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

Optimal. Leaf size=204 \[ -\frac{b d^2 \left (-6 a^2 b^2 c+a^4-3 b^4 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{4 c^{3/2} \left (a^2-b^2 c\right )^3}+\frac{a b^4 d^2 \log (x)}{\left (a^2-b^2 c\right )^3}-\frac{2 a b^4 d^2 \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^3}-\frac{a-b \sqrt{c+d x}}{2 x^2 \left (a^2-b^2 c\right )}-\frac{b d \left (4 a b c-\left (a^2+3 b^2 c\right ) \sqrt{c+d x}\right )}{4 c x \left (a^2-b^2 c\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.277508, antiderivative size = 204, 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} \[ -\frac{b d^2 \left (-6 a^2 b^2 c+a^4-3 b^4 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{4 c^{3/2} \left (a^2-b^2 c\right )^3}+\frac{a b^4 d^2 \log (x)}{\left (a^2-b^2 c\right )^3}-\frac{2 a b^4 d^2 \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^3}-\frac{a-b \sqrt{c+d x}}{2 x^2 \left (a^2-b^2 c\right )}-\frac{b d \left (4 a b c-\left (a^2+3 b^2 c\right ) \sqrt{c+d x}\right )}{4 c x \left (a^2-b^2 c\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*Sqrt[c + d*x])),x]

[Out]

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

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

Mathematica [A]  time = 0.433237, size = 228, normalized size = 1.12 \[ \frac{b d^2 x^2 \left (-6 a^2 b^2 c+a^4-3 b^4 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )+\sqrt{c} \left (4 a b^4 c d^2 x^2 \log \left (a^2-b^2 (c+d x)\right )+\left (a^2-b^2 c\right ) \left (-a^2 b \sqrt{c+d x} (2 c+d x)+2 a^3 c-2 a b^2 c (c-2 d x)+b^3 c (2 c-3 d x) \sqrt{c+d x}\right )-4 a b^4 c d^2 x^2 \log (x)\right )+8 a b^4 c^{3/2} d^2 x^2 \tanh ^{-1}\left (\frac{b \sqrt{c+d x}}{a}\right )}{4 c^{3/2} x^2 \left (b^2 c-a^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*b^4*c^(3/2)*d^2*x^2*ArcTanh[(b*Sqrt[c + d*x])/a] + b*(a^4 - 6*a^2*b^2*c - 3*b^4*c^2)*d^2*x^2*ArcTanh[Sqrt
[c + d*x]/Sqrt[c]] + Sqrt[c]*((a^2 - b^2*c)*(2*a^3*c - 2*a*b^2*c*(c - 2*d*x) + b^3*c*(2*c - 3*d*x)*Sqrt[c + d*
x] - a^2*b*Sqrt[c + d*x]*(2*c + d*x)) - 4*a*b^4*c*d^2*x^2*Log[x] + 4*a*b^4*c*d^2*x^2*Log[a^2 - b^2*(c + d*x)])
)/(4*c^(3/2)*(-a^2 + b^2*c)^3*x^2)

