Integrand size = 22, antiderivative size = 251 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {2 a x^3}{b (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 c (b c+3 a d) x^2 \sqrt {a+b x}}{3 b d (b c-a d)^2 (c+d x)^{3/2}}+\frac {\sqrt {a+b x} \left (c \left (15 b^3 c^3-31 a b^2 c^2 d+9 a^2 b c d^2-9 a^3 d^3\right )+d (b c-a d) \left (5 b^2 c^2-6 a b c d+9 a^2 d^2\right ) x\right )}{3 b^2 d^3 (b c-a d)^3 \sqrt {c+d x}}-\frac {(5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} d^{7/2}} \]
-(3*a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2 )/d^(7/2)+2*a*x^3/b/(-a*d+b*c)/(d*x+c)^(3/2)/(b*x+a)^(1/2)-2/3*c*(3*a*d+b* c)*x^2*(b*x+a)^(1/2)/b/d/(-a*d+b*c)^2/(d*x+c)^(3/2)+1/3*(c*(-9*a^3*d^3+9*a ^2*b*c*d^2-31*a*b^2*c^2*d+15*b^3*c^3)+d*(-a*d+b*c)*(9*a^2*d^2-6*a*b*c*d+5* b^2*c^2)*x)*(b*x+a)^(1/2)/b^2/d^3/(-a*d+b*c)^3/(d*x+c)^(1/2)
Time = 0.39 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {9 a^4 d^3 (c+d x)^2+3 a^3 b d^2 (-3 c+d x) (c+d x)^2-b^4 c^3 x \left (15 c^2+20 c d x+3 d^2 x^2\right )+a^2 b^2 c d \left (31 c^3+33 c^2 d x-9 c d^2 x^2-9 d^3 x^3\right )+a b^3 c^2 \left (-15 c^3+11 c^2 d x+39 c d^2 x^2+9 d^3 x^3\right )}{3 b^2 d^3 (-b c+a d)^3 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {(5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{5/2} d^{7/2}} \]
(9*a^4*d^3*(c + d*x)^2 + 3*a^3*b*d^2*(-3*c + d*x)*(c + d*x)^2 - b^4*c^3*x* (15*c^2 + 20*c*d*x + 3*d^2*x^2) + a^2*b^2*c*d*(31*c^3 + 33*c^2*d*x - 9*c*d ^2*x^2 - 9*d^3*x^3) + a*b^3*c^2*(-15*c^3 + 11*c^2*d*x + 39*c*d^2*x^2 + 9*d ^3*x^3))/(3*b^2*d^3*(-(b*c) + a*d)^3*Sqrt[a + b*x]*(c + d*x)^(3/2)) - ((5* b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(b^ (5/2)*d^(7/2))
Time = 0.41 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {109, 27, 167, 27, 160, 66, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {2 \int \frac {x^2 (6 a c-(b c-3 a d) x)}{2 \sqrt {a+b x} (c+d x)^{5/2}}dx}{b (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\int \frac {x^2 (6 a c-(b c-3 a d) x)}{\sqrt {a+b x} (c+d x)^{5/2}}dx}{b (b c-a d)}\) |
| \(\Big \downarrow \) 167 |
| \(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {2 \int \frac {x \left (4 a c (b c+3 a d)+\left (5 b^2 c^2-6 a b d c+9 a^2 d^2\right ) x\right )}{2 \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 d (b c-a d)}}{b (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {\int \frac {x \left (4 a c (b c+3 a d)+\left (5 b^2 c^2-6 a b d c+9 a^2 d^2\right ) x\right )}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{3 d (b c-a d)}}{b (b c-a d)}\) |
| \(\Big \downarrow \) 160 |
| \(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\sqrt {a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (-9 a^3 d^3+9 a^2 b c d^2-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{b d^2 \sqrt {c+d x} (b c-a d)}-\frac {3 (b c-a d)^2 (3 a d+5 b c) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b d^2}}{3 d (b c-a d)}}{b (b c-a d)}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\sqrt {a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (-9 a^3 d^3+9 a^2 b c d^2-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{b d^2 \sqrt {c+d x} (b c-a d)}-\frac {3 (b c-a d)^2 (3 a d+5 b c) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b d^2}}{3 d (b c-a d)}}{b (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\sqrt {a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (-9 a^3 d^3+9 a^2 b c d^2-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{b d^2 \sqrt {c+d x} (b c-a d)}-\frac {3 (b c-a d)^2 (3 a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} d^{5/2}}}{3 d (b c-a d)}}{b (b c-a d)}\) |
(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*d*(b*c - a*d)*(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))/(b*d^2*(b*c - a*d)*Sqrt[c + d* x]) - (3*(b*c - a*d)^2*(5*b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sq rt[b]*Sqrt[c + d*x])])/(b^(3/2)*d^(5/2)))/(3*d*(b*c - a*d)))/(b*(b*c - a*d ))
3.8.85.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> 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] + Simp[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])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) )*((g_.) + (h_.)*(x_)), x_] :> 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] + Simp[(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[m, 1] || !SumSimplerQ[n, 1])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] - Simp[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] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Leaf count of result is larger than twice the leaf count of optimal. \(1713\) vs. \(2(223)=446\).
