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

Optimal. Leaf size=221 \[ -\frac {(3 a d+5 b c) (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{7/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+5 b c) (b c-a d)^2}{64 b^2 d^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (3 a d+5 b c) (b c-a d)}{96 b^2 d^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (3 a d+5 b c)}{24 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d} \]

[Out]

1/4*(b*x+a)^(5/2)*(d*x+c)^(3/2)/b/d-1/64*(-a*d+b*c)^3*(3*a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x
+c)^(1/2))/b^(5/2)/d^(7/2)-1/96*(-a*d+b*c)*(3*a*d+5*b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b^2/d^2-1/24*(3*a*d+5*b*c
)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b^2/d+1/64*(-a*d+b*c)^2*(3*a*d+5*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2/d^3

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Rubi [A]  time = 0.13, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+5 b c) (b c-a d)^2}{64 b^2 d^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (3 a d+5 b c) (b c-a d)}{96 b^2 d^2}-\frac {(3 a d+5 b c) (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{7/2}}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (3 a d+5 b c)}{24 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(5*b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^2*d^3) - ((b*c - a*d)*(5*b*c + 3*a*d)*(a + b
*x)^(3/2)*Sqrt[c + d*x])/(96*b^2*d^2) - ((5*b*c + 3*a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*b^2*d) + ((a + b*x
)^(5/2)*(c + d*x)^(3/2))/(4*b*d) - ((b*c - a*d)^3*(5*b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqr
t[c + d*x])])/(64*b^(5/2)*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} \sqrt {c+d x} \, dx &=\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d}+\frac {\left (-\frac {5 b c}{2}-\frac {3 a d}{2}\right ) \int (a+b x)^{3/2} \sqrt {c+d x} \, dx}{4 b d}\\ &=-\frac {(5 b c+3 a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d}-\frac {((b c-a d) (5 b c+3 a d)) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{48 b^2 d}\\ &=-\frac {(b c-a d) (5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b^2 d^2}-\frac {(5 b c+3 a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d}+\frac {\left ((b c-a d)^2 (5 b c+3 a d)\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{64 b^2 d^2}\\ &=\frac {(b c-a d)^2 (5 b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d^3}-\frac {(b c-a d) (5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b^2 d^2}-\frac {(5 b c+3 a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d}-\frac {\left ((b c-a d)^3 (5 b c+3 a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 b^2 d^3}\\ &=\frac {(b c-a d)^2 (5 b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d^3}-\frac {(b c-a d) (5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b^2 d^2}-\frac {(5 b c+3 a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d}-\frac {\left ((b c-a d)^3 (5 b c+3 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 )}{64 b^3 d^3}\\ &=\frac {(b c-a d)^2 (5 b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d^3}-\frac {(b c-a d) (5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b^2 d^2}-\frac {(5 b c+3 a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d}-\frac {\left ((b c-a d)^3 (5 b c+3 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 )}{64 b^3 d^3}\\ &=\frac {(b c-a d)^2 (5 b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d^3}-\frac {(b c-a d) (5 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b^2 d^2}-\frac {(5 b c+3 a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b d}-\frac {(b c-a d)^3 (5 b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.57, size = 299, normalized size = 1.35 \[ \frac {(a+b x)^{5/2} (c+d x)^{3/2} \left (5-\frac {5 (3 a d+5 b c) \left (8 b^3 d^3 (a+b x)^3 \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}-b (b c-a d) \left (-2 b^2 d^2 (a+b x)^2 \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}+3 b^2 d (a+b x) (b c-a d)^{3/2} \sqrt {\frac {b (c+d x)}{b c-a d}}-3 b^2 \sqrt {d} \sqrt {a+b x} (b c-a d)^2 \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )\right )}{48 b^3 d^3 (a+b x)^3 (b c-a d)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2}}\right )}{20 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(5/2)*(c + d*x)^(3/2)*(5 - (5*(5*b*c + 3*a*d)*(8*b^3*d^3*Sqrt[b*c - a*d]*(a + b*x)^3*Sqrt[(b*(c + d
*x))/(b*c - a*d)] - b*(b*c - a*d)*(3*b^2*d*(b*c - a*d)^(3/2)*(a + b*x)*Sqrt[(b*(c + d*x))/(b*c - a*d)] - 2*b^2
*d^2*Sqrt[b*c - a*d]*(a + b*x)^2*Sqrt[(b*(c + d*x))/(b*c - a*d)] - 3*b^2*Sqrt[d]*(b*c - a*d)^2*Sqrt[a + b*x]*A
rcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])))/(48*b^3*d^3*(b*c - a*d)^(3/2)*(a + b*x)^3*((b*(c + d*x))/(b
*c - a*d))^(3/2))))/(20*b*d)

