Integrand size = 22, antiderivative size = 287 \[ \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}+\frac {4 c (5 b c-3 a d) (a+b x)^{5/2}}{3 d^3 (b c-a d) \sqrt {c+d x}}+\frac {5 \left (21 b^2 c^2-14 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 d^5}-\frac {5 \left (21 b^2 c^2-14 a b c d+a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^4 (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d^3}-\frac {5 (b c-a d) \left (21 b^2 c^2-14 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{11/2}} \] Output:
2/3*c^2*(b*x+a)^(7/2)/d^2/(-a*d+b*c)/(d*x+c)^(3/2)+4/3*c*(-3*a*d+5*b*c)*(b *x+a)^(5/2)/d^3/(-a*d+b*c)/(d*x+c)^(1/2)+5/8*(a^2*d^2-14*a*b*c*d+21*b^2*c^ 2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^5-5/12*(a^2*d^2-14*a*b*c*d+21*b^2*c^2)*(b *x+a)^(3/2)*(d*x+c)^(1/2)/d^4/(-a*d+b*c)+1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d ^3-5/8*(-a*d+b*c)*(a^2*d^2-14*a*b*c*d+21*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^ (1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(1/2)/d^(11/2)
Time = 10.93 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {-8 c^2 (a+b x)^4+\frac {16 c (5 b c-3 a d) (a+b x)^4 (c+d x)}{b c-a d}+\frac {15 (b c-a d)^3 \left (21 b^2 c^2-14 a b c d+a^2 d^2\right ) (c+d x)^2 \left (\frac {2 d (a+b x)}{b c-a d}-\frac {4 d^2 (a+b x)^2}{3 (b c-a d)^2}+\frac {16 d^3 (a+b x)^3}{15 (b c-a d)^3}-\frac {2 \sqrt {d} \sqrt {a+b x} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{4 d^4 (-b c+a d)}}{12 d^2 (-b c+a d) \sqrt {a+b x} (c+d x)^{3/2}} \] Input:
Integrate[(x^2*(a + b*x)^(5/2))/(c + d*x)^(5/2),x]
Output:
(-8*c^2*(a + b*x)^4 + (16*c*(5*b*c - 3*a*d)*(a + b*x)^4*(c + d*x))/(b*c - a*d) + (15*(b*c - a*d)^3*(21*b^2*c^2 - 14*a*b*c*d + a^2*d^2)*(c + d*x)^2*( (2*d*(a + b*x))/(b*c - a*d) - (4*d^2*(a + b*x)^2)/(3*(b*c - a*d)^2) + (16* d^3*(a + b*x)^3)/(15*(b*c - a*d)^3) - (2*Sqrt[d]*Sqrt[a + b*x]*ArcSinh[(Sq rt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x)) /(b*c - a*d)])))/(4*d^4*(-(b*c) + a*d)))/(12*d^2*(-(b*c) + a*d)*Sqrt[a + b *x]*(c + d*x)^(3/2))
Time = 0.35 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {100, 27, 87, 60, 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.
\(\displaystyle \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {2 \int \frac {(a+b x)^{5/2} (c (7 b c-3 a d)-3 d (b c-a d) x)}{2 (c+d x)^{3/2}}dx}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\int \frac {(a+b x)^{5/2} (c (7 b c-3 a d)-3 d (b c-a d) x)}{(c+d x)^{3/2}}dx}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{7/2} (5 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {3 \left (a^2 d^2-14 a b c d+21 b^2 c^2\right ) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}}dx}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{7/2} (5 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {3 \left (a^2 d^2-14 a b c d+21 b^2 c^2\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{7/2} (5 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {3 \left (a^2 d^2-14 a b c d+21 b^2 c^2\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{7/2} (5 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {3 \left (a^2 d^2-14 a b c d+21 b^2 c^2\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{7/2} (5 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {3 \left (a^2 d^2-14 a b c d+21 b^2 c^2\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 c^2 (a+b x)^{7/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{7/2} (5 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {3 \left (a^2 d^2-14 a b c d+21 b^2 c^2\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
Input:
Int[(x^2*(a + b*x)^(5/2))/(c + d*x)^(5/2),x]
Output:
(2*c^2*(a + b*x)^(7/2))/(3*d^2*(b*c - a*d)*(c + d*x)^(3/2)) - ((4*c*(5*b*c - 3*a*d)*(a + b*x)^(7/2))/((b*c - a*d)*Sqrt[c + d*x]) - (3*(21*b^2*c^2 - 14*a*b*c*d + a^2*d^2)*(((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*d) - (5*(b*c - a *d)*(((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)))/(6*d)))/(b*c - a*d))/(3*d^2* (b*c - a*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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> 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] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(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])))
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. \(1001\) vs. \(2(243)=486\).
