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

Optimal. Leaf size=258 \[ -\frac {3 (a d+b c) (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{7/2} d^{7/2}}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (a d+b c) (b c-a d)^3}{128 b^3 d^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+b c) (b c-a d)^2}{64 b^3 d^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (a d+b c) (b c-a d)}{16 b^3 d}-\frac {(a+b x)^{5/2} (c+d x)^{3/2} (a d+b c)}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d} \]

[Out]

-1/8*(a*d+b*c)*(b*x+a)^(5/2)*(d*x+c)^(3/2)/b^2/d+1/5*(b*x+a)^(5/2)*(d*x+c)^(5/2)/b/d-3/128*(-a*d+b*c)^4*(a*d+b
*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)/d^(7/2)-1/64*(-a*d+b*c)^2*(a*d+b*c)*(b*x+a)^(
3/2)*(d*x+c)^(1/2)/b^3/d^2-1/16*(-a*d+b*c)*(a*d+b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b^3/d+3/128*(-a*d+b*c)^3*(a*d
+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d^3

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Rubi [A]  time = 0.15, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {80, 50, 63, 217, 206} \[ \frac {3 \sqrt {a+b x} \sqrt {c+d x} (a d+b c) (b c-a d)^3}{128 b^3 d^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+b c) (b c-a d)^2}{64 b^3 d^2}-\frac {3 (a d+b c) (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{7/2} d^{7/2}}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (a d+b c) (b c-a d)}{16 b^3 d}-\frac {(a+b x)^{5/2} (c+d x)^{3/2} (a d+b c)}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*c - a*d)^3*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(128*b^3*d^3) - ((b*c - a*d)^2*(b*c + a*d)*(a + b*x)
^(3/2)*Sqrt[c + d*x])/(64*b^3*d^2) - ((b*c - a*d)*(b*c + a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(16*b^3*d) - ((b*
c + a*d)*(a + b*x)^(5/2)*(c + d*x)^(3/2))/(8*b^2*d) + ((a + b*x)^(5/2)*(c + d*x)^(5/2))/(5*b*d) - (3*(b*c - a*
d)^4*(b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(128*b^(7/2)*d^(7/2))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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 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 x (a+b x)^{3/2} (c+d x)^{3/2} \, dx &=\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {(b c+a d) \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx}{2 b d}\\ &=-\frac {(b c+a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {\left (3 \left (c^2-\frac {a^2 d^2}{b^2}\right )\right ) \int (a+b x)^{3/2} \sqrt {c+d x} \, dx}{16 d}\\ &=-\frac {(b c-a d) (b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{16 b^3 d}-\frac {(b c+a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {\left ((b c-a d)^2 (b c+a d)\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{32 b^3 d}\\ &=-\frac {(b c-a d)^2 (b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{64 b^3 d^2}-\frac {(b c-a d) (b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{16 b^3 d}-\frac {(b c+a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}+\frac {\left (3 (b c-a d)^3 (b c+a d)\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{128 b^3 d^2}\\ &=\frac {3 (b c-a d)^3 (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{128 b^3 d^3}-\frac {(b c-a d)^2 (b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{64 b^3 d^2}-\frac {(b c-a d) (b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{16 b^3 d}-\frac {(b c+a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {\left (3 (b c-a d)^4 (b c+a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{256 b^3 d^3}\\ &=\frac {3 (b c-a d)^3 (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{128 b^3 d^3}-\frac {(b c-a d)^2 (b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{64 b^3 d^2}-\frac {(b c-a d) (b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{16 b^3 d}-\frac {(b c+a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {\left (3 (b c-a d)^4 (b c+a d)\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 )}{128 b^4 d^3}\\ &=\frac {3 (b c-a d)^3 (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{128 b^3 d^3}-\frac {(b c-a d)^2 (b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{64 b^3 d^2}-\frac {(b c-a d) (b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{16 b^3 d}-\frac {(b c+a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {\left (3 (b c-a d)^4 (b c+a d)\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 )}{128 b^4 d^3}\\ &=\frac {3 (b c-a d)^3 (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{128 b^3 d^3}-\frac {(b c-a d)^2 (b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{64 b^3 d^2}-\frac {(b c-a d) (b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{16 b^3 d}-\frac {(b c+a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {3 (b c-a d)^4 (b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{7/2} d^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.77, size = 250, normalized size = 0.97 \[ \frac {b \sqrt {d} \sqrt {a+b x} (c+d x) \left (15 a^4 d^4-10 a^3 b d^3 (4 c+d x)+2 a^2 b^2 d^2 \left (9 c^2+13 c d x+4 d^2 x^2\right )+2 a b^3 d \left (-20 c^3+13 c^2 d x+136 c d^2 x^2+88 d^3 x^3\right )+b^4 \left (15 c^4-10 c^3 d x+8 c^2 d^2 x^2+176 c d^3 x^3+128 d^4 x^4\right )\right )-15 (b c-a d)^{9/2} (a d+b c) \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{640 b^4 d^{7/2} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(15*a^4*d^4 - 10*a^3*b*d^3*(4*c + d*x) + 2*a^2*b^2*d^2*(9*c^2 + 13*c*d*x +
4*d^2*x^2) + 2*a*b^3*d*(-20*c^3 + 13*c^2*d*x + 136*c*d^2*x^2 + 88*d^3*x^3) + b^4*(15*c^4 - 10*c^3*d*x + 8*c^2*
d^2*x^2 + 176*c*d^3*x^3 + 128*d^4*x^4)) - 15*(b*c - a*d)^(9/2)*(b*c + a*d)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*Arc
Sinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(640*b^4*d^(7/2)*Sqrt[c + d*x])

