3.598 \(\int x^2 (a+b x)^{3/2} \sqrt{c+d x} \, dx\)

Optimal. Leaf size=315 \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right ) (b c-a d)^2}{128 b^3 d^4}+\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right ) (b c-a d)}{192 b^3 d^3}+\frac{(a+b x)^{5/2} \sqrt{c+d x} \left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right )}{48 b^3 d^2}+\frac{\left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right ) (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{128 b^{7/2} d^{9/2}}-\frac{(a+b x)^{5/2} (c+d x)^{3/2} (5 a d+7 b c)}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d} \]

[Out]

-((b*c - a*d)^2*(7*b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(128*b^3*d^4) + ((b*c - a*d)*
(7*b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(192*b^3*d^3) + ((7*b^2*c^2 + 6*a*b*c*d + 3
*a^2*d^2)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(48*b^3*d^2) - ((7*b*c + 5*a*d)*(a + b*x)^(5/2)*(c + d*x)^(3/2))/(40*
b^2*d^2) + (x*(a + b*x)^(5/2)*(c + d*x)^(3/2))/(5*b*d) + ((b*c - a*d)^3*(7*b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*Ar
cTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(128*b^(7/2)*d^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.30816, antiderivative size = 315, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {90, 80, 50, 63, 217, 206} \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right ) (b c-a d)^2}{128 b^3 d^4}+\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right ) (b c-a d)}{192 b^3 d^3}+\frac{(a+b x)^{5/2} \sqrt{c+d x} \left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right )}{48 b^3 d^2}+\frac{\left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right ) (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{128 b^{7/2} d^{9/2}}-\frac{(a+b x)^{5/2} (c+d x)^{3/2} (5 a d+7 b c)}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c - a*d)^2*(7*b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(128*b^3*d^4) + ((b*c - a*d)*
(7*b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(192*b^3*d^3) + ((7*b^2*c^2 + 6*a*b*c*d + 3
*a^2*d^2)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(48*b^3*d^2) - ((7*b*c + 5*a*d)*(a + b*x)^(5/2)*(c + d*x)^(3/2))/(40*
b^2*d^2) + (x*(a + b*x)^(5/2)*(c + d*x)^(3/2))/(5*b*d) + ((b*c - a*d)^3*(7*b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*Ar
cTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(128*b^(7/2)*d^(9/2))

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

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

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 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]

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])

Rubi steps

\begin{align*} \int x^2 (a+b x)^{3/2} \sqrt{c+d x} \, dx &=\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d}+\frac{\int (a+b x)^{3/2} \sqrt{c+d x} \left (-a c-\frac{1}{2} (7 b c+5 a d) x\right ) \, dx}{5 b d}\\ &=-\frac{(7 b c+5 a d) (a+b x)^{5/2} (c+d x)^{3/2}}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d}+\frac{\left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) \int (a+b x)^{3/2} \sqrt{c+d x} \, dx}{16 b^2 d^2}\\ &=\frac{\left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{48 b^3 d^2}-\frac{(7 b c+5 a d) (a+b x)^{5/2} (c+d x)^{3/2}}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d}+\frac{\left ((b c-a d) \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right )\right ) \int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx}{96 b^3 d^2}\\ &=\frac{(b c-a d) \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{192 b^3 d^3}+\frac{\left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{48 b^3 d^2}-\frac{(7 b c+5 a d) (a+b x)^{5/2} (c+d x)^{3/2}}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d}-\frac{\left ((b c-a d)^2 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right )\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{128 b^3 d^3}\\ &=-\frac{(b c-a d)^2 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{128 b^3 d^4}+\frac{(b c-a d) \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{192 b^3 d^3}+\frac{\left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{48 b^3 d^2}-\frac{(7 b c+5 a d) (a+b x)^{5/2} (c+d x)^{3/2}}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d}+\frac{\left ((b c-a d)^3 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{256 b^3 d^4}\\ &=-\frac{(b c-a d)^2 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{128 b^3 d^4}+\frac{(b c-a d) \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{192 b^3 d^3}+\frac{\left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{48 b^3 d^2}-\frac{(7 b c+5 a d) (a+b x)^{5/2} (c+d x)^{3/2}}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d}+\frac{\left ((b c-a d)^3 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{128 b^4 d^4}\\ &=-\frac{(b c-a d)^2 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{128 b^3 d^4}+\frac{(b c-a d) \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{192 b^3 d^3}+\frac{\left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{48 b^3 d^2}-\frac{(7 b c+5 a d) (a+b x)^{5/2} (c+d x)^{3/2}}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d}+\frac{\left ((b c-a d)^3 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{128 b^4 d^4}\\ &=-\frac{(b c-a d)^2 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{128 b^3 d^4}+\frac{(b c-a d) \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{192 b^3 d^3}+\frac{\left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{48 b^3 d^2}-\frac{(7 b c+5 a d) (a+b x)^{5/2} (c+d x)^{3/2}}{40 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{3/2}}{5 b d}+\frac{(b c-a d)^3 \left (7 b^2 c^2+6 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{128 b^{7/2} d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.66689, size = 265, normalized size = 0.84 \[ \frac{15 (b c-a d)^{7/2} \left (3 a^2 d^2+6 a b c d+7 b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )-b \sqrt{d} \sqrt{a+b x} (c+d x) \left (-6 a^2 b^2 d^2 \left (-6 c^2+3 c d x+4 d^2 x^2\right )+30 a^3 b d^3 (c+d x)-45 a^4 d^4-2 a b^3 d \left (-61 c^2 d x+95 c^3+48 c d^2 x^2+264 d^3 x^3\right )+b^4 \left (56 c^2 d^2 x^2-70 c^3 d x+105 c^4-48 c d^3 x^3-384 d^4 x^4\right )\right )}{1920 b^4 d^{9/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(-45*a^4*d^4 + 30*a^3*b*d^3*(c + d*x) - 6*a^2*b^2*d^2*(-6*c^2 + 3*c*d*x +
4*d^2*x^2) - 2*a*b^3*d*(95*c^3 - 61*c^2*d*x + 48*c*d^2*x^2 + 264*d^3*x^3) + b^4*(105*c^4 - 70*c^3*d*x + 56*c^
2*d^2*x^2 - 48*c*d^3*x^3 - 384*d^4*x^4))) + 15*(b*c - a*d)^(7/2)*(7*b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*Sqrt[(b*(
c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(1920*b^4*d^(9/2)*Sqrt[c + d*x])

