Integrand size = 22, antiderivative size = 376 \[ \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=-\frac {2 c^3 (a+b x)^{7/2}}{3 d^3 (b c-a d) (c+d x)^{3/2}}-\frac {2 c^2 (13 b c-9 a d) (a+b x)^{5/2}}{3 d^4 (b c-a d) \sqrt {c+d x}}-\frac {5 \left (231 b^3 c^3-189 a b^2 c^2 d+21 a^2 b c d^2+a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^6}+\frac {5 \left (231 b^3 c^3-189 a b^2 c^2 d+21 a^2 b c d^2+a^3 d^3\right ) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^5 (b c-a d)}-\frac {(23 b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^4}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^3}+\frac {5 (b c-a d) \left (231 b^3 c^3-189 a b^2 c^2 d+21 a^2 b c d^2+a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{13/2}} \] Output:
-2/3*c^3*(b*x+a)^(7/2)/d^3/(-a*d+b*c)/(d*x+c)^(3/2)-2/3*c^2*(-9*a*d+13*b*c )*(b*x+a)^(5/2)/d^4/(-a*d+b*c)/(d*x+c)^(1/2)-5/64*(a^3*d^3+21*a^2*b*c*d^2- 189*a*b^2*c^2*d+231*b^3*c^3)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d^6+5/96*(a^3*d ^3+21*a^2*b*c*d^2-189*a*b^2*c^2*d+231*b^3*c^3)*(b*x+a)^(3/2)*(d*x+c)^(1/2) /b/d^5/(-a*d+b*c)-1/24*(a*d+23*b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b/d^4+1/4* (b*x+a)^(7/2)*(d*x+c)^(1/2)/b/d^3+5/64*(-a*d+b*c)*(a^3*d^3+21*a^2*b*c*d^2- 189*a*b^2*c^2*d+231*b^3*c^3)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c) ^(1/2))/b^(3/2)/d^(13/2)
Time = 11.58 (sec) , antiderivative size = 331, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {48 x^2 (a+b x)^4+\frac {16 c (a+b x)^4 \left (3 a^2 d^2 (c+2 d x)-14 a b c d (5 c+6 d x)+11 b^2 c^2 (9 c+10 d x)\right )}{d^2 (b c-a d)^2}-\frac {\left (231 b^3 c^3-189 a b^2 c^2 d+21 a^2 b c d^2+a^3 d^3\right ) (c+d x)^2 \left (\sqrt {d} \sqrt {b c-a d} (a+b x) \sqrt {\frac {b (c+d x)}{b c-a d}} \left (33 a^2 d^2+2 a b d (-20 c+13 d x)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )-15 (b c-a d)^3 \sqrt {a+b x} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )}{d^{11/2} (b c-a d)^{5/2} \sqrt {\frac {b (c+d x)}{b c-a d}}}}{192 b d \sqrt {a+b x} (c+d x)^{3/2}} \] Input:
Integrate[(x^3*(a + b*x)^(5/2))/(c + d*x)^(5/2),x]
Output:
(48*x^2*(a + b*x)^4 + (16*c*(a + b*x)^4*(3*a^2*d^2*(c + 2*d*x) - 14*a*b*c* d*(5*c + 6*d*x) + 11*b^2*c^2*(9*c + 10*d*x)))/(d^2*(b*c - a*d)^2) - ((231* b^3*c^3 - 189*a*b^2*c^2*d + 21*a^2*b*c*d^2 + a^3*d^3)*(c + d*x)^2*(Sqrt[d] *Sqrt[b*c - a*d]*(a + b*x)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*(33*a^2*d^2 + 2 *a*b*d*(-20*c + 13*d*x) + b^2*(15*c^2 - 10*c*d*x + 8*d^2*x^2)) - 15*(b*c - a*d)^3*Sqrt[a + b*x]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]]))/( d^(11/2)*(b*c - a*d)^(5/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]))/(192*b*d*Sqrt [a + b*x]*(c + d*x)^(3/2))
Time = 0.39 (sec) , antiderivative size = 322, normalized size of antiderivative = 0.86, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {108, 27, 167, 27, 164, 60, 60, 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.
