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

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

[Out]

-1/2*a*b*(-15*b^4*c^2-10*a^2*b^2*c+a^4)*d^2*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(3/2)/(-b^2*c+a^2)^4+b^4*(b^2*c+5
*a^2)*d^2*ln(x)/(-b^2*c+a^2)^4-2*b^4*(b^2*c+5*a^2)*d^2*ln(a+b*(d*x+c)^(1/2))/(-b^2*c+a^2)^4+1/2*a*b^2*(11*b^2*
c+a^2)*d^2/c/(-b^2*c+a^2)^3/(a+b*(d*x+c)^(1/2))+1/2*(-a+b*(d*x+c)^(1/2))/(-b^2*c+a^2)/x^2/(a+b*(d*x+c)^(1/2))-
1/2*b*d*(3*a*b*c-(2*b^2*c+a^2)*(d*x+c)^(1/2))/c/(-b^2*c+a^2)^2/x/(a+b*(d*x+c)^(1/2))

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Rubi [A]  time = 0.40, antiderivative size = 306, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.83, 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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fricas [B]  time = 2.02, size = 1252, normalized size = 4.09 \[ \text {result too large to display} \]

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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giac [A]  time = 0.50, size = 521, normalized size = 1.70 \[ \frac {{\left (b^{6} c d^{2} + 5 \, a^{2} b^{4} d^{2}\right )} \log \left (-d x\right )}{b^{8} c^{4} - 4 \, a^{2} b^{6} c^{3} + 6 \, a^{4} b^{4} c^{2} - 4 \, a^{6} b^{2} c + a^{8}} - \frac {2 \, {\left (b^{7} c d^{2} + 5 \, a^{2} b^{5} d^{2}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{9} c^{4} - 4 \, a^{2} b^{7} c^{3} + 6 \, a^{4} b^{5} c^{2} - 4 \, a^{6} b^{3} c + a^{8} b} - \frac {{\left (15 \, a b^{5} c^{2} d^{2} + 10 \, a^{3} b^{3} c d^{2} - a^{5} b d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{2 \, {\left (b^{8} c^{5} - 4 \, a^{2} b^{6} c^{4} + 6 \, a^{4} b^{4} c^{3} - 4 \, a^{6} b^{2} c^{2} + a^{8} c\right )} \sqrt {-c}} - \frac {7 \, a b^{6} c^{4} d^{2} - a^{3} b^{4} c^{3} d^{2} - 7 \, a^{5} b^{2} c^{2} d^{2} + a^{7} c d^{2} + {\left (11 \, a b^{6} c^{2} d^{2} - 10 \, a^{3} b^{4} c d^{2} - a^{5} b^{2} d^{2}\right )} {\left (d x + c\right )}^{2} - {\left (2 \, b^{7} c^{3} d^{2} - 3 \, a^{2} b^{5} c^{2} d^{2} + a^{6} b d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - {\left (19 \, a b^{6} c^{3} d^{2} - 14 \, a^{3} b^{4} c^{2} d^{2} - 5 \, a^{5} b^{2} c d^{2}\right )} {\left (d x + c\right )} + 3 \, {\left (b^{7} c^{4} d^{2} - 2 \, a^{2} b^{5} c^{3} d^{2} + a^{4} b^{3} c^{2} d^{2}\right )} \sqrt {d x + c}}{2 \, {\left (b^{2} c - a^{2}\right )}^{4} {\left (\sqrt {d x + c} b + a\right )} c d^{2} x^{2}} \]

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]