________________________________________________________________________________________

Maple [B]  time = 0.015, size = 942, normalized size = 3. \begin{align*} -{\frac{1}{3840\,{b}^{3}{d}^{4}}\sqrt{bx+a}\sqrt{dx+c} \left ( -768\,{x}^{4}{b}^{4}{d}^{4}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-1056\,{x}^{3}a{b}^{3}{d}^{4}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-96\,{x}^{3}{b}^{4}c{d}^{3}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-48\,{x}^{2}{a}^{2}{b}^{2}{d}^{4}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-192\,{x}^{2}a{b}^{3}c{d}^{3}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+112\,{x}^{2}{b}^{4}{c}^{2}{d}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+45\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{5}{d}^{5}-45\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{4}bc{d}^{4}-30\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}{b}^{2}{c}^{2}{d}^{3}-90\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}{b}^{3}{c}^{3}{d}^{2}+225\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{b}^{4}{c}^{4}d-105\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{5}{c}^{5}+60\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{3}b{d}^{4}-36\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{2}{b}^{2}c{d}^{3}+244\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}xa{b}^{3}{c}^{2}{d}^{2}-140\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}x{b}^{4}{c}^{3}d-90\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{4}{d}^{4}+60\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{3}bc{d}^{3}+72\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{2}{b}^{2}{c}^{2}{d}^{2}-380\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}a{b}^{3}{c}^{3}d+210\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{b}^{4}{c}^{4} \right ){\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)^(3/2)*(d*x+c)^(1/2),x)

[Out]

