3.645 \(\int \frac{1}{x^3 (a+b \sqrt{c+d x})^2} \, dx\)

Optimal. Leaf size=306 \[ -\frac{a b d^2 \left (-10 a^2 b^2 c+a^4-15 b^4 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{2 c^{3/2} \left (a^2-b^2 c\right )^4}+\frac{a b^2 d^2 \left (a^2+11 b^2 c\right )}{2 c \left (a^2-b^2 c\right )^3 \left (a+b \sqrt{c+d x}\right )}+\frac{b^4 d^2 \log (x) \left (5 a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^4}-\frac{2 b^4 d^2 \left (5 a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^4}-\frac{a-b \sqrt{c+d x}}{2 x^2 \left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}-\frac{b d \left (3 a b c-\left (a^2+2 b^2 c\right ) \sqrt{c+d x}\right )}{2 c x \left (a^2-b^2 c\right )^2 \left (a+b \sqrt{c+d x}\right )} \]

[Out]

(a*b^2*(a^2 + 11*b^2*c)*d^2)/(2*c*(a^2 - b^2*c)^3*(a + b*Sqrt[c + d*x])) - (a - b*Sqrt[c + d*x])/(2*(a^2 - b^2
*c)*x^2*(a + b*Sqrt[c + d*x])) - (b*d*(3*a*b*c - (a^2 + 2*b^2*c)*Sqrt[c + d*x]))/(2*c*(a^2 - b^2*c)^2*x*(a + b
*Sqrt[c + d*x])) - (a*b*(a^4 - 10*a^2*b^2*c - 15*b^4*c^2)*d^2*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(2*c^(3/2)*(a^2
- b^2*c)^4) + (b^4*(5*a^2 + b^2*c)*d^2*Log[x])/(a^2 - b^2*c)^4 - (2*b^4*(5*a^2 + b^2*c)*d^2*Log[a + b*Sqrt[c +
d*x]])/(a^2 - b^2*c)^4

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Rubi [A]  time = 0.402982, antiderivative size = 306, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {371, 1398, 823, 801, 635, 206, 260} \[ -\frac{a b d^2 \left (-10 a^2 b^2 c+a^4-15 b^4 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{2 c^{3/2} \left (a^2-b^2 c\right )^4}+\frac{a b^2 d^2 \left (a^2+11 b^2 c\right )}{2 c \left (a^2-b^2 c\right )^3 \left (a+b \sqrt{c+d x}\right )}+\frac{b^4 d^2 \log (x) \left (5 a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^4}-\frac{2 b^4 d^2 \left (5 a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^4}-\frac{a-b \sqrt{c+d x}}{2 x^2 \left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}-\frac{b d \left (3 a b c-\left (a^2+2 b^2 c\right ) \sqrt{c+d x}\right )}{2 c x \left (a^2-b^2 c\right )^2 \left (a+b \sqrt{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*Sqrt[c + d*x])^2),x]

[Out]

(a*b^2*(a^2 + 11*b^2*c)*d^2)/(2*c*(a^2 - b^2*c)^3*(a + b*Sqrt[c + d*x])) - (a - b*Sqrt[c + d*x])/(2*(a^2 - b^2
*c)*x^2*(a + b*Sqrt[c + d*x])) - (b*d*(3*a*b*c - (a^2 + 2*b^2*c)*Sqrt[c + d*x]))/(2*c*(a^2 - b^2*c)^2*x*(a + b
*Sqrt[c + d*x])) - (a*b*(a^4 - 10*a^2*b^2*c - 15*b^4*c^2)*d^2*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(2*c^(3/2)*(a^2
- b^2*c)^4) + (b^4*(5*a^2 + b^2*c)*d^2*Log[x])/(a^2 - b^2*c)^4 - (2*b^4*(5*a^2 + b^2*c)*d^2*Log[a + b*Sqrt[c +
d*x]])/(a^2 - b^2*c)^4