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fricas [A]  time = 0.83, size = 544, normalized size = 2.46 \[ \left [-\frac {3 \, {\left (5 \, b^{4} c^{4} - 12 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a^{3} b c d^{3} - 3 \, a^{4} d^{4}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d - 31 \, a b^{3} c^{2} d^{2} + 9 \, a^{2} b^{2} c d^{3} - 9 \, a^{3} b d^{4} + 8 \, {\left (b^{4} c d^{3} + 9 \, a b^{3} d^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{4} c^{2} d^{2} - 10 \, a b^{3} c d^{3} - 3 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b^{3} d^{4}}, \frac {3 \, {\left (5 \, b^{4} c^{4} - 12 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a^{3} b c d^{3} - 3 \, a^{4} d^{4}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d - 31 \, a b^{3} c^{2} d^{2} + 9 \, a^{2} b^{2} c d^{3} - 9 \, a^{3} b d^{4} + 8 \, {\left (b^{4} c d^{3} + 9 \, a b^{3} d^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{4} c^{2} d^{2} - 10 \, a b^{3} c d^{3} - 3 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b^{3} d^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(3*(5*b^4*c^4 - 12*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - 3*a^4*d^4)*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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d - 31*a*b^3*c^2*d^2 + 9*a^2*b^2*c*d^3 - 9*a^3*b*d^4 + 8*(
b^4*c*d^3 + 9*a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 10*a*b^3*c*d^3 - 3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x +
c))/(b^3*d^4), 1/384*(3*(5*b^4*c^4 - 12*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - 3*a^4*d^4)*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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d - 31*a*b^3*c^2*d^2 + 9*a^2*b^2*c*d^3 - 9*a^3*b*d^4 + 8*(b^4*c*
d^3 + 9*a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 10*a*b^3*c*d^3 - 3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b
^3*d^4)]

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giac [B]  time = 1.58, size = 632, normalized size = 2.86 \[ \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{b^{3}} + \frac {b^{12} c d^{5} - 25 \, a b^{11} d^{6}}{b^{14} d^{6}}\right )} - \frac {5 \, b^{13} c^{2} d^{4} + 14 \, a b^{12} c d^{5} - 163 \, a^{2} b^{11} d^{6}}{b^{14} d^{6}}\right )} + \frac {3 \, {\left (5 \, b^{14} c^{3} d^{3} + 9 \, a b^{13} c^{2} d^{4} + 15 \, a^{2} b^{12} c d^{5} - 93 \, a^{3} b^{11} d^{6}\right )}}{b^{14} d^{6}}\right )} \sqrt {b x + a} + \frac {3 \, {\left (5 \, b^{4} c^{4} + 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 20 \, a^{3} b c d^{3} - 35 \, a^{4} d^{4}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{2} d^{3}}\right )} {\left | b \right |} + \frac {16 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} + \frac {b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} a {\left | b \right |}}{b} + \frac {48 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} a^{2} {\left | b \right |}}{b^{3}}}{192 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*((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))*abs(b) + 16*(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*abs(b)/b + 48*(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*abs(b)/b^3)/b

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maple [B]  time = 0.02, size = 686, normalized size = 3.10 \[ \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (9 a^{4} d^{4} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-12 a^{3} b c \,d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-18 a^{2} b^{2} c^{2} d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+36 a \,b^{3} c^{3} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-15 b^{4} c^{4} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+96 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{3} d^{3} x^{3}+144 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a \,b^{2} d^{3} x^{2}+16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{3} c \,d^{2} x^{2}+12 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a^{2} b \,d^{3} x +40 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a \,b^{2} c \,d^{2} x -20 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{3} c^{2} d x -18 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a^{3} d^{3}+18 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a^{2} b c \,d^{2}-62 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, a \,b^{2} c^{2} d +30 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{3} c^{3}\right )}{384 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, b^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(96*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*b^3*d^3*x^3+144*(b*d*x^2+a*d
*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a*b^2*d^3*x^2+16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*b^3*c*d^2*x^2+9*a
^4*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))-12*a^3*b*c*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))-18*a^2*b^2*c^2*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))+36*a*b^3*c^3*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^4*c^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))+12*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a^2*b*d^3*x+40*(
b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a*b^2*c*d^2*x-20*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*b^3*c^
2*d*x-18*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a^3*d^3+18*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a^
2*b*c*d^2-62*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a*b^2*c^2*d+30*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^
(1/2)*b^3*c^3)/b^2/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/d^3/(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)^(1/2),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 \[ \int x\,{\left (a+b\,x\right )}^{3/2}\,\sqrt {c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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