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

Optimal. Leaf size=175 \[ \frac {3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{7/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (a d+3 b c)-2 b d x (5 b c-a d))}{4 b^2 d^3 (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {98, 147, 63, 217, 206} \[ \frac {3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{7/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (a d+3 b c)-2 b d x (5 b c-a d))}{4 b^2 d^3 (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(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 - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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 \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx &=-\frac {2 c x^2 \sqrt {a+b x}}{d (b c-a d) \sqrt {c+d x}}+\frac {2 \int \frac {x \left (2 a c+\frac {1}{2} (5 b c-a d) x\right )}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{d (b c-a d)}\\ &=-\frac {2 c x^2 \sqrt {a+b x}}{d (b c-a d) \sqrt {c+d x}}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (3 b c+a d)-2 b d (5 b c-a d) x)}{4 b^2 d^3 (b c-a d)}+\frac {\left (3 \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 b^2 d^3}\\ &=-\frac {2 c x^2 \sqrt {a+b x}}{d (b c-a d) \sqrt {c+d x}}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (3 b c+a d)-2 b d (5 b c-a d) x)}{4 b^2 d^3 (b c-a d)}+\frac {\left (3 \left (5 b^2 c^2+2 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 )}{4 b^3 d^3}\\ &=-\frac {2 c x^2 \sqrt {a+b x}}{d (b c-a d) \sqrt {c+d x}}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (3 b c+a d)-2 b d (5 b c-a d) x)}{4 b^2 d^3 (b c-a d)}+\frac {\left (3 \left (5 b^2 c^2+2 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 )}{4 b^3 d^3}\\ &=-\frac {2 c x^2 \sqrt {a+b x}}{d (b c-a d) \sqrt {c+d x}}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (3 b c+a d)-2 b d (5 b c-a d) x)}{4 b^2 d^3 (b c-a d)}+\frac {3 \left (5 b^2 c^2+2 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 )}{4 b^{5/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 210, normalized size = 1.20 \[ \frac {-b \sqrt {d} \sqrt {a+b x} \left (3 a^2 d^2 (c+d x)+2 a b d \left (2 c^2+c d x-d^2 x^2\right )+b^2 c \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )-3 \sqrt {b c-a d} \left (-a^3 d^3-a^2 b c d^2-3 a b^2 c^2 d+5 b^3 c^3\right ) \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{4 b^3 d^{7/2} \sqrt {c+d x} (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*Sqrt[d]*Sqrt[a + b*x]*(3*a^2*d^2*(c + d*x) + 2*a*b*d*(2*c^2 + c*d*x - d^2*x^2) + b^2*c*(-15*c^2 - 5*c*d*x
+ 2*d^2*x^2))) - 3*Sqrt[b*c - a*d]*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*Sqrt[(b*(c + d*x))/(b*
c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(4*b^3*d^(7/2)*(-(b*c) + a*d)*Sqrt[c + d*x])

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fricas [B]  time = 1.29, size = 630, normalized size = 3.60 \[ \left [\frac {3 \, {\left (5 \, b^{3} c^{4} - 3 \, a b^{2} c^{3} d - a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (5 \, b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} - a^{2} b c d^{3} - a^{3} d^{4}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (15 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} - 3 \, a^{2} b c d^{3} - 2 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{2} + {\left (5 \, b^{3} c^{2} d^{2} - 2 \, a b^{2} c d^{3} - 3 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b^{4} c^{2} d^{4} - a b^{3} c d^{5} + {\left (b^{4} c d^{5} - a b^{3} d^{6}\right )} x\right )}}, -\frac {3 \, {\left (5 \, b^{3} c^{4} - 3 \, a b^{2} c^{3} d - a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (5 \, b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} - a^{2} b c d^{3} - a^{3} d^{4}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (15 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} - 3 \, a^{2} b c d^{3} - 2 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{2} + {\left (5 \, b^{3} c^{2} d^{2} - 2 \, a b^{2} c d^{3} - 3 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b^{4} c^{2} d^{4} - a b^{3} c d^{5} + {\left (b^{4} c d^{5} - a b^{3} d^{6}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(5*b^3*c^4 - 3*a*b^2*c^3*d - a^2*b*c^2*d^2 - a^3*c*d^3 + (5*b^3*c^3*d - 3*a*b^2*c^2*d^2 - a^2*b*c*d^3
- a^3*d^4)*x)*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*(15*b^3*c^3*d - 4*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 -
2*(b^3*c*d^3 - a*b^2*d^4)*x^2 + (5*b^3*c^2*d^2 - 2*a*b^2*c*d^3 - 3*a^2*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/
(b^4*c^2*d^4 - a*b^3*c*d^5 + (b^4*c*d^5 - a*b^3*d^6)*x), -1/8*(3*(5*b^3*c^4 - 3*a*b^2*c^3*d - a^2*b*c^2*d^2 -
a^3*c*d^3 + (5*b^3*c^3*d - 3*a*b^2*c^2*d^2 - a^2*b*c*d^3 - a^3*d^4)*x)*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*(15*b^3*c^3*d
- 4*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 - 2*(b^3*c*d^3 - a*b^2*d^4)*x^2 + (5*b^3*c^2*d^2 - 2*a*b^2*c*d^3 - 3*a^2*b*
d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*c^2*d^4 - a*b^3*c*d^5 + (b^4*c*d^5 - a*b^3*d^6)*x)]