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

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

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (a+b \sqrt{c+d x}\right )^2} \, dx &=d^2 \operatorname{Subst}\left (\int \frac{1}{\left (a+b \sqrt{x}\right )^2 (-c+x)^3} \, dx,x,c+d x\right )\\ &=\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{x}{(a+b x)^2 \left (-c+x^2\right )^3} \, dx,x,\sqrt{c+d x}\right )\\ &=-\frac{a-b \sqrt{c+d x}}{2 \left (a^2-b^2 c\right ) x^2 \left (a+b \sqrt{c+d x}\right )}+\frac{d^2 \operatorname{Subst}\left (\int \frac{-2 a b c+4 b^2 c x}{(a+b x)^2 \left (-c+x^2\right )^2} \, dx,x,\sqrt{c+d x}\right )}{2 c \left (a^2-b^2 c\right )}\\ &=-\frac{a-b \sqrt{c+d x}}{2 \left (a^2-b^2 c\right ) x^2 \left (a+b \sqrt{c+d x}\right )}-\frac{b d \left (3 a b c-\left (a^2+2 b^2 c\right ) \sqrt{c+d x}\right )}{2 c \left (a^2-b^2 c\right )^2 x \left (a+b \sqrt{c+d x}\right )}+\frac{d^2 \operatorname{Subst}\left (\int \frac{2 a b c \left (a^2-7 b^2 c\right )+4 b^2 c \left (a^2+2 b^2 c\right ) x}{(a+b x)^2 \left (-c+x^2\right )} \, dx,x,\sqrt{c+d x}\right )}{4 c^2 \left (a^2-b^2 c\right )^2}\\ &=-\frac{a-b \sqrt{c+d x}}{2 \left (a^2-b^2 c\right ) x^2 \left (a+b \sqrt{c+d x}\right )}-\frac{b d \left (3 a b c-\left (a^2+2 b^2 c\right ) \sqrt{c+d x}\right )}{2 c \left (a^2-b^2 c\right )^2 x \left (a+b \sqrt{c+d x}\right )}+\frac{d^2 \operatorname{Subst}\left (\int \left (-\frac{2 a b^3 c \left (a^2+11 b^2 c\right )}{\left (a^2-b^2 c\right ) (a+b x)^2}-\frac{8 b^5 c^2 \left (5 a^2+b^2 c\right )}{\left (-a^2+b^2 c\right )^2 (a+b x)}+\frac{2 b c \left (-a \left (a^4-10 a^2 b^2 c-15 b^4 c^2\right )-4 b^3 c \left (5 a^2+b^2 c\right ) x\right )}{\left (a^2-b^2 c\right )^2 \left (c-x^2\right )}\right ) \, dx,x,\sqrt{c+d x}\right )}{4 c^2 \left (a^2-b^2 c\right )^2}\\ &=\frac{a b^2 \left (a^2+11 b^2 c\right ) d^2}{2 c \left (a^2-b^2 c\right )^3 \left (a+b \sqrt{c+d x}\right )}-\frac{a-b \sqrt{c+d x}}{2 \left (a^2-b^2 c\right ) x^2 \left (a+b \sqrt{c+d x}\right )}-\frac{b d \left (3 a b c-\left (a^2+2 b^2 c\right ) \sqrt{c+d x}\right )}{2 c \left (a^2-b^2 c\right )^2 x \left (a+b \sqrt{c+d x}\right )}-\frac{2 b^4 \left (5 a^2+b^2 c\right ) d^2 \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^4}+\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{-a \left (a^4-10 a^2 b^2 c-15 b^4 c^2\right )-4 b^3 c \left (5 a^2+b^2 c\right ) x}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{2 c \left (a^2-b^2 c\right )^4}\\ &=\frac{a b^2 \left (a^2+11 b^2 c\right ) d^2}{2 c \left (a^2-b^2 c\right )^3 \left (a+b \sqrt{c+d x}\right )}-\frac{a-b \sqrt{c+d x}}{2 \left (a^2-b^2 c\right ) x^2 \left (a+b \sqrt{c+d x}\right )}-\frac{b d \left (3 a b c-\left (a^2+2 b^2 c\right ) \sqrt{c+d x}\right )}{2 c \left (a^2-b^2 c\right )^2 x \left (a+b \sqrt{c+d x}\right )}-\frac{2 b^4 \left (5 a^2+b^2 c\right ) d^2 \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^4}-\frac{\left (2 b^4 \left (5 a^2+b^2 c\right ) d^2\right ) \operatorname{Subst}\left (\int \frac{x}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^4}-\frac{\left (a b \left (a^4-10 a^2 b^2 c-15 b^4 c^2\right ) d^2\right ) \operatorname{Subst}\left (\int \frac{1}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{2 c \left (a^2-b^2 c\right )^4}\\ &=\frac{a b^2 \left (a^2+11 b^2 c\right ) d^2}{2 c \left (a^2-b^2 c\right )^3 \left (a+b \sqrt{c+d x}\right )}-\frac{a-b \sqrt{c+d x}}{2 \left (a^2-b^2 c\right ) x^2 \left (a+b \sqrt{c+d x}\right )}-\frac{b d \left (3 a b c-\left (a^2+2 b^2 c\right ) \sqrt{c+d x}\right )}{2 c \left (a^2-b^2 c\right )^2 x \left (a+b \sqrt{c+d x}\right )}-\frac{a b \left (a^4-10 a^2 b^2 c-15 b^4 c^2\right ) d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{2 c^{3/2} \left (a^2-b^2 c\right )^4}+\frac{b^4 \left (5 a^2+b^2 c\right ) d^2 \log (x)}{\left (a^2-b^2 c\right )^4}-\frac{2 b^4 \left (5 a^2+b^2 c\right ) d^2 \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^4}\\ \end{align*}