(b^6*c*d^2 + 5*a^2*b^4*d^2)*log(-d*x)/(b^8*c^4 - 4*a^2*b^6*c^3 + 6*a^4*b^4*c^2 - 4*a^6*b^2*c + a^8) - 2*(b^7*c
*d^2 + 5*a^2*b^5*d^2)*log(abs(sqrt(d*x + c)*b + a))/(b^9*c^4 - 4*a^2*b^7*c^3 + 6*a^4*b^5*c^2 - 4*a^6*b^3*c + a
^8*b) - 1/2*(15*a*b^5*c^2*d^2 + 10*a^3*b^3*c*d^2 - a^5*b*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/((b^8*c^5 - 4*a^2
*b^6*c^4 + 6*a^4*b^4*c^3 - 4*a^6*b^2*c^2 + a^8*c)*sqrt(-c)) - 1/2*(7*a*b^6*c^4*d^2 - a^3*b^4*c^3*d^2 - 7*a^5*b
^2*c^2*d^2 + a^7*c*d^2 + (11*a*b^6*c^2*d^2 - 10*a^3*b^4*c*d^2 - a^5*b^2*d^2)*(d*x + c)^2 - (2*b^7*c^3*d^2 - 3*
a^2*b^5*c^2*d^2 + a^6*b*d^2)*(d*x + c)^(3/2) - (19*a*b^6*c^3*d^2 - 14*a^3*b^4*c^2*d^2 - 5*a^5*b^2*c*d^2)*(d*x
+ c) + 3*(b^7*c^4*d^2 - 2*a^2*b^5*c^3*d^2 + a^4*b^3*c^2*d^2)*sqrt(d*x + c))/((b^2*c - a^2)^4*(sqrt(d*x + c)*b
+ a)*c*d^2*x^2)

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maple [B]  time = 0.02, size = 610, normalized size = 1.99 \[ \frac {b^{6} c \,d^{2} \ln \left (d x \right )}{\left (-b^{2} c +a^{2}\right )^{4}}-\frac {2 b^{6} c \,d^{2} \ln \left (a +\sqrt {d x +c}\, b \right )}{\left (-b^{2} c +a^{2}\right )^{4}}+\frac {15 a \,b^{5} \sqrt {c}\, d^{2} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \left (-b^{2} c +a^{2}\right )^{4}}+\frac {5 a^{2} b^{4} d^{2} \ln \left (d x \right )}{\left (-b^{2} c +a^{2}\right )^{4}}-\frac {10 a^{2} b^{4} d^{2} \ln \left (a +\sqrt {d x +c}\, b \right )}{\left (-b^{2} c +a^{2}\right )^{4}}+\frac {5 a^{3} b^{3} d^{2} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\left (-b^{2} c +a^{2}\right )^{4} \sqrt {c}}+\frac {b^{6} c^{2} d}{\left (-b^{2} c +a^{2}\right )^{4} x}-\frac {a^{5} b \,d^{2} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \left (-b^{2} c +a^{2}\right )^{4} c^{\frac {3}{2}}}+\frac {2 a^{2} b^{4} c d}{\left (-b^{2} c +a^{2}\right )^{4} x}+\frac {2 a \,b^{4} d^{2}}{\left (-b^{2} c +a^{2}\right )^{3} \left (a +\sqrt {d x +c}\, b \right )}-\frac {b^{6} c^{3}}{2 \left (-b^{2} c +a^{2}\right )^{4} x^{2}}-\frac {3 a^{4} b^{2} d}{\left (-b^{2} c +a^{2}\right )^{4} x}+\frac {a^{2} b^{4} c^{2}}{2 \left (-b^{2} c +a^{2}\right )^{4} x^{2}}+\frac {9 \sqrt {d x +c}\, a \,b^{5} c^{2}}{2 \left (-b^{2} c +a^{2}\right )^{4} x^{2}}+\frac {a^{4} b^{2} c}{2 \left (-b^{2} c +a^{2}\right )^{4} x^{2}}-\frac {5 \sqrt {d x +c}\, a^{3} b^{3} c}{\left (-b^{2} c +a^{2}\right )^{4} x^{2}}-\frac {7 \left (d x +c \right )^{\frac {3}{2}} a \,b^{5} c}{2 \left (-b^{2} c +a^{2}\right )^{4} x^{2}}-\frac {a^{6}}{2 \left (-b^{2} c +a^{2}\right )^{4} x^{2}}+\frac {\sqrt {d x +c}\, a^{5} b}{2 \left (-b^{2} c +a^{2}\right )^{4} x^{2}}+\frac {3 \left (d x +c \right )^{\frac {3}{2}} a^{3} b^{3}}{\left (-b^{2} c +a^{2}\right )^{4} x^{2}}+\frac {\left (d x +c \right )^{\frac {3}{2}} a^{5} b}{2 \left (-b^{2} c +a^{2}\right )^{4} c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*b^4*a^2*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*b^2*a^4*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)*b^3*a^3*c+1/2/(-b^2*c+a^2)^4/x^2*(d*x+
c)^(1/2)*b*a^5-1/2/(-b^2*c+a^2)^4/x^2*a^6+d^2/(-b^2*c+a^2)^4*b^6*c*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*b^5*c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1/2))*a+5*d^2/(-b^2*c+a^2)^4*b^3/c^(1/2)*arctan
h((d*x+c)^(1/2)/c^(1/2))*a^3-1/2*d^2/(-b^2*c+a^2)^4*b/c^(3/2)*arctanh((d*x+c)^(1/2)/c^(1/2))*a^5+2*d^2*b^4/(-b
^2*c+a^2)^3*a/(a+(d*x+c)^(1/2)*b)-2*d^2*b^6/(-b^2*c+a^2)^4*ln(a+(d*x+c)^(1/2)*b)*c-10*d^2*b^4/(-b^2*c+a^2)^4*l
n(a+(d*x+c)^(1/2)*b)*a^2

