3.682 \(\int \frac{x^2 (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=309 \[ \frac{5 (a+b x)^{3/2} \sqrt{c+d x} \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right )}{96 b d^4}-\frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d) \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right )}{64 b d^5}+\frac{5 (b c-a d)^2 \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{3/2} d^{11/2}}+\frac{(a+b x)^{5/2} \sqrt{c+d x} \left (\frac{a^2 d}{b}+14 a c-\frac{63 b c^2}{d}\right )}{24 d^2 (b c-a d)}+\frac{2 c^2 (a+b x)^{7/2}}{d^2 \sqrt{c+d x} (b c-a d)}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2} \]

[Out]

(2*c^2*(a + b*x)^(7/2))/(d^2*(b*c - a*d)*Sqrt[c + d*x]) - (5*(b*c - a*d)*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*S
qrt[a + b*x]*Sqrt[c + d*x])/(64*b*d^5) + (5*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])
/(96*b*d^4) + ((14*a*c - (63*b*c^2)/d + (a^2*d)/b)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*d^2*(b*c - a*d)) + ((a +
b*x)^(7/2)*Sqrt[c + d*x])/(4*b*d^2) + (5*(b*c - a*d)^2*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*S
qrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*d^(11/2))

________________________________________________________________________________________

Rubi [A]  time = 0.332356, antiderivative size = 309, 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 = {89, 80, 50, 63, 217, 206} \[ \frac{5 (a+b x)^{3/2} \sqrt{c+d x} \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right )}{96 b d^4}-\frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d) \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right )}{64 b d^5}+\frac{5 (b c-a d)^2 \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{3/2} d^{11/2}}+\frac{(a+b x)^{5/2} \sqrt{c+d x} \left (\frac{a^2 d}{b}+14 a c-\frac{63 b c^2}{d}\right )}{24 d^2 (b c-a d)}+\frac{2 c^2 (a+b x)^{7/2}}{d^2 \sqrt{c+d x} (b c-a d)}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^(5/2))/(c + d*x)^(3/2),x]

[Out]

(2*c^2*(a + b*x)^(7/2))/(d^2*(b*c - a*d)*Sqrt[c + d*x]) - (5*(b*c - a*d)*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*S
qrt[a + b*x]*Sqrt[c + d*x])/(64*b*d^5) + (5*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])
/(96*b*d^4) + ((14*a*c - (63*b*c^2)/d + (a^2*d)/b)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*d^2*(b*c - a*d)) + ((a +
b*x)^(7/2)*Sqrt[c + d*x])/(4*b*d^2) + (5*(b*c - a*d)^2*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*S
qrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*d^(11/2))

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 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 \frac{x^2 (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx &=\frac{2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{2 \int \frac{(a+b x)^{5/2} \left (\frac{1}{2} c (7 b c-a d)-\frac{1}{2} d (b c-a d) x\right )}{\sqrt{c+d x}} \, dx}{d^2 (b c-a d)}\\ &=\frac{2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt{c+d x}}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2}-\frac{\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \int \frac{(a+b x)^{5/2}}{\sqrt{c+d x}} \, dx}{8 b d^2 (b c-a d)}\\ &=\frac{2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d^3 (b c-a d)}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2}+\frac{\left (5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right )\right ) \int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx}{48 b d^3}\\ &=\frac{2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt{c+d x}}+\frac{5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^4}-\frac{\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d^3 (b c-a d)}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2}-\frac{\left (5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right )\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{64 b d^4}\\ &=\frac{2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b d^5}+\frac{5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^4}-\frac{\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d^3 (b c-a d)}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2}+\frac{\left (5 (b c-a d)^2 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{128 b d^5}\\ &=\frac{2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b d^5}+\frac{5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^4}-\frac{\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d^3 (b c-a d)}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2}+\frac{\left (5 (b c-a d)^2 \left (63 b^2 c^2-14 a b c d-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 )}{64 b^2 d^5}\\ &=\frac{2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b d^5}+\frac{5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^4}-\frac{\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d^3 (b c-a d)}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2}+\frac{\left (5 (b c-a d)^2 \left (63 b^2 c^2-14 a b c d-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 )}{64 b^2 d^5}\\ &=\frac{2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt{c+d x}}-\frac{5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \sqrt{a+b x} \sqrt{c+d x}}{64 b d^5}+\frac{5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^4}-\frac{\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d^3 (b c-a d)}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b d^2}+\frac{5 (b c-a d)^2 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{3/2} d^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.638232, size = 296, normalized size = 0.96 \[ \frac{\frac{b \sqrt{d} \left (a^2 b^2 d \left (-202 c^2 d x+1785 c^3-581 c d^2 x^2+254 d^3 x^3\right )+a^3 b d^2 \left (-839 c^2-322 c d x+133 d^2 x^2\right )+15 a^4 d^3 (c+d x)+a b^3 \left (763 c^2 d^2 x^2+1470 c^3 d x-945 c^4-316 c d^3 x^3+184 d^4 x^4\right )+3 b^4 x \left (42 c^2 d^2 x^2-105 c^3 d x-315 c^4-24 c d^3 x^3+16 d^4 x^4\right )\right )}{\sqrt{a+b x}}+15 \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right ) (b c-a d)^{5/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{192 b^2 d^{11/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x)^(5/2))/(c + d*x)^(3/2),x]