Mathematica [A]  time = 0.909832, size = 401, normalized size = 1.31 \[ \frac{\frac{d^2 \left (\frac{2 b \sqrt{c} \left (a^2+2 b^2 c\right ) \left (\left (b \sqrt{c}-a\right ) \log \left (\sqrt{c}-\sqrt{c+d x}\right )+\left (a+b \sqrt{c}\right ) \log \left (\sqrt{c+d x}+\sqrt{c}\right )-2 b \sqrt{c} \log \left (a+b \sqrt{c+d x}\right )\right )}{b^2 c-a^2}-a b c \left (a^2+11 b^2 c\right ) \left (\frac{2 b \left (\frac{b^2 c-a^2}{a+b \sqrt{c+d x}}+2 a \log \left (a+b \sqrt{c+d x}\right )\right )}{\left (a^2-b^2 c\right )^2}+\frac{\log \left (\sqrt{c}-\sqrt{c+d x}\right )}{\sqrt{c} \left (a+b \sqrt{c}\right )^2}-\frac{\log \left (\sqrt{c+d x}+\sqrt{c}\right )}{\sqrt{c} \left (a-b \sqrt{c}\right )^2}\right )\right )}{2 c \left (a^2-b^2 c\right )}+\frac{b d \left (a^2 \sqrt{c+d x}-3 a b c+2 b^2 c \sqrt{c+d x}\right )}{x \left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}-\frac{c \left (a-b \sqrt{c+d x}\right )}{x^2 \left (a+b \sqrt{c+d x}\right )}}{2 c \left (a^2-b^2 c\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*Sqrt[c + d*x])^2),x]

[Out]