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maxima [B]  time = 2.09, size = 659, normalized size = 2.15 \[ \frac {1}{4} \, d^{2} {\left (\frac {4 \, {\left (b^{6} c + 5 \, a^{2} b^{4}\right )} \log \left (d x\right )}{b^{8} c^{4} - 4 \, a^{2} b^{6} c^{3} + 6 \, a^{4} b^{4} c^{2} - 4 \, a^{6} b^{2} c + a^{8}} - \frac {8 \, {\left (b^{6} c + 5 \, a^{2} b^{4}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{8} c^{4} - 4 \, a^{2} b^{6} c^{3} + 6 \, a^{4} b^{4} c^{2} - 4 \, a^{6} b^{2} c + a^{8}} - \frac {{\left (15 \, a b^{5} c^{2} + 10 \, a^{3} b^{3} c - a^{5} b\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{{\left (b^{8} c^{5} - 4 \, a^{2} b^{6} c^{4} + 6 \, a^{4} b^{4} c^{3} - 4 \, a^{6} b^{2} c^{2} + a^{8} c\right )} \sqrt {c}} - \frac {2 \, {\left (7 \, a b^{4} c^{3} + 6 \, a^{3} b^{2} c^{2} - a^{5} c + {\left (11 \, a b^{4} c + a^{3} b^{2}\right )} {\left (d x + c\right )}^{2} - {\left (2 \, b^{5} c^{2} - a^{2} b^{3} c - a^{4} b\right )} {\left (d x + c\right )}^{\frac {3}{2}} - {\left (19 \, a b^{4} c^{2} + 5 \, a^{3} b^{2} c\right )} {\left (d x + c\right )} + 3 \, {\left (b^{5} c^{3} - a^{2} b^{3} c^{2}\right )} \sqrt {d x + c}\right )}}{a b^{6} c^{6} - 3 \, a^{3} b^{4} c^{5} + 3 \, a^{5} b^{2} c^{4} - a^{7} c^{3} + {\left (b^{7} c^{4} - 3 \, a^{2} b^{5} c^{3} + 3 \, a^{4} b^{3} c^{2} - a^{6} b c\right )} {\left (d x + c\right )}^{\frac {5}{2}} + {\left (a b^{6} c^{4} - 3 \, a^{3} b^{4} c^{3} + 3 \, a^{5} b^{2} c^{2} - a^{7} c\right )} {\left (d x + c\right )}^{2} - 2 \, {\left (b^{7} c^{5} - 3 \, a^{2} b^{5} c^{4} + 3 \, a^{4} b^{3} c^{3} - a^{6} b c^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 2 \, {\left (a b^{6} c^{5} - 3 \, a^{3} b^{4} c^{4} + 3 \, a^{5} b^{2} c^{3} - a^{7} c^{2}\right )} {\left (d x + c\right )} + {\left (b^{7} c^{6} - 3 \, a^{2} b^{5} c^{5} + 3 \, a^{4} b^{3} c^{4} - a^{6} b c^{3}\right )} \sqrt {d x + c}}\right )} \]

