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

Optimal. Leaf size=193 \[ \frac {\sqrt {\frac {a x+b}{c x+d}} \left (8 a^2 c^3 x^3+10 a^2 c^2 d x^2-a^2 c d^2 x-3 a^2 d^3-10 a b c^3 x^2-14 a b c^2 d x-4 a b c d^2+15 b^2 c^3 x+15 b^2 c^2 d\right )}{24 a^3 c^2}+\frac {\left (a^3 d^3+a^2 b c d^2+3 a b^2 c^2 d-5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {a x+b}{c x+d}}}{\sqrt {a}}\right )}{8 a^{7/2} c^{5/2}} \]

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Rubi [A]  time = 0.28, antiderivative size = 266, normalized size of antiderivative = 1.38, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1960, 413, 385, 199, 208} \begin {gather*} -\frac {(3 a d+5 b c) (b c-a d)^2 \sqrt {\frac {a x+b}{c x+d}}}{12 a^2 c^2 \left (a-\frac {c (a x+b)}{c x+d}\right )^2}-\frac {\left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {a x+b}{c x+d}}}{\sqrt {a}}\right )}{8 a^{7/2} c^{5/2}}+\frac {(c x+d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \sqrt {\frac {a x+b}{c x+d}}}{8 a^3 c^2}-\frac {(b c-a d)^2 \sqrt {\frac {a x+b}{c x+d}} \left (b-\frac {d (a x+b)}{c x+d}\right )}{3 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[(b + a*x)/(d + c*x)],x]

[Out]

((5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[(b + a*x)/(d + c*x)]*(d + c*x))/(8*a^3*c^2) - ((b*c - a*d)^2*(5*b*c +
3*a*d)*Sqrt[(b + a*x)/(d + c*x)])/(12*a^2*c^2*(a - (c*(b + a*x))/(d + c*x))^2) - ((b*c - a*d)^2*Sqrt[(b + a*x)
/(d + c*x)]*(b - (d*(b + a*x))/(d + c*x)))/(3*a*c*(a - (c*(b + a*x))/(d + c*x))^3) - ((b*c - a*d)*(5*b^2*c^2 +
2*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[(b + a*x)/(d + c*x)])/Sqrt[a]])/(8*a^(7/2)*c^(5/2))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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]

Rule 385

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

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {\frac {b+a x}{d+c x}}} \, dx &=-\left ((2 (b c-a d)) \operatorname {Subst}\left (\int \frac {\left (-b+d x^2\right )^2}{\left (a-c x^2\right )^4} \, dx,x,\sqrt {\frac {b+a x}{d+c x}}\right )\right )\\ &=-\frac {(b c-a d)^2 \sqrt {\frac {b+a x}{d+c x}} \left (b-\frac {d (b+a x)}{d+c x}\right )}{3 a c \left (a-\frac {c (b+a x)}{d+c x}\right )^3}+\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {-b (5 b c+a d)+3 d (b c+a d) x^2}{\left (a-c x^2\right )^3} \, dx,x,\sqrt {\frac {b+a x}{d+c x}}\right )}{3 a c}\\ &=-\frac {(b c-a d)^2 (5 b c+3 a d) \sqrt {\frac {b+a x}{d+c x}}}{12 a^2 c^2 \left (a-\frac {c (b+a x)}{d+c x}\right )^2}-\frac {(b c-a d)^2 \sqrt {\frac {b+a x}{d+c x}} \left (b-\frac {d (b+a x)}{d+c x}\right )}{3 a c \left (a-\frac {c (b+a x)}{d+c x}\right )^3}-\frac {\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a-c x^2\right )^2} \, dx,x,\sqrt {\frac {b+a x}{d+c x}}\right )}{4 a^2 c^2}\\ &=\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {\frac {b+a x}{d+c x}} (d+c x)}{8 a^3 c^2}-\frac {(b c-a d)^2 (5 b c+3 a d) \sqrt {\frac {b+a x}{d+c x}}}{12 a^2 c^2 \left (a-\frac {c (b+a x)}{d+c x}\right )^2}-\frac {(b c-a d)^2 \sqrt {\frac {b+a x}{d+c x}} \left (b-\frac {d (b+a x)}{d+c x}\right )}{3 a c \left (a-\frac {c (b+a x)}{d+c x}\right )^3}-\frac {\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-c x^2} \, dx,x,\sqrt {\frac {b+a x}{d+c x}}\right )}{8 a^3 c^2}\\ &=\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {\frac {b+a x}{d+c x}} (d+c x)}{8 a^3 c^2}-\frac {(b c-a d)^2 (5 b c+3 a d) \sqrt {\frac {b+a x}{d+c x}}}{12 a^2 c^2 \left (a-\frac {c (b+a x)}{d+c x}\right )^2}-\frac {(b c-a d)^2 \sqrt {\frac {b+a x}{d+c x}} \left (b-\frac {d (b+a x)}{d+c x}\right )}{3 a c \left (a-\frac {c (b+a x)}{d+c x}\right )^3}-\frac {(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {b+a x}{d+c x}}}{\sqrt {a}}\right )}{8 a^{7/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 201, normalized size = 1.04 \begin {gather*} \frac {\sqrt {c} (a x+b) \sqrt {\frac {a (c x+d)}{a d-b c}} \left (a^2 \left (8 c^2 x^2+2 c d x-3 d^2\right )-2 a b c (5 c x+2 d)+15 b^2 c^2\right )+3 \sqrt {a x+b} \sqrt {a d-b c} \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {a x+b}}{\sqrt {a d-b c}}\right )}{24 a^3 c^{5/2} \sqrt {\frac {a x+b}{c x+d}} \sqrt {\frac {a (c x+d)}{a d-b c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(b + a*x)*Sqrt[(a*(d + c*x))/(-(b*c) + a*d)]*(15*b^2*c^2 - 2*a*b*c*(2*d + 5*c*x) + a^2*(-3*d^2 + 2*c*
d*x + 8*c^2*x^2)) + 3*Sqrt[-(b*c) + a*d]*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[b + a*x]*ArcSinh[(Sqrt[c]*Sqrt
[b + a*x])/Sqrt[-(b*c) + a*d]])/(24*a^3*c^(5/2)*Sqrt[(b + a*x)/(d + c*x)]*Sqrt[(a*(d + c*x))/(-(b*c) + a*d)])

