3.6.88 \(\int \frac {(a+b x)^{3/2} (c+d x)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=222 \[ \frac {(a d+b c) \left (a^2 d^2-10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{3/2} c^{3/2}}+2 b^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (\frac {b^2 c}{a}-\frac {a d^2}{c}+8 b d\right )}{8 x}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+b c)}{4 c x^2} \]

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Rubi [A]  time = 0.19, antiderivative size = 222, normalized size of antiderivative = 1.00, 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, 149, 157, 63, 217, 206, 93, 208} \begin {gather*} \frac {(a d+b c) \left (a^2 d^2-10 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 )}{8 a^{3/2} c^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (\frac {b^2 c}{a}-\frac {a d^2}{c}+8 b d\right )}{8 x}+2 b^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+b c)}{4 c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^4,x]

[Out]

-(((b^2*c)/a + 8*b*d - (a*d^2)/c)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*x) - ((b*c + a*d)*Sqrt[a + b*x]*(c + d*x)^(3
/2))/(4*c*x^2) - ((a + b*x)^(3/2)*(c + d*x)^(3/2))/(3*x^3) + ((b*c + a*d)*(b^2*c^2 - 10*a*b*c*d + a^2*d^2)*Arc
Tanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(3/2)*c^(3/2)) + 2*b^(3/2)*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 {(a+b x)^{3/2} (c+d x)^{3/2}}{x^4} \, dx &=-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}+\frac {1}{3} \int \frac {\sqrt {a+b x} \sqrt {c+d x} \left (\frac {3}{2} (b c+a d)+3 b d x\right )}{x^3} \, dx\\ &=-\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}+\frac {\int \frac {\sqrt {c+d x} \left (\frac {3}{4} \left (b^2 c^2+8 a b c d-a^2 d^2\right )+6 b^2 c d x\right )}{x^2 \sqrt {a+b x}} \, dx}{6 c}\\ &=-\frac {\left (\frac {b^2 c}{a}+8 b d-\frac {a d^2}{c}\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 x}-\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}+\frac {\int \frac {-\frac {3}{8} (b c+a d) \left (b^2 c^2-10 a b c d+a^2 d^2\right )+6 a b^2 c d^2 x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 a c}\\ &=-\frac {\left (\frac {b^2 c}{a}+8 b d-\frac {a d^2}{c}\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 x}-\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}+\left (b^2 d^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx-\frac {\left ((b c+a d) \left (b^2 c^2-10 a b c d+a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 a c}\\ &=-\frac {\left (\frac {b^2 c}{a}+8 b d-\frac {a d^2}{c}\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 x}-\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}+\left (2 b 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 c+a d) \left (b^2 c^2-10 a b c d+a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 a c}\\ &=-\frac {\left (\frac {b^2 c}{a}+8 b d-\frac {a d^2}{c}\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 x}-\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}+\frac {(b c+a d) \left (b^2 c^2-10 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{3/2} c^{3/2}}+\left (2 b 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 {\left (\frac {b^2 c}{a}+8 b d-\frac {a d^2}{c}\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 x}-\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 x^3}+\frac {(b c+a d) \left (b^2 c^2-10 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{3/2} c^{3/2}}+2 b^{3/2} 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 = 2.50, size = 237, normalized size = 1.07 \begin {gather*} -\frac {\sqrt {a+b x} \sqrt {c+d x} \left (a^2 \left (8 c^2+14 c d x+3 d^2 x^2\right )+2 a b c x (7 c+19 d x)+3 b^2 c^2 x^2\right )}{24 a c x^3}+\frac {\left (a^3 d^3-9 a^2 b c d^2-9 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{3/2} c^{3/2}}+\frac {2 d^{3/2} (b c-a d)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{(c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^4,x]

[Out]

-1/24*(Sqrt[a + b*x]*Sqrt[c + d*x]*(3*b^2*c^2*x^2 + 2*a*b*c*x*(7*c + 19*d*x) + a^2*(8*c^2 + 14*c*d*x + 3*d^2*x
^2)))/(a*c*x^3) + (2*d^(3/2)*(b*c - a*d)^(3/2)*((b*(c + d*x))/(b*c - a*d))^(3/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x
])/Sqrt[b*c - a*d]])/(c + d*x)^(3/2) + ((b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*ArcTanh[(Sqrt[c]*S
qrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(3/2)*c^(3/2))