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]

1/4*d^2*(4*(b^6*c + 5*a^2*b^4)*log(d*x)/(b^8*c^4 - 4*a^2*b^6*c^3 + 6*a^4*b^4*c^2 - 4*a^6*b^2*c + a^8) - 8*(b^6
*c + 5*a^2*b^4)*log(sqrt(d*x + c)*b + a)/(b^8*c^4 - 4*a^2*b^6*c^3 + 6*a^4*b^4*c^2 - 4*a^6*b^2*c + a^8) - (15*a
*b^5*c^2 + 10*a^3*b^3*c - a^5*b)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/((b^8*c^5 - 4*a^2*b^
6*c^4 + 6*a^4*b^4*c^3 - 4*a^6*b^2*c^2 + a^8*c)*sqrt(c)) - 2*(7*a*b^4*c^3 + 6*a^3*b^2*c^2 - a^5*c + (11*a*b^4*c
+ a^3*b^2)*(d*x + c)^2 - (2*b^5*c^2 - a^2*b^3*c - a^4*b)*(d*x + c)^(3/2) - (19*a*b^4*c^2 + 5*a^3*b^2*c)*(d*x
+ c) + 3*(b^5*c^3 - a^2*b^3*c^2)*sqrt(d*x + c))/(a*b^6*c^6 - 3*a^3*b^4*c^5 + 3*a^5*b^2*c^4 - a^7*c^3 + (b^7*c^
4 - 3*a^2*b^5*c^3 + 3*a^4*b^3*c^2 - a^6*b*c)*(d*x + c)^(5/2) + (a*b^6*c^4 - 3*a^3*b^4*c^3 + 3*a^5*b^2*c^2 - a^
7*c)*(d*x + c)^2 - 2*(b^7*c^5 - 3*a^2*b^5*c^4 + 3*a^4*b^3*c^3 - a^6*b*c^2)*(d*x + c)^(3/2) - 2*(a*b^6*c^5 - 3*
a^3*b^4*c^4 + 3*a^5*b^2*c^3 - a^7*c^2)*(d*x + c) + (b^7*c^6 - 3*a^2*b^5*c^5 + 3*a^4*b^3*c^4 - a^6*b*c^3)*sqrt(
d*x + c)))