[Out]

((b*Sqrt[d]*(15*a^4*d^3*(c + d*x) + a^3*b*d^2*(-839*c^2 - 322*c*d*x + 133*d^2*x^2) + a^2*b^2*d*(1785*c^3 - 202
*c^2*d*x - 581*c*d^2*x^2 + 254*d^3*x^3) + 3*b^4*x*(-315*c^4 - 105*c^3*d*x + 42*c^2*d^2*x^2 - 24*c*d^3*x^3 + 16
*d^4*x^4) + a*b^3*(-945*c^4 + 1470*c^3*d*x + 763*c^2*d^2*x^2 - 316*c*d^3*x^3 + 184*d^4*x^4)))/Sqrt[a + b*x] +
15*(b*c - a*d)^(5/2)*(63*b^2*c^2 - 14*a*b*c*d - 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]])/(192*b^2*d^(11/2)*Sqrt[c + d*x])

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Maple [B]  time = 0.026, size = 961, normalized size = 3.1 \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)^(5/2)/(d*x+c)^(3/2),x)

[Out]

-1/384*(b*x+a)^(1/2)*(-96*x^4*b^3*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-272*x^3*a*b^2*d^4*((b*x+a)*(d*x+c))^
(1/2)*(b*d)^(1/2)+144*x^3*b^3*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))
^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*a^4*d^5+180*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a
*d+b*c)/(b*d)^(1/2))*x*a^3*b*c*d^4-1350*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(
1/2))*x*a^2*b^2*c^2*d^3+2100*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*a*b
^3*c^3*d^2-945*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*b^4*c^4*d-236*x^2
*a^2*b*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+488*x^2*a*b^2*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-252*x^2
*b^3*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+
b*c)/(b*d)^(1/2))*a^4*c*d^4+180*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^
3*b*c^2*d^3-1350*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b^2*c^3*d^2+2
100*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^3*c^4*d-945*ln(1/2*(2*b*d*
x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^4*c^5-30*x*a^3*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*
d)^(1/2)+674*x*a^2*b*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-1274*x*a*b^2*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b
*d)^(1/2)+630*x*b^3*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-30*a^3*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)
+1678*a^2*b*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-3570*a*b^2*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+1
890*b^3*c^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(d*x+c)^(1/2)/b/d^5

