3.7.17 \(\int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac {\sqrt {b} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{7/2}}+\frac {b \sqrt {a+b x} \sqrt {c+d x} (5 b c-3 a d)}{d^3 (b c-a d)}-\frac {2 (a+b x)^{3/2} (5 b c-3 a d)}{3 d^2 \sqrt {c+d x} (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{3 d (c+d x)^{3/2} (b c-a d)} \]

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Rubi [A]  time = 0.09, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {78, 47, 50, 63, 217, 206} \begin {gather*} -\frac {2 (a+b x)^{3/2} (5 b c-3 a d)}{3 d^2 \sqrt {c+d x} (b c-a d)}+\frac {b \sqrt {a+b x} \sqrt {c+d x} (5 b c-3 a d)}{d^3 (b c-a d)}-\frac {\sqrt {b} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{7/2}}-\frac {2 c (a+b x)^{5/2}}{3 d (c+d x)^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]

[Out]

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

Rule 47

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx &=-\frac {2 c (a+b x)^{5/2}}{3 d (b c-a d) (c+d x)^{3/2}}+\frac {(5 b c-3 a d) \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx}{3 d (b c-a d)}\\ &=-\frac {2 c (a+b x)^{5/2}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac {2 (5 b c-3 a d) (a+b x)^{3/2}}{3 d^2 (b c-a d) \sqrt {c+d x}}+\frac {(b (5 b c-3 a d)) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{d^2 (b c-a d)}\\ &=-\frac {2 c (a+b x)^{5/2}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac {2 (5 b c-3 a d) (a+b x)^{3/2}}{3 d^2 (b c-a d) \sqrt {c+d x}}+\frac {b (5 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{d^3 (b c-a d)}-\frac {(b (5 b c-3 a d)) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 d^3}\\ &=-\frac {2 c (a+b x)^{5/2}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac {2 (5 b c-3 a d) (a+b x)^{3/2}}{3 d^2 (b c-a d) \sqrt {c+d x}}+\frac {b (5 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{d^3 (b c-a d)}-\frac {(5 b c-3 a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{d^3}\\ &=-\frac {2 c (a+b x)^{5/2}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac {2 (5 b c-3 a d) (a+b x)^{3/2}}{3 d^2 (b c-a d) \sqrt {c+d x}}+\frac {b (5 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{d^3 (b c-a d)}-\frac {(5 b c-3 a d) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{d^3}\\ &=-\frac {2 c (a+b x)^{5/2}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac {2 (5 b c-3 a d) (a+b x)^{3/2}}{3 d^2 (b c-a d) \sqrt {c+d x}}+\frac {b (5 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{d^3 (b c-a d)}-\frac {\sqrt {b} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 110, normalized size = 0.63 \begin {gather*} \frac {2 (a+b x)^{5/2} \left ((c+d x) (5 b c-3 a d) \sqrt {\frac {b (c+d x)}{b c-a d}} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {d (a+b x)}{a d-b c}\right )+5 c (a d-b c)\right )}{15 d (c+d x)^{3/2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]

[Out]

(2*(a + b*x)^(5/2)*(5*c*(-(b*c) + a*d) + (5*b*c - 3*a*d)*(c + d*x)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*Hypergeomet
ric2F1[3/2, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)]))/(15*d*(b*c - a*d)^2*(c + d*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.20, size = 180, normalized size = 1.03 \begin {gather*} \frac {\left (3 a \sqrt {b} d-5 b^{3/2} c\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{d^{7/2}}+\frac {(a+b x)^{3/2} \left (-\frac {15 b^2 c (c+d x)^2}{(a+b x)^2}-\frac {6 a d^2 (c+d x)}{a+b x}+\frac {9 a b d (c+d x)^2}{(a+b x)^2}+\frac {10 b c d (c+d x)}{a+b x}+2 c d^2\right )}{3 d^3 (c+d x)^{3/2} \left (d-\frac {b (c+d x)}{a+b x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]

[Out]

((a + b*x)^(3/2)*(2*c*d^2 + (10*b*c*d*(c + d*x))/(a + b*x) - (6*a*d^2*(c + d*x))/(a + b*x) - (15*b^2*c*(c + d*
x)^2)/(a + b*x)^2 + (9*a*b*d*(c + d*x)^2)/(a + b*x)^2))/(3*d^3*(c + d*x)^(3/2)*(d - (b*(c + d*x))/(a + b*x)))
+ ((-5*b^(3/2)*c + 3*a*Sqrt[b]*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/d^(7/2)