(-((c*(a - b*Sqrt[c + d*x]))/(x^2*(a + b*Sqrt[c + d*x]))) + (b*d*(-3*a*b*c + a^2*Sqrt[c + d*x] + 2*b^2*c*Sqrt[
c + d*x]))/((a^2 - b^2*c)*x*(a + b*Sqrt[c + d*x])) + (d^2*((2*b*Sqrt[c]*(a^2 + 2*b^2*c)*((-a + b*Sqrt[c])*Log[
Sqrt[c] - Sqrt[c + d*x]] + (a + b*Sqrt[c])*Log[Sqrt[c] + Sqrt[c + d*x]] - 2*b*Sqrt[c]*Log[a + b*Sqrt[c + d*x]]
))/(-a^2 + b^2*c) - a*b*c*(a^2 + 11*b^2*c)*(Log[Sqrt[c] - Sqrt[c + d*x]]/((a + b*Sqrt[c])^2*Sqrt[c]) - Log[Sqr
t[c] + Sqrt[c + d*x]]/((a - b*Sqrt[c])^2*Sqrt[c]) + (2*b*((-a^2 + b^2*c)/(a + b*Sqrt[c + d*x]) + 2*a*Log[a + b
*Sqrt[c + d*x]]))/(a^2 - b^2*c)^2)))/(2*c*(a^2 - b^2*c)))/(2*c*(a^2 - b^2*c))

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Maple [B]  time = 0.018, size = 610, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b*(d*x+c)^(1/2))^2,x)

[Out]

2*d^2*b^4/(-b^2*c+a^2)^3*a/(a+b*(d*x+c)^(1/2))-2*d^2*b^6/(-b^2*c+a^2)^4*ln(a+b*(d*x+c)^(1/2))*c-10*d^2*b^4/(-b
^2*c+a^2)^4*ln(a+b*(d*x+c)^(1/2))*a^2-7/2/(-b^2*c+a^2)^4/x^2*a*b^5*c*(d*x+c)^(3/2)+3/(-b^2*c+a^2)^4/x^2*a^3*b^
3*(d*x+c)^(3/2)+1/2/(-b^2*c+a^2)^4/x^2*a^5*b/c*(d*x+c)^(3/2)+d/(-b^2*c+a^2)^4/x*b^6*c^2-1/2/(-b^2*c+a^2)^4/x^2
*c^3*b^6+2*d/(-b^2*c+a^2)^4/x*a^2*b^4*c+1/2/(-b^2*c+a^2)^4/x^2*a^2*b^4*c^2-3*d/(-b^2*c+a^2)^4/x*a^4*b^2+1/2/(-
b^2*c+a^2)^4/x^2*a^4*b^2*c+9/2/(-b^2*c+a^2)^4/x^2*(d*x+c)^(1/2)*a*b^5*c^2-5/(-b^2*c+a^2)^4/x^2*(d*x+c)^(1/2)*a
^3*b^3*c+1/2/(-b^2*c+a^2)^4/x^2*(d*x+c)^(1/2)*a^5*b-1/2/(-b^2*c+a^2)^4/x^2*a^6+d^2/(-b^2*c+a^2)^4*c*b^6*ln(d*x
)+5*d^2/(-b^2*c+a^2)^4*b^4*ln(d*x)*a^2+15/2*d^2/(-b^2*c+a^2)^4*c^(1/2)*b^5*arctanh((d*x+c)^(1/2)/c^(1/2))*a+5*
d^2/(-b^2*c+a^2)^4/c^(1/2)*b^3*arctanh((d*x+c)^(1/2)/c^(1/2))*a^3-1/2*d^2/(-b^2*c+a^2)^4/c^(3/2)*b*arctanh((d*
x+c)^(1/2)/c^(1/2))*a^5

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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(1/x^3/(a+b*(d*x+c)^(1/2))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 10.6694, size = 2507, normalized size = 8.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*(d*x+c)^(1/2))^2,x, algorithm="fricas")

[Out]