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IntegrateAlgebraic [A]  time = 0.30, size = 193, normalized size = 1.00 \begin {gather*} \frac {\sqrt {\frac {b+a x}{d+c x}} \left (15 b^2 c^2 d-4 a b c d^2-3 a^2 d^3+15 b^2 c^3 x-14 a b c^2 d x-a^2 c d^2 x-10 a b c^3 x^2+10 a^2 c^2 d x^2+8 a^2 c^3 x^3\right )}{24 a^3 c^2}+\frac {\left (-5 b^3 c^3+3 a b^2 c^2 d+a^2 b c d^2+a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {b+a x}{d+c x}}}{\sqrt {a}}\right )}{8 a^{7/2} c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*x)/(d + c*x)]*(15*b^2*c^2*d - 4*a*b*c*d^2 - 3*a^2*d^3 + 15*b^2*c^3*x - 14*a*b*c^2*d*x - a^2*c*d^2
*x - 10*a*b*c^3*x^2 + 10*a^2*c^2*d*x^2 + 8*a^2*c^3*x^3))/(24*a^3*c^2) + ((-5*b^3*c^3 + 3*a*b^2*c^2*d + a^2*b*c
*d^2 + a^3*d^3)*ArcTanh[(Sqrt[c]*Sqrt[(b + a*x)/(d + c*x)])/Sqrt[a]])/(8*a^(7/2)*c^(5/2))

