3.776 \(\int \frac{x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=250 \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left ((a d+b c) \left (15 a^2 d^2-22 a b c d+15 b^2 c^2\right )-2 b d x \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right )\right )}{4 b^3 d^3 (b c-a d)^2}+\frac{3 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{7/2} d^{7/2}}+\frac{2 a x^3}{b \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}-\frac{2 c x^2 \sqrt{a+b x} (a d+b c)}{b d \sqrt{c+d x} (b c-a d)^2} \]

[Out]

(2*a*x^3)/(b*(b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x]) - (2*c*(b*c + a*d)*x^2*Sqrt[a + b*x])/(b*d*(b*c - a*d)^2
*Sqrt[c + d*x]) - (Sqrt[a + b*x]*Sqrt[c + d*x]*((b*c + a*d)*(15*b^2*c^2 - 22*a*b*c*d + 15*a^2*d^2) - 2*b*d*(5*
b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*x))/(4*b^3*d^3*(b*c - a*d)^2) + (3*(5*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*ArcTan
h[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(7/2)*d^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.216225, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {98, 150, 147, 63, 217, 206} \[ -\frac{\sqrt{a+b x} \sqrt{c+d x} \left ((a d+b c) \left (15 a^2 d^2-22 a b c d+15 b^2 c^2\right )-2 b d x \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right )\right )}{4 b^3 d^3 (b c-a d)^2}+\frac{3 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{7/2} d^{7/2}}+\frac{2 a x^3}{b \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}-\frac{2 c x^2 \sqrt{a+b x} (a d+b c)}{b d \sqrt{c+d x} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((a + b*x)^(3/2)*(c + d*x)^(3/2)),x]

[Out]

(2*a*x^3)/(b*(b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x]) - (2*c*(b*c + a*d)*x^2*Sqrt[a + b*x])/(b*d*(b*c - a*d)^2
*Sqrt[c + d*x]) - (Sqrt[a + b*x]*Sqrt[c + d*x]*((b*c + a*d)*(15*b^2*c^2 - 22*a*b*c*d + 15*a^2*d^2) - 2*b*d*(5*
b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*x))/(4*b^3*d^3*(b*c - a*d)^2) + (3*(5*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*ArcTan
h[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(7/2)*d^(7/2))

Rule 98

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

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 147

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

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^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{2 \int \frac{x^2 \left (3 a c+\frac{1}{2} (-b c+5 a d) x\right )}{\sqrt{a+b x} (c+d x)^{3/2}} \, dx}{b (b c-a d)}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{2 c (b c+a d) x^2 \sqrt{a+b x}}{b d (b c-a d)^2 \sqrt{c+d x}}+\frac{4 \int \frac{x \left (a c (b c+a d)+\frac{1}{4} \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) x\right )}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{b d (b c-a d)^2}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{2 c (b c+a d) x^2 \sqrt{a+b x}}{b d (b c-a d)^2 \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left ((b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right )-2 b d \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) x\right )}{4 b^3 d^3 (b c-a d)^2}+\frac{\left (3 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 b^3 d^3}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{2 c (b c+a d) x^2 \sqrt{a+b x}}{b d (b c-a d)^2 \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left ((b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right )-2 b d \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) x\right )}{4 b^3 d^3 (b c-a d)^2}+\frac{\left (3 \left (5 b^2 c^2+6 a b c d+5 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 )}{4 b^4 d^3}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{2 c (b c+a d) x^2 \sqrt{a+b x}}{b d (b c-a d)^2 \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left ((b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right )-2 b d \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) x\right )}{4 b^3 d^3 (b c-a d)^2}+\frac{\left (3 \left (5 b^2 c^2+6 a b c d+5 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 )}{4 b^4 d^3}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}-\frac{2 c (b c+a d) x^2 \sqrt{a+b x}}{b d (b c-a d)^2 \sqrt{c+d x}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left ((b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right )-2 b d \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) x\right )}{4 b^3 d^3 (b c-a d)^2}+\frac{3 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{7/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.888843, size = 285, normalized size = 1.14 \[ \frac{b \sqrt{d} \left (a^2 b^2 d \left (10 c^2 d x+7 c^3+5 c d^2 x^2+2 d^3 x^3\right )+a^3 b d^2 \left (7 c^2+2 c d x-5 d^2 x^2\right )-15 a^4 d^3 (c+d x)+a b^3 c \left (2 c^2 d x-15 c^3+5 c d^2 x^2-4 d^3 x^3\right )+b^4 c^2 x \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )+3 \sqrt{a+b x} \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) (b c-a d)^{5/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{4 b^4 d^{7/2} \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((a + b*x)^(3/2)*(c + d*x)^(3/2)),x]

[Out]