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giac [B]  time = 1.54, size = 305, normalized size = 1.74 \[ \frac {{\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{6} c d^{4} {\left | b \right |} - a b^{5} d^{5} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{9} c d^{5} - a b^{8} d^{6}} - \frac {5 \, b^{7} c^{2} d^{3} {\left | b \right |} + 2 \, a b^{6} c d^{4} {\left | b \right |} - 7 \, a^{2} b^{5} d^{5} {\left | b \right |}}{b^{9} c d^{5} - a b^{8} d^{6}}\right )} - \frac {15 \, b^{8} c^{3} d^{2} {\left | b \right |} - 9 \, a b^{7} c^{2} d^{3} {\left | b \right |} - 3 \, a^{2} b^{6} c d^{4} {\left | b \right |} + 5 \, a^{3} b^{5} d^{5} {\left | b \right |}}{b^{9} c d^{5} - a b^{8} d^{6}}\right )} \sqrt {b x + a}}{4 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {3 \, {\left (5 \, b^{2} c^{2} {\left | b \right |} + 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} b^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((b*x + a)*(2*(b^6*c*d^4*abs(b) - a*b^5*d^5*abs(b))*(b*x + a)/(b^9*c*d^5 - a*b^8*d^6) - (5*b^7*c^2*d^3*abs
(b) + 2*a*b^6*c*d^4*abs(b) - 7*a^2*b^5*d^5*abs(b))/(b^9*c*d^5 - a*b^8*d^6)) - (15*b^8*c^3*d^2*abs(b) - 9*a*b^7
*c^2*d^3*abs(b) - 3*a^2*b^6*c*d^4*abs(b) + 5*a^3*b^5*d^5*abs(b))/(b^9*c*d^5 - a*b^8*d^6))*sqrt(b*x + a)/sqrt(b
^2*c + (b*x + a)*b*d - a*b*d) - 3/4*(5*b^2*c^2*abs(b) + 2*a*b*c*d*abs(b) + a^2*d^2*abs(b))*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^3)

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maple [B]  time = 0.03, size = 673, normalized size = 3.85 \[ \frac {\sqrt {b x +a}\, \left (3 a^{3} d^{4} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a^{2} b c \,d^{3} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+9 a \,b^{2} c^{2} d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-15 b^{3} c^{3} d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a^{3} c \,d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a^{2} b \,c^{2} d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+9 a \,b^{2} c^{3} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-15 b^{3} c^{4} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b \,d^{3} x^{2}-4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c \,d^{2} x^{2}-6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} d^{3} x -4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b c \,d^{2} x +10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c^{2} d x -6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} c \,d^{2}-8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b \,c^{2} d +30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c^{3}\right )}{8 \left (a d -b c \right ) \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {d x +c}\, b^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(3*a^3*d^4*x*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))+3*a
^2*b*c*d^3*x*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))+9*a*b^2*c^2*d^2*x*ln(
1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))-15*b^3*c^3*d*x*ln(1/2*(2*b*d*x+a*d+b*
c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))+4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b*d^3*x^2-4*((b*
x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^2*c*d^2*x^2+3*a^3*c*d^3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b
*d)^(1/2))/(b*d)^(1/2))+3*a^2*b*c^2*d^2*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(
1/2))+9*a*b^2*c^3*d*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))-15*b^3*c^4*ln(
1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))-6*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)
*a^2*d^3*x-4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b*c*d^2*x+10*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^2*c^2*d*
x-6*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*c*d^2-8*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b*c^2*d+30*((b*x+a)*
(d*x+c))^(1/2)*(b*d)^(1/2)*b^2*c^3)/(a*d-b*c)/(b*d)^(1/2)/b^2/((b*x+a)*(d*x+c))^(1/2)/d^3/(d*x+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(x^3/(d*x+c)^(3/2)/(b*x+a)^(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 = -0.01 \[ \int \frac {x^3}{\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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