\(\int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx\) [401]

Optimal result

Integrand size = 22, antiderivative size = 374 \[ \int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^6 d}-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^5 d (b c-a d)}-\frac {2 a^2 (9 b c-13 a d) (c+d x)^{5/2}}{3 b^4 (b c-a d) \sqrt {a+b x}}+\frac {(5 b c-29 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 b^4 d}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b^4}+\frac {2 a^3 (c+d x)^{7/2}}{3 b^3 (b c-a d) (a+b x)^{3/2}}-\frac {5 (b c-a d) \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{13/2} d^{3/2}} \] Output:

-5/64*(231*a^3*d^3-189*a^2*b*c*d^2+21*a*b^2*c^2*d+b^3*c^3)*(b*x+a)^(1/2)*( 
d*x+c)^(1/2)/b^6/d-5/96*(231*a^3*d^3-189*a^2*b*c*d^2+21*a*b^2*c^2*d+b^3*c^ 
3)*(b*x+a)^(1/2)*(d*x+c)^(3/2)/b^5/d/(-a*d+b*c)-2/3*a^2*(-13*a*d+9*b*c)*(d 
*x+c)^(5/2)/b^4/(-a*d+b*c)/(b*x+a)^(1/2)+1/24*(-29*a*d+5*b*c)*(b*x+a)^(1/2 
)*(d*x+c)^(5/2)/b^4/d+1/4*(b*x+a)^(3/2)*(d*x+c)^(5/2)/b^4+2/3*a^3*(d*x+c)^ 
(7/2)/b^3/(-a*d+b*c)/(b*x+a)^(3/2)-5/64*(-a*d+b*c)*(231*a^3*d^3-189*a^2*b* 
c*d^2+21*a*b^2*c^2*d+b^3*c^3)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c 
)^(1/2))/b^(13/2)/d^(3/2)
 

Mathematica [A] (verified)

Time = 10.67 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.81 \[ \int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\frac {\sqrt {c+d x} \left (\frac {\sqrt {d} \left (-3465 a^5 d^3+105 a^4 b d^2 (49 c-44 d x)-21 a^3 b^2 d \left (83 c^2-334 c d x+33 d^2 x^2\right )+b^5 x^2 \left (15 c^3+118 c^2 d x+136 c d^2 x^2+48 d^3 x^3\right )+3 a^2 b^3 \left (5 c^3-824 c^2 d x+387 c d^2 x^2+66 d^3 x^3\right )-a b^4 x \left (-30 c^3+483 c^2 d x+316 c d^2 x^2+88 d^3 x^3\right )\right )}{(a+b x)^{3/2}}-\frac {15 \sqrt {b c-a d} \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{192 b^6 d^{3/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*((Sqrt[d]*(-3465*a^5*d^3 + 105*a^4*b*d^2*(49*c - 44*d*x) - 
21*a^3*b^2*d*(83*c^2 - 334*c*d*x + 33*d^2*x^2) + b^5*x^2*(15*c^3 + 118*c^2 
*d*x + 136*c*d^2*x^2 + 48*d^3*x^3) + 3*a^2*b^3*(5*c^3 - 824*c^2*d*x + 387* 
c*d^2*x^2 + 66*d^3*x^3) - a*b^4*x*(-30*c^3 + 483*c^2*d*x + 316*c*d^2*x^2 + 
 88*d^3*x^3)))/(a + b*x)^(3/2) - (15*Sqrt[b*c - a*d]*(b^3*c^3 + 21*a*b^2*c 
^2*d - 189*a^2*b*c*d^2 + 231*a^3*d^3)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt 
[b*c - a*d]])/Sqrt[(b*(c + d*x))/(b*c - a*d)]))/(192*b^6*d^(3/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 321, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {2 \int \frac {x^2 (c+d x)^{3/2} (6 c+11 d x)}{2 (a+b x)^{3/2}}dx}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {x^2 (c+d x)^{3/2} (6 c+11 d x)}{(a+b x)^{3/2}}dx}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 167

\(\displaystyle \frac {\frac {2 \int \frac {x (c+d x)^{3/2} (4 c (6 b c-11 a d)+d (59 b c-99 a d) x)}{2 \sqrt {a+b x}}dx}{b (b c-a d)}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{b \sqrt {a+b x} (b c-a d)}}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {x (c+d x)^{3/2} (4 c (6 b c-11 a d)+d (59 b c-99 a d) x)}{\sqrt {a+b x}}dx}{b (b c-a d)}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{b \sqrt {a+b x} (b c-a d)}}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (231 a^2 d^2+2 b d x (59 b c-99 a d)-156 a b c d+5 b^2 c^2\right )}{8 b^2 d}-\frac {5 \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}}dx}{16 b^2 d}}{b (b c-a d)}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{b \sqrt {a+b x} (b c-a d)}}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (231 a^2 d^2+2 b d x (59 b c-99 a d)-156 a b c d+5 b^2 c^2\right )}{8 b^2 d}-\frac {5 \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right ) \left (\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{16 b^2 d}}{b (b c-a d)}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{b \sqrt {a+b x} (b c-a d)}}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (231 a^2 d^2+2 b d x (59 b c-99 a d)-156 a b c d+5 b^2 c^2\right )}{8 b^2 d}-\frac {5 \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{16 b^2 d}}{b (b c-a d)}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{b \sqrt {a+b x} (b c-a d)}}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (231 a^2 d^2+2 b d x (59 b c-99 a d)-156 a b c d+5 b^2 c^2\right )}{8 b^2 d}-\frac {5 \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{16 b^2 d}}{b (b c-a d)}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{b \sqrt {a+b x} (b c-a d)}}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (231 a^2 d^2+2 b d x (59 b c-99 a d)-156 a b c d+5 b^2 c^2\right )}{8 b^2 d}-\frac {5 \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{16 b^2 d}}{b (b c-a d)}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{b \sqrt {a+b x} (b c-a d)}}{3 b}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}\)

