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

Optimal. Leaf size=149 \[ -\frac {\sqrt {a} (3 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{7/2}}+\frac {\sqrt {a+b x} (3 b c-5 a d)}{c^3 \sqrt {c+d x}}+\frac {(a+b x)^{3/2} (3 b c-5 a d)}{3 a c^2 (c+d x)^{3/2}}-\frac {(a+b x)^{5/2}}{a c x (c+d x)^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \begin {gather*} \frac {(a+b x)^{3/2} (3 b c-5 a d)}{3 a c^2 (c+d x)^{3/2}}+\frac {\sqrt {a+b x} (3 b c-5 a d)}{c^3 \sqrt {c+d x}}-\frac {\sqrt {a} (3 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{7/2}}-\frac {(a+b x)^{5/2}}{a c x (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 94

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

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 128, normalized size = 0.86 \begin {gather*} \frac {x (3 b c-5 a d) \left (\sqrt {c} \sqrt {a+b x} (4 a c+3 a d x+b c x)-3 a^{3/2} (c+d x)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )-3 c^{5/2} (a+b x)^{5/2}}{3 a c^{7/2} x (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 180, normalized size = 1.21 \begin {gather*} \frac {\left (5 a^{3/2} d-3 \sqrt {a} b c\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{c^{7/2}}+\frac {(a+b x)^{3/2} \left (\frac {15 a^2 d (c+d x)^2}{(a+b x)^2}+\frac {6 b c^2 (c+d x)}{a+b x}-\frac {9 a b c (c+d x)^2}{(a+b x)^2}-\frac {10 a c d (c+d x)}{a+b x}-2 c^2 d\right )}{3 c^3 (c+d x)^{3/2} \left (c-\frac {a (c+d x)}{a+b x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 3.89, size = 463, normalized size = 3.11 \begin {gather*} \left [-\frac {3 \, {\left ({\left (3 \, b c d^{2} - 5 \, a d^{3}\right )} x^{3} + 2 \, {\left (3 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2} + {\left (3 \, b c^{3} - 5 \, a c^{2} d\right )} x\right )} \sqrt {\frac {a}{c}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (3 \, a c^{2} - {\left (4 \, b c d - 15 \, a d^{2}\right )} x^{2} - 2 \, {\left (3 \, b c^{2} - 10 \, a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left (c^{3} d^{2} x^{3} + 2 \, c^{4} d x^{2} + c^{5} x\right )}}, \frac {3 \, {\left ({\left (3 \, b c d^{2} - 5 \, a d^{3}\right )} x^{3} + 2 \, {\left (3 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2} + {\left (3 \, b c^{3} - 5 \, a c^{2} d\right )} x\right )} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) - 2 \, {\left (3 \, a c^{2} - {\left (4 \, b c d - 15 \, a d^{2}\right )} x^{2} - 2 \, {\left (3 \, b c^{2} - 10 \, a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (c^{3} d^{2} x^{3} + 2 \, c^{4} d x^{2} + c^{5} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 8.56, size = 612, normalized size = 4.11 \begin {gather*} \frac {2 \, \sqrt {b x + a} {\left (\frac {2 \, {\left (b^{5} c^{5} d^{2} {\left | b \right |} - 4 \, a b^{4} c^{4} d^{3} {\left | b \right |} + 3 \, a^{2} b^{3} c^{3} d^{4} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{3} c^{7} d - a b^{2} c^{6} d^{2}} + \frac {3 \, {\left (b^{6} c^{6} d {\left | b \right |} - 4 \, a b^{5} c^{5} d^{2} {\left | b \right |} + 5 \, a^{2} b^{4} c^{4} d^{3} {\left | b \right |} - 2 \, a^{3} b^{3} c^{3} d^{4} {\left | b \right |}\right )}}{b^{3} c^{7} d - a b^{2} c^{6} d^{2}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {{\left (3 \, \sqrt {b d} a b^{3} c - 5 \, \sqrt {b d} a^{2} b^{2} d\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c^{3} {\left | b \right |}} - \frac {2 \, {\left (\sqrt {b d} a b^{5} c^{2} - 2 \, \sqrt {b d} a^{2} b^{4} c d + \sqrt {b d} a^{3} b^{3} d^{2} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{3} c - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} c^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(b*x + a)*(2*(b^5*c^5*d^2*abs(b) - 4*a*b^4*c^4*d^3*abs(b) + 3*a^2*b^3*c^3*d^4*abs(b))*(b*x + a)/(b^3*c
^7*d - a*b^2*c^6*d^2) + 3*(b^6*c^6*d*abs(b) - 4*a*b^5*c^5*d^2*abs(b) + 5*a^2*b^4*c^4*d^3*abs(b) - 2*a^3*b^3*c^
3*d^4*abs(b))/(b^3*c^7*d - a*b^2*c^6*d^2))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - (3*sqrt(b*d)*a*b^3*c - 5*sq
rt(b*d)*a^2*b^2*d)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)
)^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c^3*abs(b)) - 2*(sqrt(b*d)*a*b^5*c^2 - 2*sqrt(b*d)*a^2*b^4*c*d + sq
rt(b*d)*a^3*b^3*d^2 - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*c - sq
rt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^2*d)/((b^4*c^2 - 2*a*b^3*c*d +
a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^4)*c^3*abs(b))

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maple [B]  time = 0.03, size = 459, normalized size = 3.08 \begin {gather*} \frac {\sqrt {b x +a}\, \left (15 a^{2} d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-9 a b c \,d^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+30 a^{2} c \,d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-18 a b \,c^{2} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+15 a^{2} c^{2} d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-9 a b \,c^{3} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-30 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,d^{2} x^{2}+8 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c d \,x^{2}-40 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a c d x +12 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{2} x -6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,c^{2}\right )}{6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (d x +c \right )^{\frac {3}{2}} c^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x+a)^(1/2)/c^3*(15*a^2*d^3*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-9*a*b*c*
d^2*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)+30*a^2*c*d^2*x^2*ln((a*d*x+b*c*x+2*a*c
+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-18*a*b*c^2*d*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+
c))^(1/2))/x)+15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x*a^2*c^2*d-9*ln((a*d*x+b*c*x
+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x*a*b*c^3-30*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*d^2*x^2+8*
(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b*c*d*x^2-40*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*c*d*x+12*(a*c)^(1/2)*((
b*x+a)*(d*x+c))^(1/2)*b*c^2*x-6*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*c^2)/(a*c)^(1/2)/x/((b*x+a)*(d*x+c))^(1/
2)/(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((b*x+a)^(3/2)/x^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 {{\left (a+b\,x\right )}^{3/2}}{x^2\,{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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