3.785 \(\int \frac{x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=251 \[ \frac{\sqrt{a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (9 a^2 b c d^2-9 a^3 d^3-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{3 b^2 d^3 \sqrt{c+d x} (b c-a d)^3}-\frac{(3 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{5/2} d^{7/2}}+\frac{2 a x^3}{b \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}-\frac{2 c x^2 \sqrt{a+b x} (3 a d+b c)}{3 b d (c+d x)^{3/2} (b c-a d)^2} \]

[Out]

(2*a*x^3)/(b*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - (2*c*(b*c + 3*a*d)*x^2*Sqrt[a + b*x])/(3*b*d*(b*c -
a*d)^2*(c + d*x)^(3/2)) + (Sqrt[a + b*x]*(c*(15*b^3*c^3 - 31*a*b^2*c^2*d + 9*a^2*b*c*d^2 - 9*a^3*d^3) + d*(b*c
- a*d)*(5*b^2*c^2 - 6*a*b*c*d + 9*a^2*d^2)*x))/(3*b^2*d^3*(b*c - a*d)^3*Sqrt[c + d*x]) - ((5*b*c + 3*a*d)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(5/2)*d^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.205274, antiderivative size = 251, 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, 143, 63, 217, 206} \[ \frac{\sqrt{a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (9 a^2 b c d^2-9 a^3 d^3-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{3 b^2 d^3 \sqrt{c+d x} (b c-a d)^3}-\frac{(3 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{5/2} d^{7/2}}+\frac{2 a x^3}{b \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}-\frac{2 c x^2 \sqrt{a+b x} (3 a d+b c)}{3 b d (c+d x)^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]

[Out]

(2*a*x^3)/(b*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - (2*c*(b*c + 3*a*d)*x^2*Sqrt[a + b*x])/(3*b*d*(b*c -
a*d)^2*(c + d*x)^(3/2)) + (Sqrt[a + b*x]*(c*(15*b^3*c^3 - 31*a*b^2*c^2*d + 9*a^2*b*c*d^2 - 9*a^3*d^3) + d*(b*c
- a*d)*(5*b^2*c^2 - 6*a*b*c*d + 9*a^2*d^2)*x))/(3*b^2*d^3*(b*c - a*d)^3*Sqrt[c + d*x]) - ((5*b*c + 3*a*d)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(5/2)*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 143

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

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)^{5/2}} \, dx &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}-\frac{2 \int \frac{x^2 \left (3 a c+\frac{1}{2} (-b c+3 a d) x\right )}{\sqrt{a+b x} (c+d x)^{5/2}} \, dx}{b (b c-a d)}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}-\frac{2 c (b c+3 a d) x^2 \sqrt{a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac{4 \int \frac{x \left (a c (b c+3 a d)+\frac{1}{4} \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{\sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 b d (b c-a d)^2}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}-\frac{2 c (b c+3 a d) x^2 \sqrt{a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac{\sqrt{a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt{c+d x}}-\frac{(5 b c+3 a d) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 b^2 d^3}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}-\frac{2 c (b c+3 a d) x^2 \sqrt{a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac{\sqrt{a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt{c+d x}}-\frac{(5 b c+3 a d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{b^3 d^3}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}-\frac{2 c (b c+3 a d) x^2 \sqrt{a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac{\sqrt{a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt{c+d x}}-\frac{(5 b c+3 a d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{b^3 d^3}\\ &=\frac{2 a x^3}{b (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}-\frac{2 c (b c+3 a d) x^2 \sqrt{a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac{\sqrt{a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt{c+d x}}-\frac{(5 b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{5/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.692098, size = 285, normalized size = 1.14 \[ \frac{-a b d^{5/2} (3 a d-b c) \left (a^2 \left (8 c^2+12 c d x+3 d^2 x^2\right )+2 a b c x (2 c+3 d x)-b^2 c^2 x^2\right )+3 b^2 d^{5/2} x^3 (b c-a d)^3-\sqrt{a+b x} (b c-a d) (3 a d+5 b c) \left (b c^2 \sqrt{d} \sqrt{a+b x} (b c-a d)-(c+d x) \left (2 b c \sqrt{d} \sqrt{a+b x} (2 b c-3 a d)-3 (b c-a d)^{5/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )\right )\right )}{3 b^3 d^{7/2} \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]

[Out]

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

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Maple [B]  time = 0.028, size = 1714, normalized size = 6.8 \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)^(5/2),x)

[Out]