-1/3840*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-768*x^4*b^4*d^4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-1056*x^3*a*b
^3*d^4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-96*x^3*b^4*c*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2
)-48*x^2*a^2*b^2*d^4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-192*x^2*a*b^3*c*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*(b*d)^(1/2)+112*x^2*b^4*c^2*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+45*ln(1/2*(2*b*d*x+2*(b*d*x
^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^5*d^5-45*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+
a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^4*b*c*d^4-30*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*
(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*b^2*c^2*d^3-90*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(
1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b^3*c^3*d^2+225*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a
*d+b*c)/(b*d)^(1/2))*a*b^4*c^4*d-105*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b
*d)^(1/2))*b^5*c^5+60*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^3*b*d^4-36*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*
c*x+a*c)^(1/2)*x*a^2*b^2*c*d^3+244*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a*b^3*c^2*d^2-140*(b*d)^(1/2)
*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*b^4*c^3*d-90*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^4*d^4+60*(b*d)^(
1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*b*c*d^3+72*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*b^2*c^2*d^
2-380*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*b^3*c^3*d+210*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*
b^4*c^4)/b^3/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/d^4/(b*d)^(1/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(3/2)*(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.49663, size = 1554, normalized size = 4.93 \begin{align*} \left [-\frac{15 \,{\left (7 \, b^{5} c^{5} - 15 \, a b^{4} c^{4} d + 6 \, a^{2} b^{3} c^{3} d^{2} + 2 \, a^{3} b^{2} c^{2} d^{3} + 3 \, a^{4} b c d^{4} - 3 \, a^{5} d^{5}\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 (384 \, b^{5} d^{5} x^{4} - 105 \, b^{5} c^{4} d + 190 \, a b^{4} c^{3} d^{2} - 36 \, a^{2} b^{3} c^{2} d^{3} - 30 \, a^{3} b^{2} c d^{4} + 45 \, a^{4} b d^{5} + 48 \,{\left (b^{5} c d^{4} + 11 \, a b^{4} d^{5}\right )} x^{3} - 8 \,{\left (7 \, b^{5} c^{2} d^{3} - 12 \, a b^{4} c d^{4} - 3 \, a^{2} b^{3} d^{5}\right )} x^{2} + 2 \,{\left (35 \, b^{5} c^{3} d^{2} - 61 \, a b^{4} c^{2} d^{3} + 9 \, a^{2} b^{3} c d^{4} - 15 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{7680 \, b^{4} d^{5}}, -\frac{15 \,{\left (7 \, b^{5} c^{5} - 15 \, a b^{4} c^{4} d + 6 \, a^{2} b^{3} c^{3} d^{2} + 2 \, a^{3} b^{2} c^{2} d^{3} + 3 \, a^{4} b c d^{4} - 3 \, a^{5} d^{5}\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 (384 \, b^{5} d^{5} x^{4} - 105 \, b^{5} c^{4} d + 190 \, a b^{4} c^{3} d^{2} - 36 \, a^{2} b^{3} c^{2} d^{3} - 30 \, a^{3} b^{2} c d^{4} + 45 \, a^{4} b d^{5} + 48 \,{\left (b^{5} c d^{4} + 11 \, a b^{4} d^{5}\right )} x^{3} - 8 \,{\left (7 \, b^{5} c^{2} d^{3} - 12 \, a b^{4} c d^{4} - 3 \, a^{2} b^{3} d^{5}\right )} x^{2} + 2 \,{\left (35 \, b^{5} c^{3} d^{2} - 61 \, a b^{4} c^{2} d^{3} + 9 \, a^{2} b^{3} c d^{4} - 15 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{3840 \, b^{4} d^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(3/2)*(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/7680*(15*(7*b^5*c^5 - 15*a*b^4*c^4*d + 6*a^2*b^3*c^3*d^2 + 2*a^3*b^2*c^2*d^3 + 3*a^4*b*c*d^4 - 3*a^5*d^5)*
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*(384*b^5*d^5*x^4 - 105*b^5*c^4*d + 190*a*b^4*c^3*d^2 - 36*a^2*b^3