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fricas [A]  time = 1.15, size = 419, normalized size = 2.17 \begin {gather*} \left [-\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a c} \log \left (-2 \, a c x - b c - a d - 2 \, \sqrt {a c} {\left (c x + d\right )} \sqrt {\frac {a x + b}{c x + d}}\right ) - 2 \, {\left (8 \, a^{3} c^{4} x^{3} + 15 \, a b^{2} c^{3} d - 4 \, a^{2} b c^{2} d^{2} - 3 \, a^{3} c d^{3} - 10 \, {\left (a^{2} b c^{4} - a^{3} c^{3} d\right )} x^{2} + {\left (15 \, a b^{2} c^{4} - 14 \, a^{2} b c^{3} d - a^{3} c^{2} d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{c x + d}}}{48 \, a^{4} c^{3}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} {\left (c x + d\right )} \sqrt {\frac {a x + b}{c x + d}}}{a c x + b c}\right ) + {\left (8 \, a^{3} c^{4} x^{3} + 15 \, a b^{2} c^{3} d - 4 \, a^{2} b c^{2} d^{2} - 3 \, a^{3} c d^{3} - 10 \, {\left (a^{2} b c^{4} - a^{3} c^{3} d\right )} x^{2} + {\left (15 \, a b^{2} c^{4} - 14 \, a^{2} b c^{3} d - a^{3} c^{2} d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{c x + d}}}{24 \, a^{4} c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(a*c)*log(-2*a*c*x - b*c - a*d - 2*sqrt(a*c)
*(c*x + d)*sqrt((a*x + b)/(c*x + d))) - 2*(8*a^3*c^4*x^3 + 15*a*b^2*c^3*d - 4*a^2*b*c^2*d^2 - 3*a^3*c*d^3 - 10
*(a^2*b*c^4 - a^3*c^3*d)*x^2 + (15*a*b^2*c^4 - 14*a^2*b*c^3*d - a^3*c^2*d^2)*x)*sqrt((a*x + b)/(c*x + d)))/(a^
4*c^3), 1/24*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(-a*c)*arctan(sqrt(-a*c)*(c*x + d)*sqr
t((a*x + b)/(c*x + d))/(a*c*x + b*c)) + (8*a^3*c^4*x^3 + 15*a*b^2*c^3*d - 4*a^2*b*c^2*d^2 - 3*a^3*c*d^3 - 10*(
a^2*b*c^4 - a^3*c^3*d)*x^2 + (15*a*b^2*c^4 - 14*a^2*b*c^3*d - a^3*c^2*d^2)*x)*sqrt((a*x + b)/(c*x + d)))/(a^4*
c^3)]

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giac [B]  time = 0.36, size = 551, normalized size = 2.85 \begin {gather*} \frac {1}{24} \, {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} {\left (\frac {3 \, {\left (5 \, b^{4} c^{4} - 8 \, a b^{3} c^{3} d + 2 \, a^{2} b^{2} c^{2} d^{2} + a^{4} d^{4}\right )} \arctan \left (\frac {c \sqrt {\frac {a x + b}{c x + d}}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a^{3} c^{2}} - \frac {33 \, a^{2} b^{4} c^{4} \sqrt {\frac {a x + b}{c x + d}} - \frac {40 \, {\left (a x + b\right )} a b^{4} c^{5} \sqrt {\frac {a x + b}{c x + d}}}{c x + d} + \frac {15 \, {\left (a x + b\right )}^{2} b^{4} c^{6} \sqrt {\frac {a x + b}{c x + d}}}{{\left (c x + d\right )}^{2}} - 72 \, a^{3} b^{3} c^{3} d \sqrt {\frac {a x + b}{c x + d}} + \frac {64 \, {\left (a x + b\right )} a^{2} b^{3} c^{4} d \sqrt {\frac {a x + b}{c x + d}}}{c x + d} - \frac {24 \, {\left (a x + b\right )}^{2} a b^{3} c^{5} d \sqrt {\frac {a x + b}{c x + d}}}{{\left (c x + d\right )}^{2}} + 42 \, a^{4} b^{2} c^{2} d^{2} \sqrt {\frac {a x + b}{c x + d}} + \frac {6 \, {\left (a x + b\right )}^{2} a^{2} b^{2} c^{4} d^{2} \sqrt {\frac {a x + b}{c x + d}}}{{\left (c x + d\right )}^{2}} - \frac {32 \, {\left (a x + b\right )} a^{4} b c^{2} d^{3} \sqrt {\frac {a x + b}{c x + d}}}{c x + d} - 3 \, a^{6} d^{4} \sqrt {\frac {a x + b}{c x + d}} + \frac {8 \, {\left (a x + b\right )} a^{5} c d^{4} \sqrt {\frac {a x + b}{c x + d}}}{c x + d} + \frac {3 \, {\left (a x + b\right )}^{2} a^{4} c^{2} d^{4} \sqrt {\frac {a x + b}{c x + d}}}{{\left (c x + d\right )}^{2}}}{{\left (a - \frac {{\left (a x + b\right )} c}{c x + d}\right )}^{3} a^{3} c^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)*(3*(5*b^4*c^4 - 8*a*b^3*c^3*d + 2*a^2*b^2*c^2*d^2 + a^4*d^4)*arct
an(c*sqrt((a*x + b)/(c*x + d))/sqrt(-a*c))/(sqrt(-a*c)*a^3*c^2) - (33*a^2*b^4*c^4*sqrt((a*x + b)/(c*x + d)) -
40*(a*x + b)*a*b^4*c^5*sqrt((a*x + b)/(c*x + d))/(c*x + d) + 15*(a*x + b)^2*b^4*c^6*sqrt((a*x + b)/(c*x + d))/
(c*x + d)^2 - 72*a^3*b^3*c^3*d*sqrt((a*x + b)/(c*x + d)) + 64*(a*x + b)*a^2*b^3*c^4*d*sqrt((a*x + b)/(c*x + d)
)/(c*x + d) - 24*(a*x + b)^2*a*b^3*c^5*d*sqrt((a*x + b)/(c*x + d))/(c*x + d)^2 + 42*a^4*b^2*c^2*d^2*sqrt((a*x
+ b)/(c*x + d)) + 6*(a*x + b)^2*a^2*b^2*c^4*d^2*sqrt((a*x + b)/(c*x + d))/(c*x + d)^2 - 32*(a*x + b)*a^4*b*c^2
*d^3*sqrt((a*x + b)/(c*x + d))/(c*x + d) - 3*a^6*d^4*sqrt((a*x + b)/(c*x + d)) + 8*(a*x + b)*a^5*c*d^4*sqrt((a
*x + b)/(c*x + d))/(c*x + d) + 3*(a*x + b)^2*a^4*c^2*d^4*sqrt((a*x + b)/(c*x + d))/(c*x + d)^2)/((a - (a*x + b
)*c/(c*x + d))^3*a^3*c^2))