-1/6*(-30*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^5*d+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^5*c*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))*a^4*b*c^3*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))*a^3*b^2*c^4*d^2+36*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^5*d-18*x^2*a^4*d^5*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+30*x*b^4*c^5*((b*x+a)*(d*x+
c))^(1/2)*(b*d)^(1/2)-18*a^4*c^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+30*a*b^3*c^5*((b*x+a)*(d*x+c))^(1/2)*
(b*d)^(1/2)+9*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^3*a^4*b*d^6+18*x^3
*a^2*b^2*c*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-18*x^3*a*b^3*c^2*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^3*b^5*c^4*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^6+9*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^2*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^6+9*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^5*d^6+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*a*b^4*c^5*d-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^3*a^2*b^3*c^2*d^4+36*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^3*a*b^4*c^3*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))*x^2*a^4*b*c*d^5-42*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^2*d^4+57*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*b^4*c^4*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*a^4*b*c^2*d^4-48*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^3*d^3+54*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^4*d^2-6*
x^3*a^3*b*d^5*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+6*x^3*b^4*c^3*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+40*x^2
*b^4*c^4*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-36*x*a^4*c*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+18*a^3*b*c^3
*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-62*a^2*b^2*c^4*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-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))*x^3*a^3*b^2*c*d^5+6*x^2*a^3*b*c*d^4*((b*x+a)*(d
*x+c))^(1/2)*(b*d)^(1/2)+18*x^2*a^2*b^2*c^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-78*x^2*a*b^3*c^3*d^2*((b*x
+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+30*x*a^3*b*c^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-66*x*a^2*b^2*c^3*d^2*((b
*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-22*x*a*b^3*c^4*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2)/(a*d-b*c)^3
/((b*x+a)*(d*x+c))^(1/2)/b^2/d^3/(d*x+c)^(3/2)/(b*x+a)^(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^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.5286, size = 3308, normalized size = 13.18 \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)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(5*a*b^4*c^6 - 12*a^2*b^3*c^5*d + 6*a^3*b^2*c^4*d^2 + 4*a^4*b*c^3*d^3 - 3*a^5*c^2*d^4 + (5*b^5*c^4*d^
2 - 12*a*b^4*c^3*d^3 + 6*a^2*b^3*c^2*d^4 + 4*a^3*b^2*c*d^5 - 3*a^4*b*d^6)*x^3 + (10*b^5*c^5*d - 19*a*b^4*c^4*d
^2 + 14*a^3*b^2*c^2*d^4 - 2*a^4*b*c*d^5 - 3*a^5*d^6)*x^2 + (5*b^5*c^6 - 2*a*b^4*c^5*d - 18*a^2*b^3*c^4*d^2 + 1
6*a^3*b^2*c^3*d^3 + 5*a^4*b*c^2*d^4 - 6*a^5*c*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
^5*d - 31*a^2*b^3*c^4*d^2 + 9*a^3*b^2*c^3*d^3 - 9*a^4*b*c^2*d^4 + 3*(b^5*c^3*d^3 - 3*a*b^4*c^2*d^4 + 3*a^2*b^3
*c*d^5 - a^3*b^2*d^6)*x^3 + (20*b^5*c^4*d^2 - 39*a*b^4*c^3*d^3 + 9*a^2*b^3*c^2*d^4 + 3*a^3*b^2*c*d^5 - 9*a^4*b
*d^6)*x^2 + (15*b^5*c^5*d - 11*a*b^4*c^4*d^2 - 33*a^2*b^3*c^3*d^3 + 15*a^3*b^2*c^2*d^4 - 18*a^4*b*c*d^5)*x)*sq
rt(b*x + a)*sqrt(d*x + c))/(a*b^6*c^5*d^4 - 3*a^2*b^5*c^4*d^5 + 3*a^3*b^4*c^3*d^6 - a^4*b^3*c^2*d^7 + (b^7*c^3
*d^6 - 3*a*b^6*c^2*d^7 + 3*a^2*b^5*c*d^8 - a^3*b^4*d^9)*x^3 + (2*b^7*c^4*d^5 - 5*a*b^6*c^3*d^6 + 3*a^2*b^5*c^2
*d^7 + a^3*b^4*c*d^8 - a^4*b^3*d^9)*x^2 + (b^7*c^5*d^4 - a*b^6*c^4*d^5 - 3*a^2*b^5*c^3*d^6 + 5*a^3*b^4*c^2*d^7
- 2*a^4*b^3*c*d^8)*x), 1/6*(3*(5*a*b^4*c^6 - 12*a^2*b^3*c^5*d + 6*a^3*b^2*c^4*d^2 + 4*a^4*b*c^3*d^3 - 3*a^5*c
^2*d^4 + (5*b^5*c^4*d^2 - 12*a*b^4*c^3*d^3 + 6*a^2*b^3*c^2*d^4 + 4*a^3*b^2*c*d^5 - 3*a^4*b*d^6)*x^3 + (10*b^5*
c^5*d - 19*a*b^4*c^4*d^2 + 14*a^3*b^2*c^2*d^4 - 2*a^4*b*c*d^5 - 3*a^5*d^6)*x^2 + (5*b^5*c^6 - 2*a*b^4*c^5*d -
18*a^2*b^3*c^4*d^2 + 16*a^3*b^2*c^3*d^3 + 5*a^4*b*c^2*d^4 - 6*a^5*c*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^5*d - 31*a^2*b^3*c^4*d^2 + 9*a^3*b^2*c^3*d^3 - 9*a^4*b*c^2*d^4 + 3*(b^5*c^3*d^3 - 3*a*b^4*c^2*d^4 + 3*a^2*
b^3*c*d^5 - a^3*b^2*d^6)*x^3 + (20*b^5*c^4*d^2 - 39*a*b^4*c^3*d^3 + 9*a^2*b^3*c^2*d^4 + 3*a^3*b^2*c*d^5 - 9*a^
4*b*d^6)*x^2 + (15*b^5*c^5*d - 11*a*b^4*c^4*d^2 - 33*a^2*b^3*c^3*d^3 + 15*a^3*b^2*c^2*d^4 - 18*a^4*b*c*d^5)*x)
*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^6*c^5*d^4 - 3*a^2*b^5*c^4*d^5 + 3*a^3*b^4*c^3*d^6 - a^4*b^3*c^2*d^7 + (b^7*
c^3*d^6 - 3*a*b^6*c^2*d^7 + 3*a^2*b^5*c*d^8 - a^3*b^4*d^9)*x^3 + (2*b^7*c^4*d^5 - 5*a*b^6*c^3*d^6 + 3*a^2*b^5*
c^2*d^7 + a^3*b^4*c*d^8 - a^4*b^3*d^9)*x^2 + (b^7*c^5*d^4 - a*b^6*c^4*d^5 - 3*a^2*b^5*c^3*d^6 + 5*a^3*b^4*c^2*
d^7 - 2*a^4*b^3*c*d^8)*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{5}{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)**(5/2),x)

