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

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

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Rubi [A]  time = 0.09, 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} \begin {gather*} \frac {2 c^2 \sqrt {a+b x}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {4 c \sqrt {a+b x} (2 b c-3 a d)}{3 d^2 \sqrt {c+d 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 )}{\sqrt {b} d^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.32, size = 148, normalized size = 1.17 \begin {gather*} \frac {2 \left (-\frac {3 (b c-a d)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{b^2 \sqrt {d}}-c^2 \sqrt {a+b x}+\frac {2 c \sqrt {a+b x} (c+d x) (2 b c-3 a d)}{b c-a d}\right )}{3 d^2 (c+d x)^{3/2} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 103, normalized size = 0.82 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{5/2}}-\frac {2 c \sqrt {a+b x} \left (\frac {c d (a+b x)}{c+d x}-6 a d+3 b c\right )}{3 d^2 \sqrt {c+d x} (a d-b c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*Sqrt[a + b*x]*(3*b*c - 6*a*d + (c*d*(a + b*x))/(c + d*x)))/(3*d^2*(-(b*c) + a*d)^2*Sqrt[c + d*x]) + (2*A
rcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(5/2))

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fricas [B]  time = 1.85, size = 670, normalized size = 5.32 \begin {gather*} \left [\frac {3 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\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 (3 \, b^{2} c^{3} d - 5 \, a b c^{2} d^{2} + 2 \, {\left (2 \, b^{2} c^{2} d^{2} - 3 \, a b c d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (b^{3} c^{4} d^{3} - 2 \, a b^{2} c^{3} d^{4} + a^{2} b c^{2} d^{5} + {\left (b^{3} c^{2} d^{5} - 2 \, a b^{2} c d^{6} + a^{2} b d^{7}\right )} x^{2} + 2 \, {\left (b^{3} c^{3} d^{4} - 2 \, a b^{2} c^{2} d^{5} + a^{2} b c d^{6}\right )} x\right )}}, -\frac {3 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\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 (3 \, b^{2} c^{3} d - 5 \, a b c^{2} d^{2} + 2 \, {\left (2 \, b^{2} c^{2} d^{2} - 3 \, a b c d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (b^{3} c^{4} d^{3} - 2 \, a b^{2} c^{3} d^{4} + a^{2} b c^{2} d^{5} + {\left (b^{3} c^{2} d^{5} - 2 \, a b^{2} c d^{6} + a^{2} b d^{7}\right )} x^{2} + 2 \, {\left (b^{3} c^{3} d^{4} - 2 \, a b^{2} c^{2} d^{5} + a^{2} b c d^{6}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.33, size = 227, normalized size = 1.80 \begin {gather*} -\frac {2 \, \sqrt {b x + a} {\left (\frac {2 \, {\left (2 \, b^{6} c^{2} d^{2} - 3 \, a b^{5} c d^{3}\right )} {\left (b x + a\right )}}{b^{4} c^{2} d^{3} {\left | b \right |} - 2 \, a b^{3} c d^{4} {\left | b \right |} + a^{2} b^{2} d^{5} {\left | b \right |}} + \frac {3 \, {\left (b^{7} c^{3} d - 3 \, a b^{6} c^{2} d^{2} + 2 \, a^{2} b^{5} c d^{3}\right )}}{b^{4} c^{2} d^{3} {\left | b \right |} - 2 \, a b^{3} c d^{4} {\left | b \right |} + a^{2} b^{2} d^{5} {\left | b \right |}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {2 \, b \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^{2} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 604, normalized size = 4.79 \begin {gather*} \frac {\sqrt {b x +a}\, \left (3 a^{2} d^{4} 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 c \,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 )+3 b^{2} c^{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^{2} 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 )-12 a b \,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 )+6 b^{2} 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^{2} 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 )-6 a b \,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 )+3 b^{2} 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 )+12 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a c \,d^{2} x -8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{2} d x +10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,c^{2} d -6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{3}\right )}{3 \sqrt {b d}\, \left (a d -b c \right )^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (d x +c \right )^{\frac {3}{2}} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x+a)^(1/2)*(3*a^2*d^4*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))-6
*a*b*c*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))+3*b^2*c^2*d^2*x^2*l
n(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*a^2*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))-12*a*b*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))+6*b^2*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))+3*a^2*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))-6*
a*b*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))+3*b^2*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))+12*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*c*d^2*x
-8*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*c^2*d*x+10*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*c^2*d-6*((b*x+a)*(d*
x+c))^(1/2)*(b*d)^(1/2)*b*c^3)/(b*d)^(1/2)/(a*d-b*c)^2/((b*x+a)*(d*x+c))^(1/2)/d^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(x^2/(d*x+c)^(5/2)/(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 \begin {gather*} \int \frac {x^2}{\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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