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

Optimal. Leaf size=349 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (4 a b c d-7 (a d+b c)^2\right ) (b c-a d)^3}{512 b^4 d^4}-\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (4 a b c d-7 (a d+b c)^2\right ) (b c-a d)^2}{768 b^4 d^3}-\frac{(a+b x)^{5/2} \sqrt{c+d x} \left (4 a b c d-7 (a d+b c)^2\right ) (b c-a d)}{192 b^4 d^2}-\frac{7 (a+b x)^{5/2} (c+d x)^{5/2} (a d+b c)}{60 b^2 d^2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2} \left (4 a b c d-7 (a d+b c)^2\right )}{96 b^3 d^2}-\frac{\left (4 a b c d-7 (a d+b c)^2\right ) (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{512 b^{9/2} d^{9/2}}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d} \]

[Out]

((b*c - a*d)^3*(4*a*b*c*d - 7*(b*c + a*d)^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(512*b^4*d^4) - ((b*c - a*d)^2*(4*a*
b*c*d - 7*(b*c + a*d)^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(768*b^4*d^3) - ((b*c - a*d)*(4*a*b*c*d - 7*(b*c + a*d
)^2)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(192*b^4*d^2) - ((4*a*b*c*d - 7*(b*c + a*d)^2)*(a + b*x)^(5/2)*(c + d*x)^(
3/2))/(96*b^3*d^2) - (7*(b*c + a*d)*(a + b*x)^(5/2)*(c + d*x)^(5/2))/(60*b^2*d^2) + (x*(a + b*x)^(5/2)*(c + d*
x)^(5/2))/(6*b*d) - ((b*c - a*d)^4*(4*a*b*c*d - 7*(b*c + a*d)^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt
[c + d*x])])/(512*b^(9/2)*d^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.306579, antiderivative size = 349, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (4 a b c d-7 (a d+b c)^2\right ) (b c-a d)^3}{512 b^4 d^4}-\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (4 a b c d-7 (a d+b c)^2\right ) (b c-a d)^2}{768 b^4 d^3}-\frac{(a+b x)^{5/2} \sqrt{c+d x} \left (4 a b c d-7 (a d+b c)^2\right ) (b c-a d)}{192 b^4 d^2}-\frac{7 (a+b x)^{5/2} (c+d x)^{5/2} (a d+b c)}{60 b^2 d^2}-\frac{(a+b x)^{5/2} (c+d x)^{3/2} \left (4 a b c d-7 (a d+b c)^2\right )}{96 b^3 d^2}-\frac{\left (4 a b c d-7 (a d+b c)^2\right ) (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{512 b^{9/2} d^{9/2}}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^3*(4*a*b*c*d - 7*(b*c + a*d)^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(512*b^4*d^4) - ((b*c - a*d)^2*(4*a*
b*c*d - 7*(b*c + a*d)^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(768*b^4*d^3) - ((b*c - a*d)*(4*a*b*c*d - 7*(b*c + a*d
)^2)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(192*b^4*d^2) - ((4*a*b*c*d - 7*(b*c + a*d)^2)*(a + b*x)^(5/2)*(c + d*x)^(
3/2))/(96*b^3*d^2) - (7*(b*c + a*d)*(a + b*x)^(5/2)*(c + d*x)^(5/2))/(60*b^2*d^2) + (x*(a + b*x)^(5/2)*(c + d*
x)^(5/2))/(6*b*d) - ((b*c - a*d)^4*(4*a*b*c*d - 7*(b*c + a*d)^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt
[c + d*x])])/(512*b^(9/2)*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} (c+d x)^{3/2} \, dx &=\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}+\frac{\int (a+b x)^{3/2} (c+d x)^{3/2} \left (-a c-\frac{7}{2} (b c+a d) x\right ) \, dx}{6 b d}\\ &=-\frac{7 (b c+a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}-\frac{\left (4 a b c d-7 (b c+a d)^2\right ) \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx}{24 b^2 d^2}\\ &=-\frac{\left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} (c+d x)^{3/2}}{96 b^3 d^2}-\frac{7 (b c+a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}-\frac{\left ((b c-a d) \left (4 a b c d-7 (b c+a d)^2\right )\right ) \int (a+b x)^{3/2} \sqrt{c+d x} \, dx}{64 b^3 d^2}\\ &=-\frac{(b c-a d) \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{192 b^4 d^2}-\frac{\left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} (c+d x)^{3/2}}{96 b^3 d^2}-\frac{7 (b c+a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}-\frac{\left ((b c-a d)^2 \left (4 a b c d-7 (b c+a d)^2\right )\right ) \int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx}{384 b^4 d^2}\\ &=-\frac{(b c-a d)^2 \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{768 b^4 d^3}-\frac{(b c-a d) \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{192 b^4 d^2}-\frac{\left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} (c+d x)^{3/2}}{96 b^3 d^2}-\frac{7 (b c+a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}+\frac{\left ((b c-a d)^3 \left (4 a b c d-7 (b c+a d)^2\right )\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{512 b^4 d^3}\\ &=\frac{(b c-a d)^3 \left (4 a b c d-7 (b c+a d)^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{512 b^4 d^4}-\frac{(b c-a d)^2 \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{768 b^4 d^3}-\frac{(b c-a d) \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{192 b^4 d^2}-\frac{\left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} (c+d x)^{3/2}}{96 b^3 d^2}-\frac{7 (b c+a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}-\frac{\left ((b c-a d)^4 \left (4 a b c d-7 (b c+a d)^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{1024 b^4 d^4}\\ &=\frac{(b c-a d)^3 \left (4 a b c d-7 (b c+a d)^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{512 b^4 d^4}-\frac{(b c-a d)^2 \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{768 b^4 d^3}-\frac{(b c-a d) \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{192 b^4 d^2}-\frac{\left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} (c+d x)^{3/2}}{96 b^3 d^2}-\frac{7 (b c+a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}-\frac{\left ((b c-a d)^4 \left (4 a b c d-7 (b c+a 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 )}{512 b^5 d^4}\\ &=\frac{(b c-a d)^3 \left (4 a b c d-7 (b c+a d)^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{512 b^4 d^4}-\frac{(b c-a d)^2 \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{768 b^4 d^3}-\frac{(b c-a d) \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{192 b^4 d^2}-\frac{\left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} (c+d x)^{3/2}}{96 b^3 d^2}-\frac{7 (b c+a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}-\frac{\left ((b c-a d)^4 \left (4 a b c d-7 (b c+a 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 )}{512 b^5 d^4}\\ &=\frac{(b c-a d)^3 \left (4 a b c d-7 (b c+a d)^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{512 b^4 d^4}-\frac{(b c-a d)^2 \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{768 b^4 d^3}-\frac{(b c-a d) \left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{192 b^4 d^2}-\frac{\left (4 a b c d-7 (b c+a d)^2\right ) (a+b x)^{5/2} (c+d x)^{3/2}}{96 b^3 d^2}-\frac{7 (b c+a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d^2}+\frac{x (a+b x)^{5/2} (c+d x)^{5/2}}{6 b d}-\frac{(b c-a d)^4 \left (4 a b c d-7 (b c+a d)^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{512 b^{9/2} d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 1.78011, size = 218, normalized size = 0.62 \[ \frac{(a+b x)^{5/2} (c+d x)^{5/2} \left (\frac{25 \left (7 a^2 d^2+10 a b c d+7 b^2 c^2\right ) \left (\frac{3 (a d-b c)^3}{d^2 (a+b x)^2}+\frac{3 (b c-a d)^{7/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{d^{5/2} (a+b x)^{5/2} \sqrt{\frac{b (c+d x)}{b c-a d}}}+\frac{2 (b c-a d)^2}{d (a+b x)}-8 a d+16 b (c+d x)+8 b c\right )}{128 b^2 (c+d x)^2}-35 (a d+b c)+50 b d x\right )}{300 b^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(5/2)*(c + d*x)^(5/2)*(-35*(b*c + a*d) + 50*b*d*x + (25*(7*b^2*c^2 + 10*a*b*c*d + 7*a^2*d^2)*(8*b*c
- 8*a*d + (3*(-(b*c) + a*d)^3)/(d^2*(a + b*x)^2) + (2*(b*c - a*d)^2)/(d*(a + b*x)) + 16*b*(c + d*x) + (3*(b*c
- a*d)^(7/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(d^(5/2)*(a + b*x)^(5/2)*Sqrt[(b*(c + d*x))/(b
*c - a*d)])))/(128*b^2*(c + d*x)^2)))/(300*b^2*d^2)