[-1/4*(2*b^8*c^6 - 4*a^2*b^6*c^5 + 4*a^6*b^2*c^3 - 2*a^8*c^2 - 4*(b^8*c^4 + 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2
*x^2 - 2*(b^8*c^5 + 3*a^2*b^6*c^4 - 9*a^4*b^4*c^3 + 5*a^6*b^2*c^2)*d*x - ((15*a*b^7*c^2 + 10*a^3*b^5*c - a^5*b
^3)*d^3*x^3 + (15*a*b^7*c^3 - 5*a^3*b^5*c^2 - 11*a^5*b^3*c + a^7*b)*d^2*x^2)*sqrt(c)*log((d*x + 2*sqrt(d*x + c
)*sqrt(c) + 2*c)/x) + 8*((b^8*c^3 + 5*a^2*b^6*c^2)*d^3*x^3 + (b^8*c^4 + 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2*x^2
)*log(sqrt(d*x + c)*b + a) - 4*((b^8*c^3 + 5*a^2*b^6*c^2)*d^3*x^3 + (b^8*c^4 + 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*
d^2*x^2)*log(x) - 2*(2*a*b^7*c^5 - 6*a^3*b^5*c^4 + 6*a^5*b^3*c^3 - 2*a^7*b*c^2 - (11*a*b^7*c^3 - 10*a^3*b^5*c^
2 - a^5*b^3*c)*d^2*x^2 - (5*a*b^7*c^4 - 9*a^3*b^5*c^3 + 3*a^5*b^3*c^2 + a^7*b*c)*d*x)*sqrt(d*x + c))/((b^10*c^
6 - 4*a^2*b^8*c^5 + 6*a^4*b^6*c^4 - 4*a^6*b^4*c^3 + a^8*b^2*c^2)*d*x^3 + (b^10*c^7 - 5*a^2*b^8*c^6 + 10*a^4*b^
6*c^5 - 10*a^6*b^4*c^4 + 5*a^8*b^2*c^3 - a^10*c^2)*x^2), -1/2*(b^8*c^6 - 2*a^2*b^6*c^5 + 2*a^6*b^2*c^3 - a^8*c
^2 - 2*(b^8*c^4 + 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2*x^2 - (b^8*c^5 + 3*a^2*b^6*c^4 - 9*a^4*b^4*c^3 + 5*a^6*b^
2*c^2)*d*x + ((15*a*b^7*c^2 + 10*a^3*b^5*c - a^5*b^3)*d^3*x^3 + (15*a*b^7*c^3 - 5*a^3*b^5*c^2 - 11*a^5*b^3*c +
a^7*b)*d^2*x^2)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + 4*((b^8*c^3 + 5*a^2*b^6*c^2)*d^3*x^3 + (b^8*c^4 +
4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2*x^2)*log(sqrt(d*x + c)*b + a) - 2*((b^8*c^3 + 5*a^2*b^6*c^2)*d^3*x^3 + (b^
8*c^4 + 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2*x^2)*log(x) - (2*a*b^7*c^5 - 6*a^3*b^5*c^4 + 6*a^5*b^3*c^3 - 2*a^7*
b*c^2 - (11*a*b^7*c^3 - 10*a^3*b^5*c^2 - a^5*b^3*c)*d^2*x^2 - (5*a*b^7*c^4 - 9*a^3*b^5*c^3 + 3*a^5*b^3*c^2 + a
^7*b*c)*d*x)*sqrt(d*x + c))/((b^10*c^6 - 4*a^2*b^8*c^5 + 6*a^4*b^6*c^4 - 4*a^6*b^4*c^3 + a^8*b^2*c^2)*d*x^3 +
(b^10*c^7 - 5*a^2*b^8*c^6 + 10*a^4*b^6*c^5 - 10*a^6*b^4*c^4 + 5*a^8*b^2*c^3 - a^10*c^2)*x^2)]

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Sympy [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(1/x**3/(a+b*(d*x+c)**(1/2))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*(d*x+c)^(1/2))^2,x, algorithm="giac")

[Out]

undef