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

Optimal. Leaf size=126 \[ -\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (a+b x)^{3/2} (b c-a d)}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} \sqrt {d}}+\frac {4 a \sqrt {c+d x} (3 b c-2 a d)}{3 b^2 \sqrt {a+b x} (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 126, 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 = {89, 78, 63, 217, 206} \[ -\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (a+b x)^{3/2} (b c-a d)}+\frac {4 a \sqrt {c+d x} (3 b c-2 a d)}{3 b^2 \sqrt {a+b x} (b c-a d)^2}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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^2}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx &=-\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {2 \int \frac {-\frac {1}{2} a (3 b c-a d)+\frac {3}{2} b (b c-a d) x}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx}{3 b^2 (b c-a d)}\\ &=-\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-2 a d) \sqrt {c+d x}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{b^2}\\ &=-\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-2 a d) \sqrt {c+d x}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{b^3}\\ &=-\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-2 a d) \sqrt {c+d x}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{b^3}\\ &=-\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-2 a d) \sqrt {c+d x}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 198, normalized size = 1.57 \[ \frac {2 \sqrt {c+d x} \left (\frac {(a+b x) \left (3 b^2 c^2-a^2 d^2\right )}{d (b c-a d)^2}+\frac {a^2}{a d-b c}-\frac {3 (a+b x) \left (\sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}-\sqrt {d} \sqrt {a+b x} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )}{d \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{3 b^2 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c + d*x]*(a^2/(-(b*c) + a*d) + ((3*b^2*c^2 - a^2*d^2)*(a + b*x))/(d*(b*c - a*d)^2) - (3*(a + b*x)*(Sqr
t[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a*d)] - Sqrt[d]*Sqrt[a + b*x]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c
- a*d]]))/(d*Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a*d)])))/(3*b^2*(a + b*x)^(3/2))

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fricas [B]  time = 1.19, size = 670, normalized size = 5.32 \[ \left [\frac {3 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (5 \, a^{2} b^{2} c d - 3 \, a^{3} b d^{2} + 2 \, {\left (3 \, a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (a^{2} b^{5} c^{2} d - 2 \, a^{3} b^{4} c d^{2} + a^{4} b^{3} d^{3} + {\left (b^{7} c^{2} d - 2 \, a b^{6} c d^{2} + a^{2} b^{5} d^{3}\right )} x^{2} + 2 \, {\left (a b^{6} c^{2} d - 2 \, a^{2} b^{5} c d^{2} + a^{3} b^{4} d^{3}\right )} x\right )}}, -\frac {3 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (5 \, a^{2} b^{2} c d - 3 \, a^{3} b d^{2} + 2 \, {\left (3 \, a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{5} c^{2} d - 2 \, a^{3} b^{4} c d^{2} + a^{4} b^{3} d^{3} + {\left (b^{7} c^{2} d - 2 \, a b^{6} c d^{2} + a^{2} b^{5} d^{3}\right )} x^{2} + 2 \, {\left (a b^{6} c^{2} d - 2 \, a^{2} b^{5} c d^{2} + a^{3} b^{4} d^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x+a)^(5/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^2 + 2*(a*b^3*c^2 - 2*a^
2*b^2*c*d + a^3*b*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*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*(5*a^2*b^2*c*d - 3*a^3*b*d^2 + 2*(3*a*b
^3*c*d - 2*a^2*b^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^5*c^2*d - 2*a^3*b^4*c*d^2 + a^4*b^3*d^3 + (b^7*
c^2*d - 2*a*b^6*c*d^2 + a^2*b^5*d^3)*x^2 + 2*(a*b^6*c^2*d - 2*a^2*b^5*c*d^2 + a^3*b^4*d^3)*x), -1/3*(3*(a^2*b^
2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^2 + 2*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3
*b*d^2)*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*(5*a^2*b^2*c*d - 3*a^3*b*d^2 + 2*(3*a*b^3*c*d - 2*a^2*b^2*d^2)*x)*sqrt(b*
x + a)*sqrt(d*x + c))/(a^2*b^5*c^2*d - 2*a^3*b^4*c*d^2 + a^4*b^3*d^3 + (b^7*c^2*d - 2*a*b^6*c*d^2 + a^2*b^5*d^
3)*x^2 + 2*(a*b^6*c^2*d - 2*a^2*b^5*c*d^2 + a^3*b^4*d^3)*x)]

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giac [B]  time = 2.43, size = 313, normalized size = 2.48 \[ -\frac {\log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{\sqrt {b d} b {\left | b \right |}} + \frac {8 \, {\left (3 \, \sqrt {b d} a b^{4} c^{2} - 5 \, \sqrt {b d} a^{2} b^{3} c d + 2 \, \sqrt {b d} a^{3} b^{2} d^{2} - 6 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{2} c + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b d + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a\right )}}{3 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3} b {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x+a)^(5/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 604, normalized size = 4.79 \[ \frac {\sqrt {d x +c}\, \left (3 a^{2} b^{2} d^{2} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-6 a \,b^{3} c d \,x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{4} c^{2} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+6 a^{3} b \,d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-12 a^{2} b^{2} c d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+6 a \,b^{3} c^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a^{4} d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-6 a^{3} b c d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a^{2} b^{2} c^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-8 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b d x +12 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,b^{2} c x -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} d +10 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b c \right )}{3 \sqrt {b d}\, \left (a d -b c \right )^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (b x +a \right )^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x+a)^(5/2)/(d*x+c)^(1/2),x)

[Out]

1/3*(d*x+c)^(1/2)*(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))*x^2*a^2*b^2*d
^2-6*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))*x^2*a*b^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))*x^2*b^4*c^2+6*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))*x*a^3*b*d^2-12*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))*x*a^2*b^2*c*d+6*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))*x*a*b^3*c^2+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))*a^4*d^2-6*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))*a^3*b*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))*a^2*b^2*c^2-8*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^2*b*d+
12*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a*b^2*c-6*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*d+10*(b*d)^(1/2)*((
b*x+a)*(d*x+c))^(1/2)*a^2*b*c)/(b*d)^(1/2)/(a*d-b*c)^2/((b*x+a)*(d*x+c))^(1/2)/b^2/(b*x+a)^(3/2)

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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^2/(b*x+a)^(5/2)/(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 = -0.01 \[ \int \frac {x^2}{{\left (a+b\,x\right )}^{5/2}\,\sqrt {c+d\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a + b*x)^(5/2)*(c + d*x)^(1/2)),x)

[Out]

int(x^2/((a + b*x)^(5/2)*(c + d*x)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a + b x\right )^{\frac {5}{2}} \sqrt {c + d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x+a)**(5/2)/(d*x+c)**(1/2),x)

[Out]

Integral(x**2/((a + b*x)**(5/2)*sqrt(c + d*x)), x)

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