3.467 \(\int \frac {(c+d x)^{3/2}}{x^2 (a+b x)^2} \, dx\)

Optimal. Leaf size=149 \[ \frac {\sqrt {c} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^3}-\frac {\sqrt {b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}}-\frac {\sqrt {c+d x} (2 b c-a d)}{a^2 (a+b x)}-\frac {c \sqrt {c+d x}}{a x (a+b x)} \]

[Out]

(-3*a*d+4*b*c)*arctanh((d*x+c)^(1/2)/c^(1/2))*c^(1/2)/a^3-(-a*d+4*b*c)*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c
)^(1/2))*(-a*d+b*c)^(1/2)/a^3/b^(1/2)-(-a*d+2*b*c)*(d*x+c)^(1/2)/a^2/(b*x+a)-c*(d*x+c)^(1/2)/a/x/(b*x+a)

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Rubi [A]  time = 0.20, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {98, 151, 156, 63, 208} \[ -\frac {\sqrt {c+d x} (2 b c-a d)}{a^2 (a+b x)}+\frac {\sqrt {c} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^3}-\frac {\sqrt {b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}}-\frac {c \sqrt {c+d x}}{a x (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 127, normalized size = 0.85 \[ \frac {\frac {a \sqrt {c+d x} (-a c+a d x-2 b c x)}{x (a+b x)}+\sqrt {c} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )-\frac {\sqrt {b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b}}}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.42, size = 770, normalized size = 5.17 \[ \left [-\frac {{\left ({\left (4 \, b^{2} c - a b d\right )} x^{2} + {\left (4 \, a b c - a^{2} d\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + {\left ({\left (4 \, b^{2} c - 3 \, a b d\right )} x^{2} + {\left (4 \, a b c - 3 \, a^{2} d\right )} x\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (a^{2} c + {\left (2 \, a b c - a^{2} d\right )} x\right )} \sqrt {d x + c}}{2 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}, -\frac {2 \, {\left ({\left (4 \, b^{2} c - a b d\right )} x^{2} + {\left (4 \, a b c - a^{2} d\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left ({\left (4 \, b^{2} c - 3 \, a b d\right )} x^{2} + {\left (4 \, a b c - 3 \, a^{2} d\right )} x\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (a^{2} c + {\left (2 \, a b c - a^{2} d\right )} x\right )} \sqrt {d x + c}}{2 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}, -\frac {2 \, {\left ({\left (4 \, b^{2} c - 3 \, a b d\right )} x^{2} + {\left (4 \, a b c - 3 \, a^{2} d\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left ({\left (4 \, b^{2} c - a b d\right )} x^{2} + {\left (4 \, a b c - a^{2} d\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (a^{2} c + {\left (2 \, a b c - a^{2} d\right )} x\right )} \sqrt {d x + c}}{2 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}, -\frac {{\left ({\left (4 \, b^{2} c - a b d\right )} x^{2} + {\left (4 \, a b c - a^{2} d\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left ({\left (4 \, b^{2} c - 3 \, a b d\right )} x^{2} + {\left (4 \, a b c - 3 \, a^{2} d\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (a^{2} c + {\left (2 \, a b c - a^{2} d\right )} x\right )} \sqrt {d x + c}}{a^{3} b x^{2} + a^{4} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.42, size = 197, normalized size = 1.32 \[ \frac {{\left (4 \, b^{2} c^{2} - 5 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{3}} - \frac {{\left (4 \, b c^{2} - 3 \, a c d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{3} \sqrt {-c}} - \frac {2 \, {\left (d x + c\right )}^{\frac {3}{2}} b c d - 2 \, \sqrt {d x + c} b c^{2} d - {\left (d x + c\right )}^{\frac {3}{2}} a d^{2} + 2 \, \sqrt {d x + c} a c d^{2}}{{\left ({\left (d x + c\right )}^{2} b - 2 \, {\left (d x + c\right )} b c + b c^{2} + {\left (d x + c\right )} a d - a c d\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*b^2*c^2 - 5*a*b*c*d + a^2*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^3) - (4
*b*c^2 - 3*a*c*d)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^3*sqrt(-c)) - (2*(d*x + c)^(3/2)*b*c*d - 2*sqrt(d*x + c)*b
*c^2*d - (d*x + c)^(3/2)*a*d^2 + 2*sqrt(d*x + c)*a*c*d^2)/(((d*x + c)^2*b - 2*(d*x + c)*b*c + b*c^2 + (d*x + c
)*a*d - a*c*d)*a^2)