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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)^(5/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 8.52985, size = 1754, normalized size = 5.68 \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)^(5/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/768*(15*(63*b^4*c^5 - 140*a*b^3*c^4*d + 90*a^2*b^2*c^3*d^2 - 12*a^3*b*c^2*d^3 - a^4*c*d^4 + (63*b^4*c^4*d
- 140*a*b^3*c^3*d^2 + 90*a^2*b^2*c^2*d^3 - 12*a^3*b*c*d^4 - a^4*d^5)*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*(48*b^4*d^5*x^4 - 945*b^4*c^4*d + 1785*a*b^3*c^3*d^2 - 839*a^2*b^2*c^2*d^3 + 15*a^3*b*c*d^4 - 8*(9*b^4*
c*d^4 - 17*a*b^3*d^5)*x^3 + 2*(63*b^4*c^2*d^3 - 122*a*b^3*c*d^4 + 59*a^2*b^2*d^5)*x^2 - (315*b^4*c^3*d^2 - 637
*a*b^3*c^2*d^3 + 337*a^2*b^2*c*d^4 - 15*a^3*b*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^7*x + b^2*c*d^6), -1
/384*(15*(63*b^4*c^5 - 140*a*b^3*c^4*d + 90*a^2*b^2*c^3*d^2 - 12*a^3*b*c^2*d^3 - a^4*c*d^4 + (63*b^4*c^4*d - 1
40*a*b^3*c^3*d^2 + 90*a^2*b^2*c^2*d^3 - 12*a^3*b*c*d^4 - a^4*d^5)*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*(48*b^4*d^5*x^4
- 945*b^4*c^4*d + 1785*a*b^3*c^3*d^2 - 839*a^2*b^2*c^2*d^3 + 15*a^3*b*c*d^4 - 8*(9*b^4*c*d^4 - 17*a*b^3*d^5)*
x^3 + 2*(63*b^4*c^2*d^3 - 122*a*b^3*c*d^4 + 59*a^2*b^2*d^5)*x^2 - (315*b^4*c^3*d^2 - 637*a*b^3*c^2*d^3 + 337*a
^2*b^2*c*d^4 - 15*a^3*b*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^7*x + b^2*c*d^6)]

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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)**(5/2)/(d*x+c)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.31332, size = 568, normalized size = 1.84 \begin{align*} \frac{{\left ({\left (2 \,{\left (4 \,{\left (\frac{6 \,{\left (b x + a\right )} b^{2} d^{8}}{b^{12} c d^{10} - a b^{11} d^{11}} - \frac{9 \, b^{3} c d^{7} + 7 \, a b^{2} d^{8}}{b^{12} c d^{10} - a b^{11} d^{11}}\right )}{\left (b x + a\right )} + \frac{63 \, b^{4} c^{2} d^{6} - 14 \, a b^{3} c d^{7} - a^{2} b^{2} d^{8}}{b^{12} c d^{10} - a b^{11} d^{11}}\right )}{\left (b x + a\right )} - \frac{5 \,{\left (63 \, b^{5} c^{3} d^{5} - 77 \, a b^{4} c^{2} d^{6} + 13 \, a^{2} b^{3} c d^{7} + a^{3} b^{2} d^{8}\right )}}{b^{12} c d^{10} - a b^{11} d^{11}}\right )}{\left (b x + a\right )} - \frac{15 \,{\left (63 \, b^{6} c^{4} d^{4} - 140 \, a b^{5} c^{3} d^{5} + 90 \, a^{2} b^{4} c^{2} d^{6} - 12 \, a^{3} b^{3} c d^{7} - a^{4} b^{2} d^{8}\right )}}{b^{12} c d^{10} - a b^{11} d^{11}}\right )} \sqrt{b x + a}}{8257536 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}} - \frac{5 \,{\left (63 \, b^{3} c^{3} - 77 \, a b^{2} c^{2} d + 13 \, a^{2} b c d^{2} + a^{3} d^{3}\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 )}{2752512 \, \sqrt{b d} b^{9} d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(5/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

1/8257536*((2*(4*(6*(b*x + a)*b^2*d^8/(b^12*c*d^10 - a*b^11*d^11) - (9*b^3*c*d^7 + 7*a*b^2*d^8)/(b^12*c*d^10 -
a*b^11*d^11))*(b*x + a) + (63*b^4*c^2*d^6 - 14*a*b^3*c*d^7 - a^2*b^2*d^8)/(b^12*c*d^10 - a*b^11*d^11))*(b*x +
a) - 5*(63*b^5*c^3*d^5 - 77*a*b^4*c^2*d^6 + 13*a^2*b^3*c*d^7 + a^3*b^2*d^8)/(b^12*c*d^10 - a*b^11*d^11))*(b*x
+ a) - 15*(63*b^6*c^4*d^4 - 140*a*b^5*c^3*d^5 + 90*a^2*b^4*c^2*d^6 - 12*a^3*b^3*c*d^7 - a^4*b^2*d^8)/(b^12*c*
d^10 - a*b^11*d^11))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) - 5/2752512*(63*b^3*c^3 - 77*a*b^2*c^2*
d + 13*a^2*b*c*d^2 + a^3*d^3)*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^9*d^6)