Time = 0.59 (sec) , antiderivative size = 1714, normalized size of antiderivative = 6.83
-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))*b^5*c^5*d*x^2-18*a^4*d^5*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/ 2)+30*b^4*c^5*x*((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)-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^2-15*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))*a^5*d ^6*x^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))*b^5*c^6*x+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^4-62*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))*a^3*b^2*c*d^5*x^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^2*b^3*c^2*d^4*x^3+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*b^ 4*c^3*d^3*x^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^4*b*c*d^5*x^2-42*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))*a^3*b^2*c^2*d^4*x^2+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))*a*b^4*c^4* d^2*x^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*b*c^2*d^4*x-48*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/...
Leaf count of result is larger than twice the leaf count of optimal. 807 vs. \(2 (223) = 446\).
Time = 0.58 (sec) , antiderivative size = 1628, normalized size of antiderivative = 6.49 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \]
[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 + 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)*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)*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), 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 -...
\[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {x^{4}}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 756 vs. \(2 (223) = 446\).
Time = 0.50 (sec) , antiderivative size = 756, normalized size of antiderivative = 3.01 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=-\frac {4 \, a^{4} d}{{\left (\sqrt {b d} b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a b c d {\left | b \right |} + \sqrt {b d} a^{2} d^{2} {\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 {3 \, {\left (b^{8} c^{5} d^{4} {\left | b \right |} - 5 \, a b^{7} c^{4} d^{5} {\left | b \right |} + 10 \, a^{2} b^{6} c^{3} d^{6} {\left | b \right |} - 10 \, a^{3} b^{5} c^{2} d^{7} {\left | b \right |} + 5 \, a^{4} b^{4} c d^{8} {\left | b \right |} - a^{5} b^{3} d^{9} {\left | b \right |}\right )} {\left (b x + a\right )}}{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}} + \frac {2 \, {\left (10 \, b^{9} c^{6} d^{3} {\left | b \right |} - 44 \, a b^{8} c^{5} d^{4} {\left | b \right |} + 76 \, a^{2} b^{7} c^{4} d^{5} {\left | b \right |} - 72 \, a^{3} b^{6} c^{3} d^{6} {\left | b \right |} + 45 \, a^{4} b^{5} c^{2} d^{7} {\left | b \right |} - 18 \, a^{5} b^{4} c d^{8} {\left | b \right |} + 3 \, a^{6} b^{3} d^{9} {\left | b \right |}\right )}}{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}}\right )} + \frac {3 \, {\left (5 \, b^{10} c^{7} d^{2} {\left | b \right |} - 27 \, a b^{9} c^{6} d^{3} {\left | b \right |} + 57 \, a^{2} b^{8} c^{5} d^{4} {\left | b \right |} - 63 \, a^{3} b^{7} c^{4} d^{5} {\left | b \right |} + 43 \, a^{4} b^{6} c^{3} d^{6} {\left | b \right |} - 21 \, a^{5} b^{5} c^{2} d^{7} {\left | b \right |} + 7 \, a^{6} b^{4} c d^{8} {\left | b \right |} - a^{7} b^{3} d^{9} {\left | b \right |}\right )}}{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}}\right )} \sqrt {b x + a}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, b c + 3 \, a d\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 )}{2 \, \sqrt {b d} b d^{3} {\left | b \right |}} \]
-4*a^4*d/((sqrt(b*d)*b^2*c^2*abs(b) - 2*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b* d)*a^2*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*abs(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*abs(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*b*c + 3*a*d)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(b*d)*b*d^3*abs(b))
Timed out. \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {x^4}{{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \]