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

Optimal. Leaf size=213 \[ -2 a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {(a d+b c) \left (a^2 d^2-10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{3/2}}+\frac {1}{8} \sqrt {a+b x} \sqrt {c+d x} \left (\frac {a^2 d}{b}+8 a c-\frac {b c^2}{d}\right )+\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+b c)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 213, 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 = {101, 154, 157, 63, 217, 206, 93, 208} \[ -\frac {(a d+b c) \left (a^2 d^2-10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{3/2}}+\frac {1}{8} \sqrt {a+b x} \sqrt {c+d x} \left (\frac {a^2 d}{b}+8 a c-\frac {b c^2}{d}\right )-2 a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+b c)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*c - (b*c^2)/d + (a^2*d)/b)*Sqrt[a + b*x]*Sqrt[c + d*x])/8 + ((b*c + a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/
(4*d) + ((a + b*x)^(3/2)*(c + d*x)^(3/2))/3 - 2*a^(3/2)*c^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[
c + d*x])] - ((b*c + a*d)*(b^2*c^2 - 10*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d
*x])])/(8*b^(3/2)*d^(3/2))

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 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[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 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} \, dx &=\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}-\frac {1}{3} \int \frac {\sqrt {a+b x} \sqrt {c+d x} \left (-3 a c-\frac {3}{2} (b c+a d) x\right )}{x} \, dx\\ &=\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 d}+\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}-\frac {\int \frac {\sqrt {c+d x} \left (-6 a^2 c d+\frac {3}{4} \left (b^2 c^2-8 a b c d-a^2 d^2\right ) x\right )}{x \sqrt {a+b x}} \, dx}{6 d}\\ &=\frac {1}{8} \left (8 a c-\frac {b c^2}{d}+\frac {a^2 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 d}+\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}-\frac {\int \frac {-6 a^2 b c^2 d+\frac {3}{8} (b c+a d) \left (b^2 c^2-10 a b c d+a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 b d}\\ &=\frac {1}{8} \left (8 a c-\frac {b c^2}{d}+\frac {a^2 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 d}+\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}+\left (a^2 c^2\right ) \int \frac {1}{x \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}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b d}\\ &=\frac {1}{8} \left (8 a c-\frac {b c^2}{d}+\frac {a^2 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 d}+\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}+\left (2 a^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d 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}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{8 b^2 d}\\ &=\frac {1}{8} \left (8 a c-\frac {b c^2}{d}+\frac {a^2 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 d}+\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}-2 a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d 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}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 b^2 d}\\ &=\frac {1}{8} \left (8 a c-\frac {b c^2}{d}+\frac {a^2 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 d}+\frac {1}{3} (a+b x)^{3/2} (c+d x)^{3/2}-2 a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\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 {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 243, normalized size = 1.14 \[ \frac {\sqrt {d} \left (\sqrt {a+b x} (c+d x) \left (3 a^2 d^2+2 a b d (19 c+7 d x)+b^2 \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )-48 a^{3/2} b c^{3/2} d \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )-\frac {3 \sqrt {b c-a d} \left (a^3 d^3-9 a^2 b c d^2-9 a b^2 c^2 d+b^3 c^3\right ) \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{b}}{24 b d^{3/2} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*Sqrt[b*c - a*d]*(b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSi
nh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/b + Sqrt[d]*(Sqrt[a + b*x]*(c + d*x)*(3*a^2*d^2 + 2*a*b*d*(19*c +
7*d*x) + b^2*(3*c^2 + 14*c*d*x + 8*d^2*x^2)) - 48*a^(3/2)*b*c^(3/2)*d*Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a +
b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(24*b*d^(3/2)*Sqrt[c + d*x])

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fricas [A]  time = 13.29, size = 1193, normalized size = 5.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(48*sqrt(a*c)*a*b^2*c*d^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) + 3*(b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*
b*c*d^2 + a^3*d^3)*sqrt(b*d)*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*(8*b^3*d^3*x^2 + 3*b^3*c^2*d + 38*a*b^2*c*d^2
+ 3*a^2*b*d^3 + 14*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^2), 1/48*(24*sqrt(a*c)*a*b^2
*c*d^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) + 3*(b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(-
b*d)*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*(8*b^3*d^3*x^2 + 3*b^3*c^2*d + 38*a*b^2*c*d^2 + 3*a^2*b*d^3 + 14*(b^3*c*d^2 + a*b^2*d^3)*x
)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^2), 1/96*(96*sqrt(-a*c)*a*b^2*c*d^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)*s
qrt(-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^3*c^3 - 9*a*b^2*
c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(b*d)*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*(8*b^3*d^3*x^2 + 3*b^3*c^2*d + 3
8*a*b^2*c*d^2 + 3*a^2*b*d^3 + 14*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^2), 1/48*(48*s
qrt(-a*c)*a*b^2*c*d^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)) + 3*(b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(-b*d)*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*(8*b^3*d^3*x^2 + 3*b^3*c^2*d + 38*a*b^2*c*d^2 + 3*a^2*b*d^3 + 14*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x +
a)*sqrt(d*x + c))/(b^2*d^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.7

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maple [B]  time = 0.02, size = 587, normalized size = 2.76 \[ -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 \sqrt {a c}\, a^{3} d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+48 \sqrt {b d}\, a^{2} b \,c^{2} d \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-27 \sqrt {a c}\, a^{2} b c \,d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-27 \sqrt {a c}\, a \,b^{2} c^{2} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 \sqrt {a c}\, b^{3} c^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} d^{2} x^{2}-28 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b \,d^{2} x -28 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c d x -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} d^{2}-76 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b c d -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c^{2}\right )}{48 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((b*x+a)^(3/2)*(d*x+c)^(3/2)/x,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 \[ \int \frac {{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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