[Out]

Integral(x**4/((a + b*x)**(3/2)*(c + d*x)**(5/2)), x)

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Giac [B]  time = 1.82163, size = 1017, normalized size = 4.05 \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)^(5/2),x, algorithm="giac")

[Out]

-4*sqrt(b*d)*a^4/((b^3*c^2*abs(b) - 2*a*b^2*c*d*abs(b) + a^2*b*d^2*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)) + 1/3*((b*x + a)*(3*(b^8*c^5*d^4*abs(b) - 5*a*b^7*c^4*d^5*ab
s(b) + 10*a^2*b^6*c^3*d^6*abs(b) - 10*a^3*b^5*c^2*d^7*abs(b) + 5*a^4*b^4*c*d^8*abs(b) - a^5*b^3*d^9*abs(b))*(b
*x + a)/(b^10*c^5*d^5 - 5*a*b^9*c^4*d^6 + 10*a^2*b^8*c^3*d^7 - 10*a^3*b^7*c^2*d^8 + 5*a^4*b^6*c*d^9 - a^5*b^5*
d^10) + 2*(10*b^9*c^6*d^3*abs(b) - 44*a*b^8*c^5*d^4*abs(b) + 76*a^2*b^7*c^4*d^5*abs(b) - 72*a^3*b^6*c^3*d^6*ab
s(b) + 45*a^4*b^5*c^2*d^7*abs(b) - 18*a^5*b^4*c*d^8*abs(b) + 3*a^6*b^3*d^9*abs(b))/(b^10*c^5*d^5 - 5*a*b^9*c^4
*d^6 + 10*a^2*b^8*c^3*d^7 - 10*a^3*b^7*c^2*d^8 + 5*a^4*b^6*c*d^9 - a^5*b^5*d^10)) + 3*(5*b^10*c^7*d^2*abs(b) -
27*a*b^9*c^6*d^3*abs(b) + 57*a^2*b^8*c^5*d^4*abs(b) - 63*a^3*b^7*c^4*d^5*abs(b) + 43*a^4*b^6*c^3*d^6*abs(b) -
21*a^5*b^5*c^2*d^7*abs(b) + 7*a^6*b^4*c*d^8*abs(b) - a^7*b^3*d^9*abs(b))/(b^10*c^5*d^5 - 5*a*b^9*c^4*d^6 + 10
*a^2*b^8*c^3*d^7 - 10*a^3*b^7*c^2*d^8 + 5*a^4*b^6*c*d^9 - a^5*b^5*d^10))*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d
- a*b*d)^(3/2) + 1/2*(5*sqrt(b*d)*b*c + 3*sqrt(b*d)*a*d)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)
*b*d - a*b*d))^2)/(b^2*d^4*abs(b))