\(\displaystyle \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 108 |
\(\displaystyle \frac {2 \int \frac {x^2 (a+b x)^{3/2} (6 a+11 b x)}{2 (c+d x)^{3/2}}dx}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x^2 (a+b x)^{3/2} (6 a+11 b x)}{(c+d x)^{3/2}}dx}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 167 |
\(\displaystyle \frac {-\frac {2 \int -\frac {x (a+b x)^{3/2} (4 a (11 b c-6 a d)+b (99 b c-59 a d) x)}{2 \sqrt {c+d x}}dx}{d (b c-a d)}-\frac {2 x^2 (a+b x)^{5/2} (11 b c-6 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {x (a+b x)^{3/2} (4 a (11 b c-6 a d)+b (99 b c-59 a d) x)}{\sqrt {c+d x}}dx}{d (b c-a d)}-\frac {2 x^2 (a+b x)^{5/2} (11 b c-6 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {\frac {\frac {5 \left (a^3 d^3+21 a^2 b c d^2-189 a b^2 c^2 d+231 b^3 c^3\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{16 b d^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x} \left (5 a^2 d^2-2 b d x (99 b c-59 a d)-156 a b c d+231 b^2 c^2\right )}{8 b d^2}}{d (b c-a d)}-\frac {2 x^2 (a+b x)^{5/2} (11 b c-6 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\frac {\frac {5 \left (a^3 d^3+21 a^2 b c d^2-189 a b^2 c^2 d+231 b^3 c^3\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{16 b d^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x} \left (5 a^2 d^2-2 b d x (99 b c-59 a d)-156 a b c d+231 b^2 c^2\right )}{8 b d^2}}{d (b c-a d)}-\frac {2 x^2 (a+b x)^{5/2} (11 b c-6 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\frac {\frac {5 \left (a^3 d^3+21 a^2 b c d^2-189 a b^2 c^2 d+231 b^3 c^3\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{16 b d^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x} \left (5 a^2 d^2-2 b d x (99 b c-59 a d)-156 a b c d+231 b^2 c^2\right )}{8 b d^2}}{d (b c-a d)}-\frac {2 x^2 (a+b x)^{5/2} (11 b c-6 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {\frac {\frac {5 \left (a^3 d^3+21 a^2 b c d^2-189 a b^2 c^2 d+231 b^3 c^3\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{16 b d^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x} \left (5 a^2 d^2-2 b d x (99 b c-59 a d)-156 a b c d+231 b^2 c^2\right )}{8 b d^2}}{d (b c-a d)}-\frac {2 x^2 (a+b x)^{5/2} (11 b c-6 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\frac {5 \left (a^3 d^3+21 a^2 b c d^2-189 a b^2 c^2 d+231 b^3 c^3\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{16 b d^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x} \left (5 a^2 d^2-2 b d x (99 b c-59 a d)-156 a b c d+231 b^2 c^2\right )}{8 b d^2}}{d (b c-a d)}-\frac {2 x^2 (a+b x)^{5/2} (11 b c-6 a d)}{d \sqrt {c+d x} (b c-a d)}}{3 d}-\frac {2 x^3 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}\) |
Input:
Int[(x^3*(a + b*x)^(5/2))/(c + d*x)^(5/2),x]
Output:
(-2*x^3*(a + b*x)^(5/2))/(3*d*(c + d*x)^(3/2)) + ((-2*(11*b*c - 6*a*d)*x^2 *(a + b*x)^(5/2))/(d*(b*c - a*d)*Sqrt[c + d*x]) + (-1/8*((a + b*x)^(5/2)*S qrt[c + d*x]*(231*b^2*c^2 - 156*a*b*c*d + 5*a^2*d^2 - 2*b*d*(99*b*c - 59*a *d)*x))/(b*d^2) + (5*(231*b^3*c^3 - 189*a*b^2*c^2*d + 21*a^2*b*c*d^2 + a^3 *d^3)*(((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (3*(b*c - a*d)*((Sqrt[a + b *x]*Sqrt[c + d*x])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[ b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(3/2))))/(4*d)))/(16*b*d^2))/(d*(b*c - a*d) ))/(3*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (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[(-(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] + Simp[(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]
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. \(1365\) vs. \(2(326)=652\).