(b*Sqrt[d]*(-15*a^4*d^3*(c + d*x) + a^3*b*d^2*(7*c^2 + 2*c*d*x - 5*d^2*x^2) + b^4*c^2*x*(-15*c^2 - 5*c*d*x + 2
*d^2*x^2) + a*b^3*c*(-15*c^3 + 2*c^2*d*x + 5*c*d^2*x^2 - 4*d^3*x^3) + a^2*b^2*d*(7*c^3 + 10*c^2*d*x + 5*c*d^2*
x^2 + 2*d^3*x^3)) + 3*(b*c - a*d)^(5/2)*(5*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*Sqrt[a + b*x]*Sqrt[(b*(c + d*x))/(
b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(4*b^4*d^(7/2)*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[
c + d*x])

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Maple [B]  time = 0.032, size = 1369, normalized size = 5.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),x)

[Out]

1/8*(-30*x*b^4*c^4*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-30*a^4*c*d^3*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-30*a*b
^3*c^4*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(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^2*a^4*b*d^5+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^2*b
^5*c^4*d-12*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*b*c^2*d^3-6*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^2*c^3*d^2-12*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^3*c^4*d-30*x*a^4*d^4*(b*d)^(1/2)*((b*x+a)*(d*x+c)
)^(1/2)-6*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^2*a^2*b^3*c^2*d^3-12*l
n(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^2*a*b^4*c^3*d^2+3*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*b*c*d^4-18*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^2*c^2*d^3-18*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^3*c^3*d^2+3*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^4*c^4*d+4*x^3*a^2*b^2*d^4*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*x^3*b^4*c^2*d^2*(b*d)^(
1/2)*((b*x+a)*(d*x+c))^(1/2)-10*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a^3*b*d^4-10*(b*d)^(1/2)*((b*x+a)*(d*x
+c))^(1/2)*x^2*b^4*c^3*d+14*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b*c^2*d^2+14*(b*d)^(1/2)*((b*x+a)*(d*x+c))
^(1/2)*a^2*b^2*c^3*d-8*x^3*a*b^3*c*d^3*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+20*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1
/2)*x*a^2*b^2*c^2*d^2+4*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a*b^3*c^3*d+10*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/
2)*x^2*a^2*b^2*c*d^3+10*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a*b^3*c^2*d^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*b^5*c^5+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^5*c*d^4+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*b^4*c^5+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^5*d^5-12*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^2*a^3*b^2*c*d^4+4*(b*d)^(1/2)*((b*x
+a)*(d*x+c))^(1/2)*x*a^3*b*c*d^3)/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(a*d-b*c)^2/(b*x+a)^(1/2)/(d*x+c)^(1/2)/
b^3/d^3

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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^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.35165, size = 2448, normalized size = 9.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*(5*a*b^4*c^5 - 4*a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 - 4*a^4*b*c^2*d^3 + 5*a^5*c*d^4 + (5*b^5*c^4*d - 4
*a*b^4*c^3*d^2 - 2*a^2*b^3*c^2*d^3 - 4*a^3*b^2*c*d^4 + 5*a^4*b*d^5)*x^2 + (5*b^5*c^5 + a*b^4*c^4*d - 6*a^2*b^3
*c^3*d^2 - 6*a^3*b^2*c^2*d^3 + a^4*b*c*d^4 + 5*a^5*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*(15*a*
b^4*c^4*d - 7*a^2*b^3*c^3*d^2 - 7*a^3*b^2*c^2*d^3 + 15*a^4*b*c*d^4 - 2*(b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*
d^5)*x^3 + 5*(b^5*c^3*d^2 - a*b^4*c^2*d^3 - a^2*b^3*c*d^4 + a^3*b^2*d^5)*x^2 + (15*b^5*c^4*d - 2*a*b^4*c^3*d^2
- 10*a^2*b^3*c^2*d^3 - 2*a^3*b^2*c*d^4 + 15*a^4*b*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^6*c^3*d^4 - 2*a^2
*b^5*c^2*d^5 + a^3*b^4*c*d^6 + (b^7*c^2*d^5 - 2*a*b^6*c*d^6 + a^2*b^5*d^7)*x^2 + (b^7*c^3*d^4 - a*b^6*c^2*d^5
- a^2*b^5*c*d^6 + a^3*b^4*d^7)*x), -1/8*(3*(5*a*b^4*c^5 - 4*a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 - 4*a^4*b*c^2*d^
3 + 5*a^5*c*d^4 + (5*b^5*c^4*d - 4*a*b^4*c^3*d^2 - 2*a^2*b^3*c^2*d^3 - 4*a^3*b^2*c*d^4 + 5*a^4*b*d^5)*x^2 + (5
*b^5*c^5 + a*b^4*c^4*d - 6*a^2*b^3*c^3*d^2 - 6*a^3*b^2*c^2*d^3 + a^4*b*c*d^4 + 5*a^5*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*(15*a*b^4*c^4*d - 7*a^2*b^3*c^3*d^2 - 7*a^3*b^2*c^2*d^3 + 15*a^4*b*c*d^4 - 2*(b^5*c^2*d^3 - 2*a*b^4*c
*d^4 + a^2*b^3*d^5)*x^3 + 5*(b^5*c^3*d^2 - a*b^4*c^2*d^3 - a^2*b^3*c*d^4 + a^3*b^2*d^5)*x^2 + (15*b^5*c^4*d -
2*a*b^4*c^3*d^2 - 10*a^2*b^3*c^2*d^3 - 2*a^3*b^2*c*d^4 + 15*a^4*b*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^6*
c^3*d^4 - 2*a^2*b^5*c^2*d^5 + a^3*b^4*c*d^6 + (b^7*c^2*d^5 - 2*a*b^6*c*d^6 + a^2*b^5*d^7)*x^2 + (b^7*c^3*d^4 -
a*b^6*c^2*d^5 - a^2*b^5*c*d^6 + a^3*b^4*d^7)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\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**4/(b*x+a)**(3/2)/(d*x+c)**(3/2),x)

