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

Optimal. Leaf size=170 \[ \frac {\left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2} \sqrt {c}}+2 \sqrt {b} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 x^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{4 a x} \]

[Out]

2*d^(3/2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)+1/4*(-3*a^2*d^2-6*a*b*c*d+b^2*c^2)*arct
anh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(3/2)/c^(1/2)-1/2*(d*x+c)^(3/2)*(b*x+a)^(1/2)/x^2-1/4*(3*a*
d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/x

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Rubi [A]  time = 0.11, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {97, 149, 157, 63, 217, 206, 93, 208} \[ \frac {\left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2} \sqrt {c}}+2 \sqrt {b} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 x^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{4 a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*a*x) - (Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*x^2) + ((b^2*c^2 - 6
*a*b*c*d - 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(3/2)*Sqrt[c]) + 2*Sqrt[b
]*d^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*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 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 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 149

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*(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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

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 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 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]

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]

Rubi steps

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

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Mathematica [A]  time = 1.38, size = 186, normalized size = 1.09 \[ \frac {\left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2} \sqrt {c}}+\frac {2 d^{3/2} \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {c+d x}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (2 a c+5 a d x+b c x)}{4 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c + b*c*x + 5*a*d*x))/(a*x^2) + (2*d^(3/2)*Sqrt[b*c - a*d]*Sqrt[(b*(c +
d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[c + d*x] + ((b^2*c^2 - 6*a*b*c*d -
3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(3/2)*Sqrt[c])

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fricas [A]  time = 2.44, size = 1023, normalized size = 6.02 \[ \left [\frac {8 \, \sqrt {b d} a^{2} c d x^{2} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {a c} x^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} + {\left (a b c^{2} + 5 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a^{2} c x^{2}}, -\frac {16 \, \sqrt {-b d} a^{2} c d x^{2} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {a c} x^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (2 \, a^{2} c^{2} + {\left (a b c^{2} + 5 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a^{2} c x^{2}}, \frac {4 \, \sqrt {b d} a^{2} c d x^{2} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {-a c} x^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} + {\left (a b c^{2} + 5 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a^{2} c x^{2}}, -\frac {8 \, \sqrt {-b d} a^{2} c d x^{2} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {-a c} x^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (2 \, a^{2} c^{2} + {\left (a b c^{2} + 5 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a^{2} c x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 14.70, size = 1127, normalized size = 6.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(b*d)*d*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2) - (sqrt(b*d)
*b^3*c^2*abs(b) - 6*sqrt(b*d)*a*b^2*c*d*abs(b) - 3*sqrt(b*d)*a^2*b*d^2*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (s
qrt(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)*a*b) + 2*
(sqrt(b*d)*b^9*c^5*abs(b) + sqrt(b*d)*a*b^8*c^4*d*abs(b) - 14*sqrt(b*d)*a^2*b^7*c^3*d^2*abs(b) + 26*sqrt(b*d)*
a^3*b^6*c^2*d^3*abs(b) - 19*sqrt(b*d)*a^4*b^5*c*d^4*abs(b) + 5*sqrt(b*d)*a^5*b^4*d^5*abs(b) - 3*sqrt(b*d)*(sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^4*abs(b) - 8*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^6*c^3*d*abs(b) + 10*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^5*c^2*d^2*abs(b) + 16*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2
*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^4*c*d^3*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*
x + a)*b*d - a*b*d))^2*a^4*b^3*d^4*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^4*b^5*c^3*abs(b) + 17*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*
b^4*c^2*d*abs(b) + 13*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^3*c*d^
2*abs(b) + 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^2*d^3*abs(b) -
sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^3*c^2*abs(b) - 10*sqrt(b*d)*(sq
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^2*c*d*abs(b) - 5*sqrt(b*d)*(sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b*d^2*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)^2*a
))/b

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maple [B]  time = 0.02, size = 400, normalized size = 2.35 \[ \frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (-3 \sqrt {b d}\, a^{2} d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-6 \sqrt {b d}\, a b c d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+8 \sqrt {a c}\, a b \,d^{2} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+\sqrt {b d}\, b^{2} c^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-10 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a d x -2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b c x -4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a c \right )}{8 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a*(8*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b
*d)^(1/2))*x^2*a*b*d^2*(a*c)^(1/2)-3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x
^2*a^2*d^2*(b*d)^(1/2)-6*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x^2*a*b*c*d*(
b*d)^(1/2)+ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x^2*b^2*c^2*(b*d)^(1/2)-10*
(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a*d-2*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a
*c)^(1/2)*x*b*c-4*(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^2/(b*d)^(1/2)/(a*c)^(1/2)

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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)*(b*x+a)^(1/2)/x^3,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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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