Time = 0.24 (sec) , antiderivative size = 1002, normalized size of antiderivative = 3.49
Input:
int(x^2*(b*x+a)^(5/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/48*(b*x+a)^(1/2)*(16*b^2*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^3*d^5*x^2-225*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*c*d^4*x^2+525*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^2*c^2*d^3*x^2-315*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))*b^3*c^3* d^2*x^2+52*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a*b*d^4*x^3-36*((b*x+a)*(d* x+c))^(1/2)*(d*b)^(1/2)*b^2*c*d^3*x^3+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^3*c*d^4*x-450*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*c^2*d^3 *x+1050*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^2*c^3*d^2*x-630*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))*b^3*c^4*d*x+66*((b*x+a)*(d*x+c))^(1/2)*(d* b)^(1/2)*a^2*d^4*x^2-192*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d^3*x^2 +126*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^2*d^2*x^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^3*c^2*d^3- 225*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*c^3*d^2+525*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^2*c^4*d-315*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))*b^3*c^5+324*((b*x+a)*(d*x+c...
Time = 0.57 (sec) , antiderivative size = 810, normalized size of antiderivative = 2.82 \[ \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(x^2*(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="fricas")
Output:
[-1/96*(15*(21*b^3*c^5 - 35*a*b^2*c^4*d + 15*a^2*b*c^3*d^2 - a^3*c^2*d^3 + (21*b^3*c^3*d^2 - 35*a*b^2*c^2*d^3 + 15*a^2*b*c*d^4 - a^3*d^5)*x^2 + 2*(2 1*b^3*c^4*d - 35*a*b^2*c^3*d^2 + 15*a^2*b*c^2*d^3 - a^3*c*d^4)*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*( 8*b^3*d^5*x^4 + 315*b^3*c^4*d - 420*a*b^2*c^3*d^2 + 113*a^2*b*c^2*d^3 - 2* (9*b^3*c*d^4 - 13*a*b^2*d^5)*x^3 + 3*(21*b^3*c^2*d^3 - 32*a*b^2*c*d^4 + 11 *a^2*b*d^5)*x^2 + 2*(210*b^3*c^3*d^2 - 287*a*b^2*c^2*d^3 + 81*a^2*b*c*d^4) *x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^8*x^2 + 2*b*c*d^7*x + b*c^2*d^6), 1/ 48*(15*(21*b^3*c^5 - 35*a*b^2*c^4*d + 15*a^2*b*c^3*d^2 - a^3*c^2*d^3 + (21 *b^3*c^3*d^2 - 35*a*b^2*c^2*d^3 + 15*a^2*b*c*d^4 - a^3*d^5)*x^2 + 2*(21*b^ 3*c^4*d - 35*a*b^2*c^3*d^2 + 15*a^2*b*c^2*d^3 - a^3*c*d^4)*x)*sqrt(-b*d)*a rctan(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*(8*b^3*d^5*x^4 + 315*b^3 *c^4*d - 420*a*b^2*c^3*d^2 + 113*a^2*b*c^2*d^3 - 2*(9*b^3*c*d^4 - 13*a*b^2 *d^5)*x^3 + 3*(21*b^3*c^2*d^3 - 32*a*b^2*c*d^4 + 11*a^2*b*d^5)*x^2 + 2*(21 0*b^3*c^3*d^2 - 287*a*b^2*c^2*d^3 + 81*a^2*b*c*d^4)*x)*sqrt(b*x + a)*sqrt( d*x + c))/(b*d^8*x^2 + 2*b*c*d^7*x + b*c^2*d^6)]
\[ \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x^{2} \left (a + b x\right )^{\frac {5}{2}}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**2*(b*x+a)**(5/2)/(d*x+c)**(5/2),x)
Output:
Integral(x**2*(a + b*x)**(5/2)/(c + d*x)**(5/2), x)
Exception generated. \[ \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^2*(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. 514 vs. \(2 (243) = 486\).
