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

Optimal. Leaf size=191 \[ \frac{3 \left (a^2 d^2+6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} \sqrt{d}}-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}+\frac{3}{2} b \sqrt{a+b x} (c+d x)^{3/2}+\frac{3}{4} \sqrt{a+b x} \sqrt{c+d x} (3 a d+b c)-3 \sqrt{a} \sqrt{c} (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.188788, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {97, 154, 157, 63, 217, 206, 93, 208} \[ \frac{3 \left (a^2 d^2+6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} \sqrt{d}}-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}+\frac{3}{2} b \sqrt{a+b x} (c+d x)^{3/2}+\frac{3}{4} \sqrt{a+b x} \sqrt{c+d x} (3 a d+b c)-3 \sqrt{a} \sqrt{c} (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 157

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

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 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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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{(a+b x)^{3/2} (c+d x)^{3/2}}{x^2} \, dx &=-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}+\int \frac{\sqrt{a+b x} \sqrt{c+d x} \left (\frac{3}{2} (b c+a d)+3 b d x\right )}{x} \, dx\\ &=\frac{3}{2} b \sqrt{a+b x} (c+d x)^{3/2}-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}+\frac{\int \frac{\sqrt{c+d x} \left (3 a d (b c+a d)+\frac{3}{2} b d (b c+3 a d) x\right )}{x \sqrt{a+b x}} \, dx}{2 d}\\ &=\frac{3}{4} (b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}+\frac{3}{2} b \sqrt{a+b x} (c+d x)^{3/2}-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}+\frac{\int \frac{3 a b c d (b c+a d)+\frac{3}{4} b d \left (b^2 c^2+6 a b c d+a^2 d^2\right ) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 b d}\\ &=\frac{3}{4} (b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}+\frac{3}{2} b \sqrt{a+b x} (c+d x)^{3/2}-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}+\frac{1}{2} (3 a c (b c+a d)) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx+\frac{1}{8} \left (3 \left (b^2 c^2+6 a b c d+a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx\\ &=\frac{3}{4} (b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}+\frac{3}{2} b \sqrt{a+b x} (c+d x)^{3/2}-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}+(3 a c (b c+a d)) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )+\frac{\left (3 \left (b^2 c^2+6 a b c d+a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{4 b}\\ &=\frac{3}{4} (b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}+\frac{3}{2} b \sqrt{a+b x} (c+d x)^{3/2}-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}-3 \sqrt{a} \sqrt{c} (b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{\left (3 \left (b^2 c^2+6 a b c d+a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{4 b}\\ &=\frac{3}{4} (b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}+\frac{3}{2} b \sqrt{a+b x} (c+d x)^{3/2}-\frac{(a+b x)^{3/2} (c+d x)^{3/2}}{x}-3 \sqrt{a} \sqrt{c} (b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{3 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 1.09538, size = 197, normalized size = 1.03 \[ \frac{3 \sqrt{c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right ) \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{4 \sqrt{d} \sqrt{b c-a d} \sqrt{\frac{b (c+d x)}{b c-a d}}}+\frac{\sqrt{a+b x} \sqrt{c+d x} (a (5 d x-4 c)+b x (5 c+2 d x))}{4 x}-3 \sqrt{a} \sqrt{c} (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.015, size = 489, normalized size = 2.6 \begin{align*}{\frac{1}{8\,x}\sqrt{bx+a}\sqrt{dx+c} \left ( 3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}x{a}^{2}{d}^{2}+18\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}xabcd+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}x{b}^{2}{c}^{2}-12\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ) x{a}^{2}cd-12\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ) xab{c}^{2}+4\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}{x}^{2}bd+10\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}xad+10\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}xbc-8\,\sqrt{d{x}^{2}b+adx+bcx+ac}ac\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(3*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d
)^(1/2))*(a*c)^(1/2)*x*a^2*d^2+18*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)