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Maple [B]  time = 0.014, size = 459, normalized size = 2.3 \begin{align*} -2\,{\frac{a{b}^{4}{d}^{2}\ln \left ( a+b\sqrt{dx+c} \right ) }{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}}-{\frac{3\,{b}^{5}c}{4\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{2}{b}^{3}}{2\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{4}b}{4\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}c} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{a{b}^{4}cd}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}x}}-{\frac{a{b}^{4}{c}^{2}}{2\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}}}-{\frac{d{a}^{3}{b}^{2}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}x}}+{\frac{{a}^{3}{b}^{2}c}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}}}-{\frac{3\,{a}^{2}{b}^{3}c}{2\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}}\sqrt{dx+c}}+{\frac{{a}^{4}b}{4\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}}\sqrt{dx+c}}+{\frac{5\,{b}^{5}{c}^{2}}{4\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}}\sqrt{dx+c}}-{\frac{{a}^{5}}{2\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}{x}^{2}}}+{\frac{a{b}^{4}{d}^{2}\ln \left ( dx \right ) }{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}}+{\frac{3\,{d}^{2}{b}^{5}}{4\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}\sqrt{c}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ) }+{\frac{3\,{b}^{3}{d}^{2}{a}^{2}}{2\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}}-{\frac{b{d}^{2}{a}^{4}}{4\, \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b^4*d^2*ln(a+b*(d*x+c)^(1/2))/(-b^2*c+a^2)^3-3/4/(-b^2*c+a^2)^3/x^2*b^5*c*(d*x+c)^(3/2)+1/2/(-b^2*c+a^2)^
3/x^2*b^3*(d*x+c)^(3/2)*a^2+1/4/(-b^2*c+a^2)^3/x^2*b/c*(d*x+c)^(3/2)*a^4+d/(-b^2*c+a^2)^3/x*a*b^4*c-1/2/(-b^2*
c+a^2)^3/x^2*a*b^4*c^2-d/(-b^2*c+a^2)^3/x*a^3*b^2+1/(-b^2*c+a^2)^3/x^2*a^3*b^2*c-3/2/(-b^2*c+a^2)^3/x^2*(d*x+c
)^(1/2)*a^2*b^3*c+1/4/(-b^2*c+a^2)^3/x^2*(d*x+c)^(1/2)*a^4*b+5/4/(-b^2*c+a^2)^3/x^2*(d*x+c)^(1/2)*b^5*c^2-1/2/
(-b^2*c+a^2)^3/x^2*a^5+d^2/(-b^2*c+a^2)^3*b^4*a*ln(d*x)+3/4*d^2/(-b^2*c+a^2)^3*c^(1/2)*b^5*arctanh((d*x+c)^(1/
2)/c^(1/2))+3/2*d^2/(-b^2*c+a^2)^3/c^(1/2)*b^3*arctanh((d*x+c)^(1/2)/c^(1/2))*a^2-1/4*d^2/(-b^2*c+a^2)^3/c^(3/
2)*b*arctanh((d*x+c)^(1/2)/c^(1/2))*a^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(1/x^3/(a+b*(d*x+c)^(1/2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.62273, size = 1112, normalized size = 5.45 \begin{align*} \left [\frac{16 \, a b^{4} c^{2} d^{2} x^{2} \log \left (\sqrt{d x + c} b + a\right ) - 8 \, a b^{4} c^{2} d^{2} x^{2} \log \left (x\right ) + 4 \, a b^{4} c^{4} - 8 \, a^{3} b^{2} c^{3} + 4 \, a^{5} c^{2} +{\left (3 \, b^{5} c^{2} + 6 \, a^{2} b^{3} c - a^{4} b\right )} \sqrt{c} d^{2} x^{2} \log \left (\frac{d x - 2 \, \sqrt{d x + c} \sqrt{c} + 2 \, c}{x}\right ) - 8 \,{\left (a b^{4} c^{3} - a^{3} b^{2} c^{2}\right )} d x - 2 \,{\left (2 \, b^{5} c^{4} - 4 \, a^{2} b^{3} c^{3} + 2 \, a^{4} b c^{2} -{\left (3 \, b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} - a^{4} b c\right )} d x\right )} \sqrt{d x + c}}{8 \,{\left (b^{6} c^{5} - 3 \, a^{2} b^{4} c^{4} + 3 \, a^{4} b^{2} c^{3} - a^{6} c^{2}\right )} x^{2}}, \frac{8 \, a b^{4} c^{2} d^{2} x^{2} \log \left (\sqrt{d x + c} b + a\right ) - 4 \, a b^{4} c^{2} d^{2} x^{2} \log \left (x\right ) + 2 \, a b^{4} c^{4} - 4 \, a^{3} b^{2} c^{3} + 2 \, a^{5} c^{2} +{\left (3 \, b^{5} c^{2} + 6 \, a^{2} b^{3} c - a^{4} b\right )} \sqrt{-c} d^{2} x^{2} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c}}{c}\right ) - 4 \,{\left (a b^{4} c^{3} - a^{3} b^{2} c^{2}\right )} d x -{\left (2 \, b^{5} c^{4} - 4 \, a^{2} b^{3} c^{3} + 2 \, a^{4} b c^{2} -{\left (3 \, b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} - a^{4} b c\right )} d x\right )} \sqrt{d x + c}}{4 \,{\left (b^{6} c^{5} - 3 \, a^{2} b^{4} c^{4} + 3 \, a^{4} b^{2} c^{3} - a^{6} c^{2}\right )} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(16*a*b^4*c^2*d^2*x^2*log(sqrt(d*x + c)*b + a) - 8*a*b^4*c^2*d^2*x^2*log(x) + 4*a*b^4*c^4 - 8*a^3*b^2*c^3
+ 4*a^5*c^2 + (3*b^5*c^2 + 6*a^2*b^3*c - a^4*b)*sqrt(c)*d^2*x^2*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x)
- 8*(a*b^4*c^3 - a^3*b^2*c^2)*d*x - 2*(2*b^5*c^4 - 4*a^2*b^3*c^3 + 2*a^4*b*c^2 - (3*b^5*c^3 - 2*a^2*b^3*c^2 -
a^4*b*c)*d*x)*sqrt(d*x + c))/((b^6*c^5 - 3*a^2*b^4*c^4 + 3*a^4*b^2*c^3 - a^6*c^2)*x^2), 1/4*(8*a*b^4*c^2*d^2*x
^2*log(sqrt(d*x + c)*b + a) - 4*a*b^4*c^2*d^2*x^2*log(x) + 2*a*b^4*c^4 - 4*a^3*b^2*c^3 + 2*a^5*c^2 + (3*b^5*c^
2 + 6*a^2*b^3*c - a^4*b)*sqrt(-c)*d^2*x^2*arctan(sqrt(d*x + c)*sqrt(-c)/c) - 4*(a*b^4*c^3 - a^3*b^2*c^2)*d*x -
(2*b^5*c^4 - 4*a^2*b^3*c^3 + 2*a^4*b*c^2 - (3*b^5*c^3 - 2*a^2*b^3*c^2 - a^4*b*c)*d*x)*sqrt(d*x + c))/((b^6*c^
5 - 3*a^2*b^4*c^4 + 3*a^4*b^2*c^3 - a^6*c^2)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (a + b \sqrt{c + d x}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*(a + b*sqrt(c + d*x))), x)