[Out]

Integral(x**4/((a + b*x)**(3/2)*(c + d*x)**(3/2)), x)

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Giac [B]  time = 2.19517, size = 653, normalized size = 2.61 \begin{align*} -\frac{4 \, \sqrt{b d} a^{4}}{{\left (b^{3} c{\left | b \right |} - a b^{2} d{\left | b \right |}\right )}{\left (b^{2} c - a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} + \frac{{\left ({\left (b x + a\right )}{\left (\frac{2 \,{\left (b^{10} c^{2} d^{4}{\left | b \right |} - 2 \, a b^{9} c d^{5}{\left | b \right |} + a^{2} b^{8} d^{6}{\left | b \right |}\right )}{\left (b x + a\right )}}{b^{14} c^{2} d^{5} - 2 \, a b^{13} c d^{6} + a^{2} b^{12} d^{7}} - \frac{5 \, b^{11} c^{3} d^{3}{\left | b \right |} + a b^{10} c^{2} d^{4}{\left | b \right |} - 17 \, a^{2} b^{9} c d^{5}{\left | b \right |} + 11 \, a^{3} b^{8} d^{6}{\left | b \right |}}{b^{14} c^{2} d^{5} - 2 \, a b^{13} c d^{6} + a^{2} b^{12} d^{7}}\right )} - \frac{15 \, b^{12} c^{4} d^{2}{\left | b \right |} - 12 \, a b^{11} c^{3} d^{3}{\left | b \right |} - 6 \, a^{2} b^{10} c^{2} d^{4}{\left | b \right |} + 20 \, a^{3} b^{9} c d^{5}{\left | b \right |} - 9 \, a^{4} b^{8} d^{6}{\left | b \right |}}{b^{14} c^{2} d^{5} - 2 \, a b^{13} c d^{6} + a^{2} b^{12} d^{7}}\right )} \sqrt{b x + a}}{4 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}} - \frac{3 \,{\left (5 \, \sqrt{b d} b^{2} c^{2} + 6 \, \sqrt{b d} a b c d + 5 \, \sqrt{b d} a^{2} d^{2}\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 )}^{2}\right )}{8 \, b^{3} d^{4}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

-4*sqrt(b*d)*a^4/((b^3*c*abs(b) - a*b^2*d*abs(b))*(b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*
x + a)*b*d - a*b*d))^2)) + 1/4*((b*x + a)*(2*(b^10*c^2*d^4*abs(b) - 2*a*b^9*c*d^5*abs(b) + a^2*b^8*d^6*abs(b))
*(b*x + a)/(b^14*c^2*d^5 - 2*a*b^13*c*d^6 + a^2*b^12*d^7) - (5*b^11*c^3*d^3*abs(b) + a*b^10*c^2*d^4*abs(b) - 1
7*a^2*b^9*c*d^5*abs(b) + 11*a^3*b^8*d^6*abs(b))/(b^14*c^2*d^5 - 2*a*b^13*c*d^6 + a^2*b^12*d^7)) - (15*b^12*c^4
*d^2*abs(b) - 12*a*b^11*c^3*d^3*abs(b) - 6*a^2*b^10*c^2*d^4*abs(b) + 20*a^3*b^9*c*d^5*abs(b) - 9*a^4*b^8*d^6*a
bs(b))/(b^14*c^2*d^5 - 2*a*b^13*c*d^6 + a^2*b^12*d^7))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) - 3/8
*(5*sqrt(b*d)*b^2*c^2 + 6*sqrt(b*d)*a*b*c*d + 5*sqrt(b*d)*a^2*d^2)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^2)/(b^3*d^4*abs(b))