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mupad [B]  time = 5.96, size = 1441, normalized size = 4.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((5*a^3*b^2*d^2 + 19*a*b^4*c*d^2)*(c + d*x))/(2*(b^2*c - a^2)*(a^4 + b^4*c^2 - 2*a^2*b^2*c)) + ((a^3*b^2*d^2
+ 11*a*b^4*c*d^2)*(c + d*x)^2)/(2*c*(a^6 - b^6*c^3 - 3*a^4*b^2*c + 3*a^2*b^4*c^2)) - (a*(7*b^4*c^2*d^2 - a^4*d
^2 + 6*a^2*b^2*c*d^2))/(2*(b^2*c - a^2)*(a^4 + b^4*c^2 - 2*a^2*b^2*c)) + (b*(a^2*d^2 + 2*b^2*c*d^2)*(c + d*x)^
(3/2))/(2*c*(a^4 + b^4*c^2 - 2*a^2*b^2*c)) - (3*b^3*c*d^2*(c + d*x)^(1/2))/(2*(a^4 + b^4*c^2 - 2*a^2*b^2*c)))/
(a*(c + d*x)^2 + b*(c + d*x)^(5/2) + a*c^2 - 2*a*c*(c + d*x) - 2*b*c*(c + d*x)^(3/2) + b*c^2*(c + d*x)^(1/2))
+ log(a + b*(c + d*x)^(1/2))*((10*b^4*d^2)/(b^2*c - a^2)^3 - (12*b^6*c*d^2)/(b^2*c - a^2)^4) + (log((a*b^4*d^4
*(a^6 - 44*b^6*c^3 + 2*a^4*b^2*c - 103*a^2*b^4*c^2))/(4*c^2*(b^2*c - a^2)^6) - (b*d^2*((b^2*d^2*(a^2*(c + d*x)
^(1/2) + 4*a*b*c + 3*b^2*c*(c + d*x)^(1/2))*(a^5*(c^3)^(1/2) + 4*b^5*c^4 + 20*a^2*b^3*c^3 - 10*a^3*b^2*c*(c^3)
^(1/2) - 15*a*b^4*c^2*(c^3)^(1/2)))/(2*c^3*(b^2*c - a^2)^4) - (b^3*d^2*(c + d*x)^(1/2)*(6*b^4*c^2 - a^4 + 19*a
^2*b^2*c))/(c*(b^2*c - a^2)^3) + (a*b^2*d^2*(7*b^2*c - a^2))/(2*c*(b^2*c - a^2)^2))*(a^5*(c^3)^(1/2) + 4*b^5*c
^4 + 20*a^2*b^3*c^3 - 10*a^3*b^2*c*(c^3)^(1/2) - 15*a*b^4*c^2*(c^3)^(1/2)))/(4*c^3*(b^2*c - a^2)^4) + (a^2*b^5
*d^4*(11*b^2*c + a^2)^2*(c + d*x)^(1/2))/(4*c^2*(b^2*c - a^2)^6))*(4*b^6*c^4*d^2 + 20*a^2*b^4*c^3*d^2 + a^5*b*
d^2*(c^3)^(1/2) - 10*a^3*b^3*c*d^2*(c^3)^(1/2) - 15*a*b^5*c^2*d^2*(c^3)^(1/2)))/(4*(a^8*c^3 + b^8*c^7 - 4*a^6*
b^2*c^4 + 6*a^4*b^4*c^5 - 4*a^2*b^6*c^6)) + (log((a*b^4*d^4*(a^6 - 44*b^6*c^3 + 2*a^4*b^2*c - 103*a^2*b^4*c^2)
)/(4*c^2*(b^2*c - a^2)^6) - (b*d^2*((b^2*d^2*(a^2*(c + d*x)^(1/2) + 4*a*b*c + 3*b^2*c*(c + d*x)^(1/2))*(4*b^5*
c^4 - a^5*(c^3)^(1/2) + 20*a^2*b^3*c^3 + 10*a^3*b^2*c*(c^3)^(1/2) + 15*a*b^4*c^2*(c^3)^(1/2)))/(2*c^3*(b^2*c -
a^2)^4) - (b^3*d^2*(c + d*x)^(1/2)*(6*b^4*c^2 - a^4 + 19*a^2*b^2*c))/(c*(b^2*c - a^2)^3) + (a*b^2*d^2*(7*b^2*
c - a^2))/(2*c*(b^2*c - a^2)^2))*(4*b^5*c^4 - a^5*(c^3)^(1/2) + 20*a^2*b^3*c^3 + 10*a^3*b^2*c*(c^3)^(1/2) + 15
*a*b^4*c^2*(c^3)^(1/2)))/(4*c^3*(b^2*c - a^2)^4) + (a^2*b^5*d^4*(11*b^2*c + a^2)^2*(c + d*x)^(1/2))/(4*c^2*(b^
2*c - a^2)^6))*(4*b^6*c^4*d^2 + 20*a^2*b^4*c^3*d^2 - a^5*b*d^2*(c^3)^(1/2) + 10*a^3*b^3*c*d^2*(c^3)^(1/2) + 15
*a*b^5*c^2*d^2*(c^3)^(1/2)))/(4*(a^8*c^3 + b^8*c^7 - 4*a^6*b^2*c^4 + 6*a^4*b^4*c^5 - 4*a^2*b^6*c^6))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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