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maple [B]  time = 0.11, size = 588, normalized size = 3.05

method	result	size
default	\(\frac {\left (a x +b \right ) \left (-12 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}\, a^{2} c d x -36 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}\, a b \,c^{2} x +3 \ln \left (\frac {2 a c x +2 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) a^{3} d^{3}+3 \ln \left (\frac {2 a c x +2 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) a^{2} b c \,d^{2}-15 \ln \left (\frac {2 a c x +2 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) a \,b^{2} c^{2} d +9 \ln \left (\frac {2 a c x +2 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) b^{3} c^{3}+24 \ln \left (\frac {2 a c x +2 \sqrt {\left (c x +d \right ) \left (a x +b \right )}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) a \,b^{2} c^{2} d -24 \ln \left (\frac {2 a c x +2 \sqrt {\left (c x +d \right ) \left (a x +b \right )}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) b^{3} c^{3}+16 \left (x^{2} a c +a d x +c x b +b d \right )^{\frac {3}{2}} a c \sqrt {a c}-6 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}\, a^{2} d^{2}-24 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}\, a b c d -18 \sqrt {x^{2} a c +a d x +c x b +b d}\, \sqrt {a c}\, b^{2} c^{2}+48 \sqrt {a c}\, \sqrt {\left (c x +d \right ) \left (a x +b \right )}\, b^{2} c^{2}\right )}{48 a^{3} \sqrt {\frac {a x +b}{c x +d}}\, \sqrt {\left (c x +d \right ) \left (a x +b \right )}\, c^{2} \sqrt {a c}}\)	\(588\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a*x+b)/(c*x+d))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/48*(a*x+b)/a^3*(-12*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)*a^2*c*d*x-36*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)
*(a*c)^(1/2)*a*b*c^2*x+3*ln(1/2*(2*a*c*x+2*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1/2))*a
^3*d^3+3*ln(1/2*(2*a*c*x+2*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1/2))*a^2*b*c*d^2-15*ln
(1/2*(2*a*c*x+2*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1/2))*a*b^2*c^2*d+9*ln(1/2*(2*a*c*
x+2*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1/2))*b^3*c^3+24*ln(1/2*(2*a*c*x+2*((c*x+d)*(a
*x+b))^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1/2))*a*b^2*c^2*d-24*ln(1/2*(2*a*c*x+2*((c*x+d)*(a*x+b))^(1/2)*(a*c)^
(1/2)+a*d+b*c)/(a*c)^(1/2))*b^3*c^3+16*(a*c*x^2+a*d*x+b*c*x+b*d)^(3/2)*a*c*(a*c)^(1/2)-6*(a*c*x^2+a*d*x+b*c*x+
b*d)^(1/2)*(a*c)^(1/2)*a^2*d^2-24*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)*a*b*c*d-18*(a*c*x^2+a*d*x+b*c*x+
b*d)^(1/2)*(a*c)^(1/2)*b^2*c^2+48*(a*c)^(1/2)*((c*x+d)*(a*x+b))^(1/2)*b^2*c^2)/((a*x+b)/(c*x+d))^(1/2)/((c*x+d
)*(a*x+b))^(1/2)/c^2/(a*c)^(1/2)