^(1/2))*(a*c)^(1/2)*x*a*b*c*d+3*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(
1/2))*(a*c)^(1/2)*x*b^2*c^2-12*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c
)/x)*x*a^2*c*d-12*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x*a*b*c^
2+4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x^2*b*d+10*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(
1/2)*(a*c)^(1/2)*x*a*d+10*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*b*c-8*(b*d*x^2+a*d*x+b*c*x
+a*c)^(1/2)*a*c*(b*d)^(1/2)*(a*c)^(1/2))/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/x/(b*d)^(1/2)/(a*c)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 13.8344, size = 2460, normalized size = 12.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(b^2*c^2 + 6*a*b*c*d + a^2*d^2)*sqrt(b*d)*x*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*
b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 12*(b^2*c*d + a*b*d^2)*s
qrt(a*c)*x*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x
+ a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(2*b^2*d^2*x^2 - 4*a*b*c*d + 5*(b^2*c*d + a*b*d^2)*x)*
sqrt(b*x + a)*sqrt(d*x + c))/(b*d*x), -1/8*(3*(b^2*c^2 + 6*a*b*c*d + a^2*d^2)*sqrt(-b*d)*x*arctan(1/2*(2*b*d*x
+ b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 6*(b^2
*c*d + a*b*d^2)*sqrt(a*c)*x*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*s
qrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 2*(2*b^2*d^2*x^2 - 4*a*b*c*d + 5*(b^2*c
*d + a*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d*x), 1/16*(24*(b^2*c*d + a*b*d^2)*sqrt(-a*c)*x*arctan(1/2*(2
*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x))
+ 3*(b^2*c^2 + 6*a*b*c*d + a^2*d^2)*sqrt(b*d)*x*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x
+ b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b^2*d^2*x^2 - 4*a*b*c*d
+ 5*(b^2*c*d + a*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d*x), 1/8*(12*(b^2*c*d + a*b*d^2)*sqrt(-a*c)*x*arct
an(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*
c*d)*x)) - 3*(b^2*c^2 + 6*a*b*c*d + a^2*d^2)*sqrt(-b*d)*x*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x
+ a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(2*b^2*d^2*x^2 - 4*a*b*c*d + 5*(b^2*c
*d + a*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{3}{2}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.72406, size = 801, normalized size = 4.19 \begin{align*} \frac{2 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )} d{\left | b \right |}}{b} + \frac{5 \, b c d^{2}{\left | b \right |} + 3 \, a d^{3}{\left | b \right |}}{b d^{2}}\right )} - \frac{24 \,{\left (\sqrt{b d} a b^{2} c^{2}{\left | b \right |} + \sqrt{b d} a^{2} b c d{\left | b \right |}\right )} \arctan \left (-\frac{b^{2} c + a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt{-a b c d} b}\right )}{\sqrt{-a b c d} b} - \frac{16 \,{\left (\sqrt{b d} a b^{4} c^{3}{\left | b \right |} - 2 \, \sqrt{b d} a^{2} b^{3} c^{2} d{\left | b \right |} + \sqrt{b d} a^{3} b^{2} c d^{2}{\left | b \right |} - \sqrt{b d}{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{2} c^{2}{\left | b \right |} - \sqrt{b d}{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b c d{\left | b \right |}\right )}}{b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a b d +{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{4}} - \frac{3 \,{\left (\sqrt{b d} b^{2} c^{2}{\left | b \right |} + 6 \, \sqrt{b d} a b c d{\left | b \right |} + \sqrt{b d} a^{2} d^{2}{\left | b \right |}\right )} \log \left ({\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{b d}}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*d*abs(b)/b + (5*b*c*d^2*abs(b) + 3*a*d^3
*abs(b))/(b*d^2)) - 24*(sqrt(b*d)*a*b^2*c^2*abs(b) + sqrt(b*d)*a^2*b*c*d*abs(b))*arctan(-1/2*(b^2*c + a*b*d -
(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b) - 16
*(sqrt(b*d)*a*b^4*c^3*abs(b) - 2*sqrt(b*d)*a^2*b^3*c^2*d*abs(b) + sqrt(b*d)*a^3*b^2*c*d^2*abs(b) - sqrt(b*d)*(
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^2*c^2*abs(b) - sqrt(b*d)*(sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b*c*d*abs(b))/(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2
*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4) - 3*
(sqrt(b*d)*b^2*c^2*abs(b) + 6*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)*a^2*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a
) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(b*d))/b