*c^2*d^3 - 30*a^3*b^2*c*d^4 + 45*a^4*b*d^5 + 48*(b^5*c*d^4 + 11*a*b^4*d^5)*x^3 - 8*(7*b^5*c^2*d^3 - 12*a*b^4*c
*d^4 - 3*a^2*b^3*d^5)*x^2 + 2*(35*b^5*c^3*d^2 - 61*a*b^4*c^2*d^3 + 9*a^2*b^3*c*d^4 - 15*a^3*b^2*d^5)*x)*sqrt(b
*x + a)*sqrt(d*x + c))/(b^4*d^5), -1/3840*(15*(7*b^5*c^5 - 15*a*b^4*c^4*d + 6*a^2*b^3*c^3*d^2 + 2*a^3*b^2*c^2*
d^3 + 3*a^4*b*c*d^4 - 3*a^5*d^5)*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*(384*b^5*d^5*x^4 - 105*b^5*c^4*d + 190*a*b^4*c^3*d^
2 - 36*a^2*b^3*c^2*d^3 - 30*a^3*b^2*c*d^4 + 45*a^4*b*d^5 + 48*(b^5*c*d^4 + 11*a*b^4*d^5)*x^3 - 8*(7*b^5*c^2*d^
3 - 12*a*b^4*c*d^4 - 3*a^2*b^3*d^5)*x^2 + 2*(35*b^5*c^3*d^2 - 61*a*b^4*c^2*d^3 + 9*a^2*b^3*c*d^4 - 15*a^3*b^2*
d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^5)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+a)**(3/2)*(d*x+c)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.40959, size = 886, normalized size = 2.81 \begin{align*} \frac{\frac{10 \,{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}{\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{6 \,{\left (b x + a\right )}}{b^{2}} + \frac{b^{7} c d^{5} - 17 \, a b^{6} d^{6}}{b^{8} d^{6}}\right )} - \frac{5 \, b^{8} c^{2} d^{4} + 6 \, a b^{7} c d^{5} - 59 \, a^{2} b^{6} d^{6}}{b^{8} d^{6}}\right )} + \frac{3 \,{\left (5 \, b^{9} c^{3} d^{3} + a b^{8} c^{2} d^{4} - a^{2} b^{7} c d^{5} - 5 \, a^{3} b^{6} d^{6}\right )}}{b^{8} d^{6}}\right )} \sqrt{b x + a} + \frac{3 \,{\left (5 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 5 \, a^{4} d^{4}\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} b d^{3}}\right )} a{\left | b \right |}}{b^{2}} + \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}{\left (2 \,{\left (4 \,{\left (b x + a\right )}{\left (6 \,{\left (b x + a\right )}{\left (\frac{8 \,{\left (b x + a\right )}}{b^{3}} + \frac{b^{13} c d^{7} - 31 \, a b^{12} d^{8}}{b^{15} d^{8}}\right )} - \frac{7 \, b^{14} c^{2} d^{6} + 16 \, a b^{13} c d^{7} - 263 \, a^{2} b^{12} d^{8}}{b^{15} d^{8}}\right )} + \frac{5 \,{\left (7 \, b^{15} c^{3} d^{5} + 9 \, a b^{14} c^{2} d^{6} + 9 \, a^{2} b^{13} c d^{7} - 121 \, a^{3} b^{12} d^{8}\right )}}{b^{15} d^{8}}\right )}{\left (b x + a\right )} - \frac{15 \,{\left (7 \, b^{16} c^{4} d^{4} + 2 \, a b^{15} c^{3} d^{5} - 2 \, a^{3} b^{13} c d^{7} - 7 \, a^{4} b^{12} d^{8}\right )}}{b^{15} d^{8}}\right )} \sqrt{b x + a} - \frac{15 \,{\left (7 \, b^{5} c^{5} - 5 \, a b^{4} c^{4} d - 2 \, a^{2} b^{3} c^{3} d^{2} - 2 \, a^{3} b^{2} c^{2} d^{3} - 5 \, a^{4} b c d^{4} + 7 \, a^{5} d^{5}\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} b^{2} d^{4}}\right )}{\left | b \right |}}{b}}{1920 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(3/2)*(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/1920*(10*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^2 + (b^7*c*d^5 - 17*a
*b^6*d^6)/(b^8*d^6)) - (5*b^8*c^2*d^4 + 6*a*b^7*c*d^5 - 59*a^2*b^6*d^6)/(b^8*d^6)) + 3*(5*b^9*c^3*d^3 + a*b^8*
c^2*d^4 - a^2*b^7*c*d^5 - 5*a^3*b^6*d^6)/(b^8*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 - 4*a*b^3*c^3*d - 2*a^2*b^2*c
^2*d^2 - 4*a^3*b*c*d^3 + 5*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(
sqrt(b*d)*b*d^3))*a*abs(b)/b^2 + (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*(b*x + a)*(6*(b*x + a)*(8*(b*x + a
)/b^3 + (b^13*c*d^7 - 31*a*b^12*d^8)/(b^15*d^8)) - (7*b^14*c^2*d^6 + 16*a*b^13*c*d^7 - 263*a^2*b^12*d^8)/(b^15
*d^8)) + 5*(7*b^15*c^3*d^5 + 9*a*b^14*c^2*d^6 + 9*a^2*b^13*c*d^7 - 121*a^3*b^12*d^8)/(b^15*d^8))*(b*x + a) - 1
5*(7*b^16*c^4*d^4 + 2*a*b^15*c^3*d^5 - 2*a^3*b^13*c*d^7 - 7*a^4*b^12*d^8)/(b^15*d^8))*sqrt(b*x + a) - 15*(7*b^
5*c^5 - 5*a*b^4*c^4*d - 2*a^2*b^3*c^3*d^2 - 2*a^3*b^2*c^2*d^3 - 5*a^4*b*c*d^4 + 7*a^5*d^5)*log(abs(-sqrt(b*d)*
sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^4))*abs(b)/b)/b