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IntegrateAlgebraic [A]  time = 0.64, size = 399, normalized size = 1.80 \begin {gather*} \frac {\left (a^3 d^3-9 a^2 b c d^2-9 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 a^{3/2} c^{3/2}}-\frac {\sqrt {c+d x} \left (\frac {3 a^5 d^3 (c+d x)^2}{(a+b x)^2}+\frac {8 a^4 c d^3 (c+d x)}{a+b x}+\frac {21 a^4 b c d^2 (c+d x)^2}{(a+b x)^2}-\frac {27 a^3 b^2 c^2 d (c+d x)^2}{(a+b x)^2}-\frac {72 a^3 b c^2 d^2 (c+d x)}{a+b x}-3 a^3 c^2 d^3+\frac {3 a^2 b^3 c^3 (c+d x)^2}{(a+b x)^2}+\frac {72 a^2 b^2 c^3 d (c+d x)}{a+b x}+27 a^2 b c^3 d^2-\frac {8 a b^3 c^4 (c+d x)}{a+b x}-21 a b^2 c^4 d-3 b^3 c^5\right )}{24 a c \sqrt {a+b x} \left (\frac {a (c+d x)}{a+b x}-c\right )^3}+2 b^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^4,x]

[Out]

-1/24*(Sqrt[c + d*x]*(-3*b^3*c^5 - 21*a*b^2*c^4*d + 27*a^2*b*c^3*d^2 - 3*a^3*c^2*d^3 - (8*a*b^3*c^4*(c + d*x))
/(a + b*x) + (72*a^2*b^2*c^3*d*(c + d*x))/(a + b*x) - (72*a^3*b*c^2*d^2*(c + d*x))/(a + b*x) + (8*a^4*c*d^3*(c
+ d*x))/(a + b*x) + (3*a^2*b^3*c^3*(c + d*x)^2)/(a + b*x)^2 - (27*a^3*b^2*c^2*d*(c + d*x)^2)/(a + b*x)^2 + (2
1*a^4*b*c*d^2*(c + d*x)^2)/(a + b*x)^2 + (3*a^5*d^3*(c + d*x)^2)/(a + b*x)^2))/(a*c*Sqrt[a + b*x]*(-c + (a*(c
+ d*x))/(a + b*x))^3) + ((b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(
Sqrt[c]*Sqrt[a + b*x])])/(8*a^(3/2)*c^(3/2)) + 2*b^(3/2)*d^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt
[a + b*x])]

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fricas [A]  time = 7.78, size = 1237, normalized size = 5.57

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(3/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(48*sqrt(b*d)*a^2*b*c^2*d*x^3*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) + 3*(b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^
2 + a^3*d^3)*sqrt(a*c)*x^3*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)*sq
rt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(8*a^3*c^3 + (3*a*b^2*c^3 + 38*a^2*b*c
^2*d + 3*a^3*c*d^2)*x^2 + 14*(a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^3), -1/96*(96*
sqrt(-b*d)*a^2*b*c^2*d*x^3*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)) - 3*(b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(a*c)*x^3*lo
g((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*(8*a^3*c^3 + (3*a*b^2*c^3 + 38*a^2*b*c^2*d + 3*a^3*c*d^2)*x^2 + 14*
(a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^3), 1/48*(24*sqrt(b*d)*a^2*b*c^2*d*x^3*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) - 3*(b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(-a*c)*x^3*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*(8*a^3*c^3 + (3*a*b^2*c^3 + 38*a^2*b*c^2*d + 3*a^3*c*d^2)*x^2 + 14*(a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x +
a)*sqrt(d*x + c))/(a^2*c^2*x^3), -1/48*(48*sqrt(-b*d)*a^2*b*c^2*d*x^3*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)) + 3*(b^3*c^3 - 9*a*b^2*c^2*d
- 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(-a*c)*x^3*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*(8*a^3*c^3 + (3*a*b^2*c^3 + 38*a^2*b*c^2*d + 3*a^3*
c*d^2)*x^2 + 14*(a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^3)]

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giac [B]  time = 97.60, size = 2269, normalized size = 10.22

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(3/2)/x^4,x, algorithm="giac")

[Out]