________________________________________________________________________________________

Maple [B]  time = 0.016, size = 1240, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15360*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(96*x^3*b^5*c^2*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+4672*x^3*a
*b^4*c*d^4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-304*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^3*b
^2*c*d^4+72*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^2*b^3*c^2*d^3-304*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x
+a*c)^(1/2)*x*a*b^4*c^3*d^2+240*x^2*a^2*b^3*c*d^4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+240*x^2*a*b^4*c^
2*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+2560*x^5*b^5*d^5*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)
-112*x^2*a^3*b^2*d^5*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-112*x^2*b^5*c^3*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*(b*d)^(1/2)+3328*x^4*a*b^4*d^5*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+3328*x^4*b^5*c*d^4*(b*d*x^2+
a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+96*x^3*a^2*b^3*d^5*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+140*(b*d)^(1
/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^4*b*d^5+140*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*b^5*c^4*d+47
0*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^4*b*c*d^4-132*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*
b^2*c^2*d^3-132*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*b^3*c^3*d^2+470*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c
*x+a*c)^(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))*a^6*d^6+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^6*c^6-21
0*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^5*d^5-210*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^5*c^5-
270*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*b*c*d^5+135*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^2*c^2*d^4+60*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^3*c^3*d^3+135*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^4*c^4*d^2-270*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^5*c^5*d)/b^4/d^4/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2
)/(b*d)^(1/2)