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fricas [A]  time = 1.46, size = 694, normalized size = 2.69 \[ \left [\frac {15 \, {\left (b^{5} c^{5} - 3 \, a b^{4} c^{4} d + 2 \, a^{2} b^{3} c^{3} d^{2} + 2 \, a^{3} b^{2} c^{2} d^{3} - 3 \, a^{4} b c d^{4} + a^{5} d^{5}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (128 \, b^{5} d^{5} x^{4} + 15 \, b^{5} c^{4} d - 40 \, a b^{4} c^{3} d^{2} + 18 \, a^{2} b^{3} c^{2} d^{3} - 40 \, a^{3} b^{2} c d^{4} + 15 \, a^{4} b d^{5} + 176 \, {\left (b^{5} c d^{4} + a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (b^{5} c^{2} d^{3} + 34 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} - 2 \, {\left (5 \, b^{5} c^{3} d^{2} - 13 \, a b^{4} c^{2} d^{3} - 13 \, a^{2} b^{3} c d^{4} + 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2560 \, b^{4} d^{4}}, \frac {15 \, {\left (b^{5} c^{5} - 3 \, a b^{4} c^{4} d + 2 \, a^{2} b^{3} c^{3} d^{2} + 2 \, a^{3} b^{2} c^{2} d^{3} - 3 \, a^{4} b c d^{4} + a^{5} d^{5}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (128 \, b^{5} d^{5} x^{4} + 15 \, b^{5} c^{4} d - 40 \, a b^{4} c^{3} d^{2} + 18 \, a^{2} b^{3} c^{2} d^{3} - 40 \, a^{3} b^{2} c d^{4} + 15 \, a^{4} b d^{5} + 176 \, {\left (b^{5} c d^{4} + a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (b^{5} c^{2} d^{3} + 34 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} - 2 \, {\left (5 \, b^{5} c^{3} d^{2} - 13 \, a b^{4} c^{2} d^{3} - 13 \, a^{2} b^{3} c d^{4} + 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{1280 \, b^{4} d^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2560*(15*(b^5*c^5 - 3*a*b^4*c^4*d + 2*a^2*b^3*c^3*d^2 + 2*a^3*b^2*c^2*d^3 - 3*a^4*b*c*d^4 + a^5*d^5)*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*(128*b^5*d^5*x^4 + 15*b^5*c^4*d - 40*a*b^4*c^3*d^2 + 18*a^2*b^3*c^2*d^3
- 40*a^3*b^2*c*d^4 + 15*a^4*b*d^5 + 176*(b^5*c*d^4 + a*b^4*d^5)*x^3 + 8*(b^5*c^2*d^3 + 34*a*b^4*c*d^4 + a^2*b
^3*d^5)*x^2 - 2*(5*b^5*c^3*d^2 - 13*a*b^4*c^2*d^3 - 13*a^2*b^3*c*d^4 + 5*a^3*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*
x + c))/(b^4*d^4), 1/1280*(15*(b^5*c^5 - 3*a*b^4*c^4*d + 2*a^2*b^3*c^3*d^2 + 2*a^3*b^2*c^2*d^3 - 3*a^4*b*c*d^4
+ a^5*d^5)*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*(128*b^5*d^5*x^4 + 15*b^5*c^4*d - 40*a*b^4*c^3*d^2 + 18*a^2*b^3*c^2*d^3
- 40*a^3*b^2*c*d^4 + 15*a^4*b*d^5 + 176*(b^5*c*d^4 + a*b^4*d^5)*x^3 + 8*(b^5*c^2*d^3 + 34*a*b^4*c*d^4 + a^2*b^
3*d^5)*x^2 - 2*(5*b^5*c^3*d^2 - 13*a*b^4*c^2*d^3 - 13*a^2*b^3*c*d^4 + 5*a^3*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x
+ c))/(b^4*d^4)]