Time = 0.25 (sec) , antiderivative size = 1366, normalized size of antiderivative = 3.63
Input:
int(x^3*(b*x+a)^(5/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/384*(b*x+a)^(1/2)*(-14028*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^3 *d^2*x+632*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a*b^2*c*d^4*x^3-60*((b*x+a) *(d*x+c))^(1/2)*(d*b)^(1/2)*a^3*c*d^4*x+300*ln(1/2*(2*b*d*x+2*((b*x+a)*(d* x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^3*b*c*d^5*x^2-3150*ln(1/2* (2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b ^2*c^2*d^4*x^2+6300*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+ a*d+b*c)/(d*b)^(1/2))*a*b^3*c^3*d^3*x^2+1386*(d*b)^(1/2)*((b*x+a)*(d*x+c)) ^(1/2)*b^3*c^3*d^2*x^2+600*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b) ^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^3*b*c^2*d^4*x-6300*ln(1/2*(2*b*d*x+2*((b*x+ a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2*c^3*d^3*x+1260 0*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/ 2))*a*b^3*c^4*d^2*x+9240*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^3*c^4*d*x+3 486*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c^3*d^2-10290*(d*b)^(1/2)*(( b*x+a)*(d*x+c))^(1/2)*a*b^2*c^4*d+4944*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2) *a^2*b*c^2*d^3*x-30*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a^3*d^5*x^2+30*ln( 1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a ^4*c*d^5*x-30*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a^3*c^2*d^3-272*a*b^2*d^ 5*x^4*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+176*b^3*c*d^4*x^4*((b*x+a)*(d*x+ c))^(1/2)*(d*b)^(1/2)+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^( 1/2)+a*d+b*c)/(d*b)^(1/2))*a^4*d^6*x^2+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d...
Time = 0.99 (sec) , antiderivative size = 1066, normalized size of antiderivative = 2.84 \[ \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(x^3*(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="fricas")
Output:
[-1/768*(15*(231*b^4*c^6 - 420*a*b^3*c^5*d + 210*a^2*b^2*c^4*d^2 - 20*a^3* b*c^3*d^3 - a^4*c^2*d^4 + (231*b^4*c^4*d^2 - 420*a*b^3*c^3*d^3 + 210*a^2*b ^2*c^2*d^4 - 20*a^3*b*c*d^5 - a^4*d^6)*x^2 + 2*(231*b^4*c^5*d - 420*a*b^3* c^4*d^2 + 210*a^2*b^2*c^3*d^3 - 20*a^3*b*c^2*d^4 - a^4*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*(4 8*b^4*d^6*x^5 - 3465*b^4*c^5*d + 5145*a*b^3*c^4*d^2 - 1743*a^2*b^2*c^3*d^3 + 15*a^3*b*c^2*d^4 - 8*(11*b^4*c*d^5 - 17*a*b^3*d^6)*x^4 + 2*(99*b^4*c^2* d^4 - 158*a*b^3*c*d^5 + 59*a^2*b^2*d^6)*x^3 - 3*(231*b^4*c^3*d^3 - 387*a*b ^3*c^2*d^4 + 161*a^2*b^2*c*d^5 - 5*a^3*b*d^6)*x^2 - 6*(770*b^4*c^4*d^2 - 1 169*a*b^3*c^3*d^3 + 412*a^2*b^2*c^2*d^4 - 5*a^3*b*c*d^5)*x)*sqrt(b*x + a)* sqrt(d*x + c))/(b^2*d^9*x^2 + 2*b^2*c*d^8*x + b^2*c^2*d^7), -1/384*(15*(23 1*b^4*c^6 - 420*a*b^3*c^5*d + 210*a^2*b^2*c^4*d^2 - 20*a^3*b*c^3*d^3 - a^4 *c^2*d^4 + (231*b^4*c^4*d^2 - 420*a*b^3*c^3*d^3 + 210*a^2*b^2*c^2*d^4 - 20 *a^3*b*c*d^5 - a^4*d^6)*x^2 + 2*(231*b^4*c^5*d - 420*a*b^3*c^4*d^2 + 210*a ^2*b^2*c^3*d^3 - 20*a^3*b*c^2*d^4 - a^4*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*(48*b^4*d^6*x^5 - 3465*b^4*c^5*d + 5 145*a*b^3*c^4*d^2 - 1743*a^2*b^2*c^3*d^3 + 15*a^3*b*c^2*d^4 - 8*(11*b^4*c* d^5 - 17*a*b^3*d^6)*x^4 + 2*(99*b^4*c^2*d^4 - 158*a*b^3*c*d^5 + 59*a^2*...