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.71322, size = 1949, normalized size = 5.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30720*(15*(7*b^6*c^6 - 18*a*b^5*c^5*d + 9*a^2*b^4*c^4*d^2 + 4*a^3*b^3*c^3*d^3 + 9*a^4*b^2*c^2*d^4 - 18*a^5*
b*c*d^5 + 7*a^6*d^6)*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)*sqr
t(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(1280*b^6*d^6*x^5 - 105*b^6*c^5*d + 235*a*b^
5*c^4*d^2 - 66*a^2*b^4*c^3*d^3 - 66*a^3*b^3*c^2*d^4 + 235*a^4*b^2*c*d^5 - 105*a^5*b*d^6 + 1664*(b^6*c*d^5 + a*
b^5*d^6)*x^4 + 16*(3*b^6*c^2*d^4 + 146*a*b^5*c*d^5 + 3*a^2*b^4*d^6)*x^3 - 8*(7*b^6*c^3*d^3 - 15*a*b^5*c^2*d^4
- 15*a^2*b^4*c*d^5 + 7*a^3*b^3*d^6)*x^2 + 2*(35*b^6*c^4*d^2 - 76*a*b^5*c^3*d^3 + 18*a^2*b^4*c^2*d^4 - 76*a^3*b
^3*c*d^5 + 35*a^4*b^2*d^6)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*d^5), -1/15360*(15*(7*b^6*c^6 - 18*a*b^5*c^5*d
+ 9*a^2*b^4*c^4*d^2 + 4*a^3*b^3*c^3*d^3 + 9*a^4*b^2*c^2*d^4 - 18*a^5*b*c*d^5 + 7*a^6*d^6)*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*(1280*b^6*d^6*x^5 - 105*b^6*c^5*d + 235*a*b^5*c^4*d^2 - 66*a^2*b^4*c^3*d^3 - 66*a^3*b^3*c^2*d^4 + 235*a
^4*b^2*c*d^5 - 105*a^5*b*d^6 + 1664*(b^6*c*d^5 + a*b^5*d^6)*x^4 + 16*(3*b^6*c^2*d^4 + 146*a*b^5*c*d^5 + 3*a^2*
b^4*d^6)*x^3 - 8*(7*b^6*c^3*d^3 - 15*a*b^5*c^2*d^4 - 15*a^2*b^4*c*d^5 + 7*a^3*b^3*d^6)*x^2 + 2*(35*b^6*c^4*d^2
- 76*a*b^5*c^3*d^3 + 18*a^2*b^4*c^2*d^4 - 76*a^3*b^3*c*d^5 + 35*a^4*b^2*d^6)*x)*sqrt(b*x + a)*sqrt(d*x + c))/
(b^5*d^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.65484, size = 2034, normalized size = 5.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7680*(40*(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*c*abs(b)/b^2 + 4*(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)
- 15*(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))*c*abs(b)/b + 4*(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) - 15*(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))*a*d*abs(b)/b^2 + (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*(2*(b*x + a)*(8*(b*x + a
)*(10*(b*x + a)/b^4 + (b^21*c*d^9 - 49*a*b^20*d^10)/(b^24*d^10)) - 3*(3*b^22*c^2*d^8 + 10*a*b^21*c*d^9 - 253*a
^2*b^20*d^10)/(b^24*d^10)) + (21*b^23*c^3*d^7 + 49*a*b^22*c^2*d^8 + 79*a^2*b^21*c*d^9 - 1429*a^3*b^20*d^10)/(b
^24*d^10))*(b*x + a) - 5*(21*b^24*c^4*d^6 + 28*a*b^23*c^3*d^7 + 30*a^2*b^22*c^2*d^8 + 28*a^3*b^21*c*d^9 - 491*
a^4*b^20*d^10)/(b^24*d^10))*(b*x + a) + 15*(21*b^25*c^5*d^5 + 7*a*b^24*c^4*d^6 + 2*a^2*b^23*c^3*d^7 - 2*a^3*b^
22*c^2*d^8 - 7*a^4*b^21*c*d^9 - 21*a^5*b^20*d^10)/(b^24*d^10))*sqrt(b*x + a) + 15*(21*b^6*c^6 - 14*a*b^5*c^5*d
- 5*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 - 5*a^4*b^2*c^2*d^4 - 14*a^5*b*c*d^5 + 21*a^6*d^6)*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^3*d^5))*d*abs(b)/b)/b