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maxima [A]  time = 0.45, size = 355, normalized size = 1.84 \begin {gather*} -\frac {3 \, {\left (5 \, b^{3} c^{5} - 3 \, a b^{2} c^{4} d - a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} \left (\frac {a x + b}{c x + d}\right )^{\frac {5}{2}} - 8 \, {\left (5 \, a b^{3} c^{4} - 3 \, a^{2} b^{2} c^{3} d - 3 \, a^{3} b c^{2} d^{2} + a^{4} c d^{3}\right )} \left (\frac {a x + b}{c x + d}\right )^{\frac {3}{2}} + 3 \, {\left (11 \, a^{2} b^{3} c^{3} - 13 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2} + a^{5} d^{3}\right )} \sqrt {\frac {a x + b}{c x + d}}}{24 \, {\left (a^{6} c^{2} - \frac {3 \, {\left (a x + b\right )} a^{5} c^{3}}{c x + d} + \frac {3 \, {\left (a x + b\right )}^{2} a^{4} c^{4}}{{\left (c x + d\right )}^{2}} - \frac {{\left (a x + b\right )}^{3} a^{3} c^{5}}{{\left (c x + d\right )}^{3}}\right )}} + \frac {{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (\frac {c \sqrt {\frac {a x + b}{c x + d}} - \sqrt {a c}}{c \sqrt {\frac {a x + b}{c x + d}} + \sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(5*b^3*c^5 - 3*a*b^2*c^4*d - a^2*b*c^3*d^2 - a^3*c^2*d^3)*((a*x + b)/(c*x + d))^(5/2) - 8*(5*a*b^3*c^
4 - 3*a^2*b^2*c^3*d - 3*a^3*b*c^2*d^2 + a^4*c*d^3)*((a*x + b)/(c*x + d))^(3/2) + 3*(11*a^2*b^3*c^3 - 13*a^3*b^
2*c^2*d + a^4*b*c*d^2 + a^5*d^3)*sqrt((a*x + b)/(c*x + d)))/(a^6*c^2 - 3*(a*x + b)*a^5*c^3/(c*x + d) + 3*(a*x
+ b)^2*a^4*c^4/(c*x + d)^2 - (a*x + b)^3*a^3*c^5/(c*x + d)^3) + 1/16*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2
- a^3*d^3)*log((c*sqrt((a*x + b)/(c*x + d)) - sqrt(a*c))/(c*sqrt((a*x + b)/(c*x + d)) + sqrt(a*c)))/(sqrt(a*c)
*a^3*c^2)

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mupad [B]  time = 0.86, size = 305, normalized size = 1.58 \begin {gather*} \frac {\frac {{\left (\frac {b+a\,x}{d+c\,x}\right )}^{5/2}\,\left (\frac {a^3\,d^3}{8}+\frac {a^2\,b\,c\,d^2}{8}+\frac {3\,a\,b^2\,c^2\,d}{8}-\frac {5\,b^3\,c^3}{8}\right )}{a^6}+\frac {{\left (\frac {b+a\,x}{d+c\,x}\right )}^{3/2}\,\left (\frac {a^3\,d^3}{3}-a^2\,b\,c\,d^2-a\,b^2\,c^2\,d+\frac {5\,b^3\,c^3}{3}\right )}{a^5\,c}-\frac {\sqrt {\frac {b+a\,x}{d+c\,x}}\,\left (\frac {a^3\,d^3}{8}+\frac {a^2\,b\,c\,d^2}{8}-\frac {13\,a\,b^2\,c^2\,d}{8}+\frac {11\,b^3\,c^3}{8}\right )}{a^4\,c^2}}{\frac {3\,c^2\,{\left (b+a\,x\right )}^2}{a^2\,{\left (d+c\,x\right )}^2}-\frac {c^3\,{\left (b+a\,x\right )}^3}{a^3\,{\left (d+c\,x\right )}^3}-\frac {3\,c\,\left (b+a\,x\right )}{a\,\left (d+c\,x\right )}+1}+\frac {\mathrm {atanh}\left (\frac {\sqrt {c}\,\sqrt {\frac {b+a\,x}{d+c\,x}}}{\sqrt {a}}\right )\,\left (a\,d-b\,c\right )\,\left (a^2\,d^2+2\,a\,b\,c\,d+5\,b^2\,c^2\right )}{8\,a^{7/2}\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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