\[ \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x^{3} \left (a + b x\right )^{\frac {5}{2}}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**3*(b*x+a)**(5/2)/(d*x+c)**(5/2),x)
Output:
Integral(x**3*(a + b*x)**(5/2)/(c + d*x)**(5/2), x)
Exception generated. \[ \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^3*(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="maxima")
Output:
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. 686 vs. \(2 (326) = 652\).
Time = 0.25 (sec) , antiderivative size = 686, normalized size of antiderivative = 1.82 \[ \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(x^3*(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="giac")
Output:
1/192*(((2*(4*(b*x + a)*(6*(b^5*c*d^10*abs(b) - a*b^4*d^11*abs(b))*(b*x + a)/(b^6*c*d^11 - a*b^5*d^12) - (11*b^6*c^2*d^9*abs(b) + 2*a*b^5*c*d^10*abs (b) - 13*a^2*b^4*d^11*abs(b))/(b^6*c*d^11 - a*b^5*d^12)) + 9*(11*b^7*c^3*d ^8*abs(b) - 9*a*b^6*c^2*d^9*abs(b) + a^2*b^5*c*d^10*abs(b) - 3*a^3*b^4*d^1 1*abs(b))/(b^6*c*d^11 - a*b^5*d^12))*(b*x + a) - 3*(231*b^8*c^4*d^7*abs(b) - 420*a*b^7*c^3*d^8*abs(b) + 210*a^2*b^6*c^2*d^9*abs(b) - 20*a^3*b^5*c*d^ 10*abs(b) - a^4*b^4*d^11*abs(b))/(b^6*c*d^11 - a*b^5*d^12))*(b*x + a) - 20 *(231*b^9*c^5*d^6*abs(b) - 651*a*b^8*c^4*d^7*abs(b) + 630*a^2*b^7*c^3*d^8* abs(b) - 230*a^3*b^6*c^2*d^9*abs(b) + 19*a^4*b^5*c*d^10*abs(b) + a^5*b^4*d ^11*abs(b))/(b^6*c*d^11 - a*b^5*d^12))*(b*x + a) - 15*(231*b^10*c^6*d^5*ab s(b) - 882*a*b^9*c^5*d^6*abs(b) + 1281*a^2*b^8*c^4*d^7*abs(b) - 860*a^3*b^ 7*c^3*d^8*abs(b) + 249*a^4*b^6*c^2*d^9*abs(b) - 18*a^5*b^5*c*d^10*abs(b) - a^6*b^4*d^11*abs(b))/(b^6*c*d^11 - a*b^5*d^12))*sqrt(b*x + a)/(b^2*c + (b *x + a)*b*d - a*b*d)^(3/2) - 5/64*(231*b^4*c^4*abs(b) - 420*a*b^3*c^3*d*ab s(b) + 210*a^2*b^2*c^2*d^2*abs(b) - 20*a^3*b*c*d^3*abs(b) - a^4*d^4*abs(b) )*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^6)
Timed out. \[ \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x^3\,{\left (a+b\,x\right )}^{5/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int((x^3*(a + b*x)^(5/2))/(c + d*x)^(5/2),x)
Output:
int((x^3*(a + b*x)^(5/2))/(c + d*x)^(5/2), x)
\[ \int \frac {x^3 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x^{3} \left (b x +a \right )^{\frac {5}{2}}}{\left (d x +c \right )^{\frac {5}{2}}}d x \] Input:
int(x^3*(b*x+a)^(5/2)/(d*x+c)^(5/2),x)
Output:
int(x^3*(b*x+a)^(5/2)/(d*x+c)^(5/2),x)