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maple [A]  time = 0.02, size = 237, normalized size = 1.59 \[ \frac {d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a}-\frac {5 b c d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{2}}+\frac {4 b^{2} c^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{3}}+\frac {\sqrt {d x +c}\, d^{2}}{\left (b d x +a d \right ) a}-\frac {\sqrt {d x +c}\, b c d}{\left (b d x +a d \right ) a^{2}}-\frac {3 \sqrt {c}\, d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2}}+\frac {4 b \,c^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{3}}-\frac {\sqrt {d x +c}\, c}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2/a*(d*x+c)^(1/2)/(b*d*x+a*d)-d/a^2*(d*x+c)^(1/2)/(b*d*x+a*d)*b*c+d^2/a/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(
1/2)/((a*d-b*c)*b)^(1/2)*b)-5*d/a^2/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*b*c+4/a^3/
((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*b^2*c^2-c/a^2*(d*x+c)^(1/2)/x-3*d*c^(1/2)/a^2*
arctanh((d*x+c)^(1/2)/c^(1/2))+4*c^(3/2)/a^3*arctanh((d*x+c)^(1/2)/c^(1/2))*b

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.62, size = 429, normalized size = 2.88 \[ -\frac {\frac {2\,\left (a\,c\,d^2-b\,c^2\,d\right )\,\sqrt {c+d\,x}}{a^2}-\frac {d\,\left (a\,d-2\,b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{a^2}}{\left (a\,d-2\,b\,c\right )\,\left (c+d\,x\right )+b\,{\left (c+d\,x\right )}^2+b\,c^2-a\,c\,d}-\frac {\sqrt {c}\,\mathrm {atanh}\left (\frac {6\,b\,\sqrt {c}\,d^7\,\sqrt {c+d\,x}}{6\,b\,c\,d^7-\frac {14\,b^2\,c^2\,d^6}{a}+\frac {8\,b^3\,c^3\,d^5}{a^2}}-\frac {14\,b^2\,c^{3/2}\,d^6\,\sqrt {c+d\,x}}{6\,a\,b\,c\,d^7-14\,b^2\,c^2\,d^6+\frac {8\,b^3\,c^3\,d^5}{a}}+\frac {8\,b^3\,c^{5/2}\,d^5\,\sqrt {c+d\,x}}{6\,a^2\,b\,c\,d^7-14\,a\,b^2\,c^2\,d^6+8\,b^3\,c^3\,d^5}\right )\,\left (3\,a\,d-4\,b\,c\right )}{a^3}-\frac {\mathrm {atanh}\left (\frac {2\,b\,c\,d^6\,\sqrt {b^2\,c-a\,b\,d}\,\sqrt {c+d\,x}}{2\,a\,b\,c\,d^7-10\,b^2\,c^2\,d^6+\frac {8\,b^3\,c^3\,d^5}{a}}-\frac {8\,b^2\,c^2\,d^5\,\sqrt {b^2\,c-a\,b\,d}\,\sqrt {c+d\,x}}{2\,a^2\,b\,c\,d^7-10\,a\,b^2\,c^2\,d^6+8\,b^3\,c^3\,d^5}\right )\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (a\,d-4\,b\,c\right )}{a^3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*(a*c*d^2 - b*c^2*d)*(c + d*x)^(1/2))/a^2 - (d*(a*d - 2*b*c)*(c + d*x)^(3/2))/a^2)/((a*d - 2*b*c)*(c + d*
x) + b*(c + d*x)^2 + b*c^2 - a*c*d) - (c^(1/2)*atanh((6*b*c^(1/2)*d^7*(c + d*x)^(1/2))/(6*b*c*d^7 - (14*b^2*c^
2*d^6)/a + (8*b^3*c^3*d^5)/a^2) - (14*b^2*c^(3/2)*d^6*(c + d*x)^(1/2))/(6*a*b*c*d^7 - 14*b^2*c^2*d^6 + (8*b^3*
c^3*d^5)/a) + (8*b^3*c^(5/2)*d^5*(c + d*x)^(1/2))/(8*b^3*c^3*d^5 - 14*a*b^2*c^2*d^6 + 6*a^2*b*c*d^7))*(3*a*d -
4*b*c))/a^3 - (atanh((2*b*c*d^6*(b^2*c - a*b*d)^(1/2)*(c + d*x)^(1/2))/(2*a*b*c*d^7 - 10*b^2*c^2*d^6 + (8*b^3
*c^3*d^5)/a) - (8*b^2*c^2*d^5*(b^2*c - a*b*d)^(1/2)*(c + d*x)^(1/2))/(8*b^3*c^3*d^5 - 10*a*b^2*c^2*d^6 + 2*a^2
*b*c*d^7))*(-b*(a*d - b*c))^(1/2)*(a*d - 4*b*c))/(a^3*b)

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

[Out]

Timed out

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