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fricas [A]  time = 2.07, size = 431, normalized size = 2.48 \begin {gather*} \left [-\frac {3 \, {\left (5 \, b c^{3} - 3 \, a c^{2} d + {\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} x^{2} + 2 \, {\left (5 \, b c^{2} d - 3 \, a c d^{2}\right )} x\right )} \sqrt {\frac {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^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (3 \, b d^{2} x^{2} + 15 \, b c^{2} - 4 \, a c d + 2 \, {\left (10 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}}, \frac {3 \, {\left (5 \, b c^{3} - 3 \, a c^{2} d + {\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} x^{2} + 2 \, {\left (5 \, b c^{2} d - 3 \, a c d^{2}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (3 \, b d^{2} x^{2} + 15 \, b c^{2} - 4 \, a c d + 2 \, {\left (10 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(5*b*c^3 - 3*a*c^2*d + (5*b*c*d^2 - 3*a*d^3)*x^2 + 2*(5*b*c^2*d - 3*a*c*d^2)*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^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d)
+ 8*(b^2*c*d + a*b*d^2)*x) - 4*(3*b*d^2*x^2 + 15*b*c^2 - 4*a*c*d + 2*(10*b*c*d - 3*a*d^2)*x)*sqrt(b*x + a)*sqr
t(d*x + c))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3), 1/6*(3*(5*b*c^3 - 3*a*c^2*d + (5*b*c*d^2 - 3*a*d^3)*x^2 + 2*(5*b*
c^2*d - 3*a*c*d^2)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*
d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) + 2*(3*b*d^2*x^2 + 15*b*c^2 - 4*a*c*d + 2*(10*b*c*d - 3*a*d^2)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)]

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giac [A]  time = 1.89, size = 286, normalized size = 1.64 \begin {gather*} \frac {{\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (b^{5} c d^{4} {\left | b \right |} - a b^{4} d^{5} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{4} c d^{5} - a b^{3} d^{6}} + \frac {4 \, {\left (5 \, b^{6} c^{2} d^{3} {\left | b \right |} - 8 \, a b^{5} c d^{4} {\left | b \right |} + 3 \, a^{2} b^{4} d^{5} {\left | b \right |}\right )}}{b^{4} c d^{5} - a b^{3} d^{6}}\right )} + \frac {3 \, {\left (5 \, b^{7} c^{3} d^{2} {\left | b \right |} - 13 \, a b^{6} c^{2} d^{3} {\left | b \right |} + 11 \, a^{2} b^{5} c d^{4} {\left | b \right |} - 3 \, a^{3} b^{4} d^{5} {\left | b \right |}\right )}}{b^{4} c d^{5} - a b^{3} d^{6}}\right )} \sqrt {b x + a}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, b c {\left | b \right |} - 3 \, a d {\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 )}{\sqrt {b d} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/3*((b*x + a)*(3*(b^5*c*d^4*abs(b) - a*b^4*d^5*abs(b))*(b*x + a)/(b^4*c*d^5 - a*b^3*d^6) + 4*(5*b^6*c^2*d^3*a
bs(b) - 8*a*b^5*c*d^4*abs(b) + 3*a^2*b^4*d^5*abs(b))/(b^4*c*d^5 - a*b^3*d^6)) + 3*(5*b^7*c^3*d^2*abs(b) - 13*a
*b^6*c^2*d^3*abs(b) + 11*a^2*b^5*c*d^4*abs(b) - 3*a^3*b^4*d^5*abs(b))/(b^4*c*d^5 - a*b^3*d^6))*sqrt(b*x + a)/(
b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + (5*b*c*abs(b) - 3*a*d*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)*d^3)

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maple [B]  time = 0.02, size = 459, normalized size = 2.64 \begin {gather*} \frac {\sqrt {b x +a}\, \left (9 a b \,d^{3} x^{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 )-15 b^{2} c \,d^{2} x^{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 )+18 a b c \,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 )-30 b^{2} c^{2} 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 )+9 a b \,c^{2} 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^{2} c^{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 )+6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,d^{2} x^{2}-12 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,d^{2} x +40 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b c d x -8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a c d +30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{2}\right )}{6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (d x +c \right )^{\frac {3}{2}} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x+a)^(3/2)/(d*x+c)^(5/2),x)

[Out]

1/6*(b*x+a)^(1/2)*(9*a*b*d^3*x^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))-1
5*b^2*c*d^2*x^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))+18*a*b*c*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))-30*b^2*c^2*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))+6*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*d^2*x^2+9*a*b*c^
2*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^2*c^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))-12*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*d^2*x+40*((b
*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*c*d*x-8*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*c*d+30*((b*x+a)*(d*x+c))^(1/2
)*(b*d)^(1/2)*b*c^2)/(b*d)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/d^3/(d*x+c)^(3/2)

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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(x*(b*x+a)^(3/2)/(d*x+c)^(5/2),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 \begin {gather*} \int \frac {x\,{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*x)^(3/2))/(c + d*x)^(5/2),x)

[Out]

int((x*(a + b*x)^(3/2))/(c + d*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)**(3/2)/(d*x+c)**(5/2),x)

[Out]

Timed out

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