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giac [B]  time = 3.53, size = 1511, normalized size = 5.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*(10*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 + (b^12*c*d^5 - 25*
a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^
3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3
*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 35*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x
+ a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3))*c*abs(b) + 160*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2
*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*d^3 - 11*a^2*
b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) +
sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d^2))*a*c*abs(b)/b + (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*
(2*(4*(b*x + a)*(6*(b*x + a)*(8*(b*x + a)/b^4 + (b^20*c*d^7 - 41*a*b^19*d^8)/(b^23*d^8)) - (7*b^21*c^2*d^6 + 2
6*a*b^20*c*d^7 - 513*a^2*b^19*d^8)/(b^23*d^8)) + 5*(7*b^22*c^3*d^5 + 19*a*b^21*c^2*d^6 + 37*a^2*b^20*c*d^7 - 4
47*a^3*b^19*d^8)/(b^23*d^8))*(b*x + a) - 15*(7*b^23*c^4*d^4 + 12*a*b^22*c^3*d^5 + 18*a^2*b^21*c^2*d^6 + 28*a^3
*b^20*c*d^7 - 193*a^4*b^19*d^8)/(b^23*d^8))*sqrt(b*x + a) - 15*(7*b^5*c^5 + 5*a*b^4*c^4*d + 6*a^2*b^3*c^3*d^2
+ 10*a^3*b^2*c^2*d^3 + 35*a^4*b*c*d^4 - 63*a^5*d^5)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*
b*d - a*b*d)))/(sqrt(b*d)*b^3*d^4))*d*abs(b) + 80*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x +
a)*(4*(b*x + a)/b^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*d^3 - 11*a^2*b^5*d^4
)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b
^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d^2))*a^2*d*abs(b)/b^2 + 20*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*
(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2*d^4 + 14*a
*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^
3*b^11*d^6)/(b^14*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 35
*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3))*a*d*ab
s(b)/b + 480*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^3*c
^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt
(b*d)*d))*a^2*c*abs(b)/b^3)/b

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maple [B]  time = 0.02, size = 942, normalized size = 3.65 \[ -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-256 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{4} d^{4} x^{4}+15 a^{5} d^{5} \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 )-45 a^{4} b c \,d^{4} \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 )+30 a^{3} b^{2} c^{2} 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 )+30 a^{2} b^{3} c^{3} 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 )-45 a \,b^{4} c^{4} 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 )-352 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a \,b^{3} d^{4} x^{3}+15 b^{5} c^{5} \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 )-352 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{4} c \,d^{3} x^{3}-16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a^{2} b^{2} d^{4} x^{2}-544 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a \,b^{3} c \,d^{3} x^{2}-16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{4} c^{2} d^{2} x^{2}+20 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{3} b \,d^{4} x -52 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} b^{2} c \,d^{3} x -52 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a \,b^{3} c^{2} d^{2} x +20 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{4} c^{3} d x -30 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{4} d^{4}+80 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{3} b c \,d^{3}-36 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} b^{2} c^{2} d^{2}+80 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a \,b^{3} c^{3} d -30 \sqrt {b d}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{4} c^{4}\right )}{1280 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1280*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-256*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*b^4*d^4*x^4-352*(b*d*x^2
+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a*b^3*d^4*x^3-352*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*b^4*c*d^3*x^
3-16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a^2*b^2*d^4*x^2-544*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/
2)*a*b^3*c*d^3*x^2-16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*b^4*c^2*d^2*x^2+15*a^5*d^5*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))-45*a^4*b*c*d^4*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))+30*a^3*b^2*c^2*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))+30*a^2*b^3*c^3*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))-45*a*b^4*c^4*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))+15*b^5*c^5*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))+20*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*b*d^4*x-52*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x
+a*c)^(1/2)*a^2*b^2*c*d^3*x-52*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*b^3*c^2*d^2*x+20*(b*d)^(1/2)*(b*d
*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^4*c^3*d*x-30*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^4*d^4+80*(b*d)^(1/2)*
(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*b*c*d^3-36*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*b^2*c^2*d^2+80*
(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*b^3*c^3*d-30*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^4*c^4
)/b^3/d^3/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/(b*d)^(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(x*(b*x+a)^(3/2)*(d*x+c)^(3/2),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 zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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