Time = 0.23 (sec) , antiderivative size = 514, normalized size of antiderivative = 1.79 \[ \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {{\left ({\left ({\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b^{6} c d^{8} - a b^{5} d^{9}\right )} {\left (b x + a\right )}}{b^{4} c d^{9} {\left | b \right |} - a b^{3} d^{10} {\left | b \right |}} - \frac {3 \, {\left (3 \, b^{7} c^{2} d^{7} - 2 \, a b^{6} c d^{8} - a^{2} b^{5} d^{9}\right )}}{b^{4} c d^{9} {\left | b \right |} - a b^{3} d^{10} {\left | b \right |}}\right )} + \frac {3 \, {\left (21 \, b^{8} c^{3} d^{6} - 35 \, a b^{7} c^{2} d^{7} + 15 \, a^{2} b^{6} c d^{8} - a^{3} b^{5} d^{9}\right )}}{b^{4} c d^{9} {\left | b \right |} - a b^{3} d^{10} {\left | b \right |}}\right )} {\left (b x + a\right )} + \frac {20 \, {\left (21 \, b^{9} c^{4} d^{5} - 56 \, a b^{8} c^{3} d^{6} + 50 \, a^{2} b^{7} c^{2} d^{7} - 16 \, a^{3} b^{6} c d^{8} + a^{4} b^{5} d^{9}\right )}}{b^{4} c d^{9} {\left | b \right |} - a b^{3} d^{10} {\left | b \right |}}\right )} {\left (b x + a\right )} + \frac {15 \, {\left (21 \, b^{10} c^{5} d^{4} - 77 \, a b^{9} c^{4} d^{5} + 106 \, a^{2} b^{8} c^{3} d^{6} - 66 \, a^{3} b^{7} c^{2} d^{7} + 17 \, a^{4} b^{6} c d^{8} - a^{5} b^{5} d^{9}\right )}}{b^{4} c d^{9} {\left | b \right |} - a b^{3} d^{10} {\left | b \right |}}\right )} \sqrt {b x + a}}{24 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {5 \, {\left (21 \, b^{4} c^{3} - 35 \, a b^{3} c^{2} d + 15 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{8 \, \sqrt {b d} d^{5} {\left | b \right |}} \] Input:
integrate(x^2*(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="giac")
Output:
1/24*(((2*(b*x + a)*(4*(b^6*c*d^8 - a*b^5*d^9)*(b*x + a)/(b^4*c*d^9*abs(b) - a*b^3*d^10*abs(b)) - 3*(3*b^7*c^2*d^7 - 2*a*b^6*c*d^8 - a^2*b^5*d^9)/(b ^4*c*d^9*abs(b) - a*b^3*d^10*abs(b))) + 3*(21*b^8*c^3*d^6 - 35*a*b^7*c^2*d ^7 + 15*a^2*b^6*c*d^8 - a^3*b^5*d^9)/(b^4*c*d^9*abs(b) - a*b^3*d^10*abs(b) ))*(b*x + a) + 20*(21*b^9*c^4*d^5 - 56*a*b^8*c^3*d^6 + 50*a^2*b^7*c^2*d^7 - 16*a^3*b^6*c*d^8 + a^4*b^5*d^9)/(b^4*c*d^9*abs(b) - a*b^3*d^10*abs(b)))* (b*x + a) + 15*(21*b^10*c^5*d^4 - 77*a*b^9*c^4*d^5 + 106*a^2*b^8*c^3*d^6 - 66*a^3*b^7*c^2*d^7 + 17*a^4*b^6*c*d^8 - a^5*b^5*d^9)/(b^4*c*d^9*abs(b) - a*b^3*d^10*abs(b)))*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + 5/8*(21*b^4*c^3 - 35*a*b^3*c^2*d + 15*a^2*b^2*c*d^2 - a^3*b*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) *d^5*abs(b))
Timed out. \[ \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x^2\,{\left (a+b\,x\right )}^{5/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int((x^2*(a + b*x)^(5/2))/(c + d*x)^(5/2),x)
Output:
int((x^2*(a + b*x)^(5/2))/(c + d*x)^(5/2), x)
\[ \int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x^{2} \left (b x +a \right )^{\frac {5}{2}}}{\left (d x +c \right )^{\frac {5}{2}}}d x \] Input:
int(x^2*(b*x+a)^(5/2)/(d*x+c)^(5/2),x)
Output:
int(x^2*(b*x+a)^(5/2)/(d*x+c)^(5/2),x)