-1/24*(24*sqrt(b*d)*b*d*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2) - 3*(sqr
t(b*d)*b^4*c^3*abs(b) - 9*sqrt(b*d)*a*b^3*c^2*d*abs(b) - 9*sqrt(b*d)*a^2*b^2*c*d^2*abs(b) + sqrt(b*d)*a^3*b*d^
3*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)/(sqr
t(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a*b*c) + 2*(3*sqrt(b*d)*b^14*c^8*abs(b) + 20*sqrt(b*d)*a*b^13*c^7*d*abs(b) - 1
80*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) + 492*sqrt(b*d)*a^3*b^11*c^5*d^3*abs(b) - 670*sqrt(b*d)*a^4*b^10*c^4*d^4*
abs(b) + 492*sqrt(b*d)*a^5*b^9*c^3*d^5*abs(b) - 180*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) + 20*sqrt(b*d)*a^7*b^7*c*
d^7*abs(b) + 3*sqrt(b*d)*a^8*b^6*d^8*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*b^12*c^7*abs(b) - 117*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^11*c^6*d*abs(b) + 441*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
^10*c^5*d^2*abs(b) - 309*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^9*c
^4*d^3*abs(b) - 309*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^8*c^3*d^
4*abs(b) + 441*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b^7*c^2*d^5*abs
(b) - 117*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^6*b^6*c*d^6*abs(b) - 1
5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^7*b^5*d^7*abs(b) + 30*sqrt(b*d
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^10*c^6*abs(b) + 300*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^9*c^5*d*abs(b) - 222*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^8*c^4*d^2*abs(b) - 216*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^7*c^3*d^3*abs(b) - 222*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^6*c^2*d^4*abs(b) + 300*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^
2*c + (b*x + a)*b*d - a*b*d))^4*a^5*b^5*c*d^5*abs(b) + 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b
*x + a)*b*d - a*b*d))^4*a^6*b^4*d^6*abs(b) - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*
d - a*b*d))^6*b^8*c^5*abs(b) - 398*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6
*a*b^7*c^4*d*abs(b) - 276*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^6*
c^3*d^2*abs(b) - 276*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b^5*c^2*d
^3*abs(b) - 398*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^4*b^4*c*d^4*abs(
b) - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^5*b^3*d^5*abs(b) + 15*sq
rt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^6*c^4*abs(b) + 264*sqrt(b*d)*(sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*b^5*c^3*d*abs(b) + 306*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^4*c^2*d^2*abs(b) + 264*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^3*b^3*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))^8*a^4*b^2*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))^10*b^4*c^3*abs(b) - 69*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b
*d - a*b*d))^10*a*b^3*c^2*d*abs(b) - 69*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^10*a^2*b^2*c*d^2*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))^10*a
^3*b*d^3*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*a*c))/b

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maple [B]  time = 0.02, size = 605, normalized size = 2.73 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 \sqrt {b d}\, a^{3} d^{3} x^{3} \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 )-27 \sqrt {b d}\, a^{2} b c \,d^{2} x^{3} \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 )-27 \sqrt {b d}\, a \,b^{2} c^{2} d \,x^{3} \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 )+48 \sqrt {a c}\, a \,b^{2} c \,d^{2} x^{3} \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 )+3 \sqrt {b d}\, b^{3} c^{3} x^{3} \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 \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} d^{2} x^{2}-76 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b c d \,x^{2}-6 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} c^{2} x^{2}-28 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c d x -28 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b \,c^{2} x -16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a c \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(d*x+c)^(3/2)/x^4,x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c*(48*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^3*a*b^2*c*d^2*(a*c)^(1/2)+3*(b*d)^(1/2)*a^3*d^3*x^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)-27*(b*d)^(1/2)*a^2*b*c*d^2*x^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)-27*(b*d)^(1/2)*a*b^2*c^2*d*x^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)+3*(b*d)^(1/2)*b^3*c^3*x^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)-6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*d^2*x^2-76*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*
(b*d)^(1/2)*(a*c)^(1/2)*a*b*c*d*x^2-6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b^2*c^2*x^2-28*(
b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*c*d*x-28*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2
)*(a*c)^(1/2)*a*b*c^2*x-16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*c^2)/(b*d*x^2+a*d*x+b*c
*x+a*c)^(1/2)/x^3/(b*d)^(1/2)/(a*c)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(3/2)/x^4,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 = -0.00 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^4,x)

[Out]

int(((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(d*x+c)**(3/2)/x**4,x)

[Out]

Integral((a + b*x)**(3/2)*(c + d*x)**(3/2)/x**4, x)

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