^(3/2)*d*e^4 + (sqrt(c)*x - sqrt(c*x^2 + a))*a^3*c*e^5)*e^(-6)/((sqrt(c)*x - sqrt(c*x^2 + a))^2*e + 2*(sqr
t(c)*x - sqrt(c*x^2 + a))*sqrt(c)*d - a*e)^2