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Giac [B]  time = 1.23144, size = 649, normalized size = 3.18 \begin{align*} \frac{2 \, a b^{5} d^{2} \log \left ({\left | \sqrt{d x + c} b + a \right |}\right )}{b^{7} c^{3} - 3 \, a^{2} b^{5} c^{2} + 3 \, a^{4} b^{3} c - a^{6} b} - \frac{a b^{4} d^{2} \log \left (d x\right )}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}} + \frac{{\left (3 \, b^{5} c^{2} d^{2} + 6 \, a^{2} b^{3} c d^{2} - a^{4} b d^{2}\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{4 \,{\left (b^{6} c^{4} - 3 \, a^{2} b^{4} c^{3} + 3 \, a^{4} b^{2} c^{2} - a^{6} c\right )} \sqrt{-c}} + \frac{2 \, a b^{4} c^{2} d^{2} \log \left (-c\right ) - 4 \, a b^{4} c^{2} d^{2} \log \left ({\left | a \right |}\right ) - 3 \, a b^{4} c^{2} d^{2} + 4 \, a^{3} b^{2} c d^{2} - a^{5} d^{2}}{2 \,{\left (b^{6} c^{5} - 3 \, a^{2} b^{4} c^{4} + 3 \, a^{4} b^{2} c^{3} - a^{6} c^{2}\right )}} + \frac{6 \, a b^{4} c^{3} d^{2} - 8 \, a^{3} b^{2} c^{2} d^{2} + 2 \, a^{5} c d^{2} +{\left (3 \, b^{5} c^{2} d^{2} - 2 \, a^{2} b^{3} c d^{2} - a^{4} b d^{2}\right )}{\left (d x + c\right )}^{\frac{3}{2}} - 4 \,{\left (a b^{4} c^{2} d^{2} - a^{3} b^{2} c d^{2}\right )}{\left (d x + c\right )} -{\left (5 \, b^{5} c^{3} d^{2} - 6 \, a^{2} b^{3} c^{2} d^{2} + a^{4} b c d^{2}\right )} \sqrt{d x + c}}{4 \,{\left (b^{2} c - a^{2}\right )}^{3} c d^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*b^5*d^2*log(abs(sqrt(d*x + c)*b + a))/(b^7*c^3 - 3*a^2*b^5*c^2 + 3*a^4*b^3*c - a^6*b) - a*b^4*d^2*log(d*x)
/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6) + 1/4*(3*b^5*c^2*d^2 + 6*a^2*b^3*c*d^2 - a^4*b*d^2)*arctan(sqrt
(d*x + c)/sqrt(-c))/((b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)*sqrt(-c)) + 1/2*(2*a*b^4*c^2*d^2*log(-c
) - 4*a*b^4*c^2*d^2*log(abs(a)) - 3*a*b^4*c^2*d^2 + 4*a^3*b^2*c*d^2 - a^5*d^2)/(b^6*c^5 - 3*a^2*b^4*c^4 + 3*a^
4*b^2*c^3 - a^6*c^2) + 1/4*(6*a*b^4*c^3*d^2 - 8*a^3*b^2*c^2*d^2 + 2*a^5*c*d^2 + (3*b^5*c^2*d^2 - 2*a^2*b^3*c*d
^2 - a^4*b*d^2)*(d*x + c)^(3/2) - 4*(a*b^4*c^2*d^2 - a^3*b^2*c*d^2)*(d*x + c) - (5*b^5*c^3*d^2 - 6*a^2*b^3*c^2
*d^2 + a^4*b*c*d^2)*sqrt(d*x + c))/((b^2*c - a^2)^3*c*d^2*x^2)