Input:

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

Output:

(-2*x^3*(c + d*x)^(5/2))/(3*b*(a + b*x)^(3/2)) + ((-2*(6*b*c - 11*a*d)*x^2 
*(c + d*x)^(5/2))/(b*(b*c - a*d)*Sqrt[a + b*x]) + ((Sqrt[a + b*x]*(c + d*x 
)^(5/2)*(5*b^2*c^2 - 156*a*b*c*d + 231*a^2*d^2 + 2*b*d*(59*b*c - 99*a*d)*x 
))/(8*b^2*d) - (5*(b^3*c^3 + 21*a*b^2*c^2*d - 189*a^2*b*c*d^2 + 231*a^3*d^ 
3)*((Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b) + (3*(b*c - a*d)*((Sqrt[a + b*x] 
*Sqrt[c + d*x])/b + ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]* 
Sqrt[c + d*x])])/(b^(3/2)*Sqrt[d])))/(4*b)))/(16*b^2*d))/(b*(b*c - a*d)))/ 
(3*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

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]
 

rule 66

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]
 

rule 108

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])
 

rule 164

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]
 

rule 167

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]
 

rule 221

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1365\) vs. \(2(324)=648\).

Time = 0.26 (sec) , antiderivative size = 1366, normalized size of antiderivative = 3.65

method	result	size
default	\(\text {Expression too large to display}\)	\(1366\)

Input:

int(x^3*(d*x+c)^(5/2)/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/384*(d*x+c)^(1/2)*(60*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^4*c^3*x-63 
00*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^5*b*c*d^3+96*b^5*d^3*x^5*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+2322*( 
d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b^3*c*d^2*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^4*b^2*c^2*d^ 
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))*b^6*c^4*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^2*b^4*c^4-6930*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/ 
2)*a^5*d^3+30*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b^3*c^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*b^5*c 
^4*x-3486*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b^2*c^2*d-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^2*b^4* 
c^3*d*x+3465*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*b^2*d^4*x^2+6930*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^5*b*d^4*x-9240*(d*b)^(1/2)*((b*x+a 
)*(d*x+c))^(1/2)*a^4*b*d^3*x+30*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^5*c^ 
3*x^2-300*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^3*c^3*d-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*b^5*c^3*d*x^2+396*(d*b)^(1/2)*((b*x+a)*( 
d*x+c))^(1/2)*a^2*b^3*d^3*x^3+236*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b...
 

Fricas [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 1060, normalized size of antiderivative = 2.83 \[ \int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^3*(d*x+c)^(5/2)/(b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

[-1/768*(15*(a^2*b^4*c^4 + 20*a^3*b^3*c^3*d - 210*a^4*b^2*c^2*d^2 + 420*a^ 
5*b*c*d^3 - 231*a^6*d^4 + (b^6*c^4 + 20*a*b^5*c^3*d - 210*a^2*b^4*c^2*d^2 
+ 420*a^3*b^3*c*d^3 - 231*a^4*b^2*d^4)*x^2 + 2*(a*b^5*c^4 + 20*a^2*b^4*c^3 
*d - 210*a^3*b^3*c^2*d^2 + 420*a^4*b^2*c*d^3 - 231*a^5*b*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*(4 
8*b^6*d^4*x^5 + 15*a^2*b^4*c^3*d - 1743*a^3*b^3*c^2*d^2 + 5145*a^4*b^2*c*d 
^3 - 3465*a^5*b*d^4 + 8*(17*b^6*c*d^3 - 11*a*b^5*d^4)*x^4 + 2*(59*b^6*c^2* 
d^2 - 158*a*b^5*c*d^3 + 99*a^2*b^4*d^4)*x^3 + 3*(5*b^6*c^3*d - 161*a*b^5*c 
^2*d^2 + 387*a^2*b^4*c*d^3 - 231*a^3*b^3*d^4)*x^2 + 6*(5*a*b^5*c^3*d - 412 
*a^2*b^4*c^2*d^2 + 1169*a^3*b^3*c*d^3 - 770*a^4*b^2*d^4)*x)*sqrt(b*x + a)* 
sqrt(d*x + c))/(b^9*d^2*x^2 + 2*a*b^8*d^2*x + a^2*b^7*d^2), 1/384*(15*(a^2 
*b^4*c^4 + 20*a^3*b^3*c^3*d - 210*a^4*b^2*c^2*d^2 + 420*a^5*b*c*d^3 - 231* 
a^6*d^4 + (b^6*c^4 + 20*a*b^5*c^3*d - 210*a^2*b^4*c^2*d^2 + 420*a^3*b^3*c* 
d^3 - 231*a^4*b^2*d^4)*x^2 + 2*(a*b^5*c^4 + 20*a^2*b^4*c^3*d - 210*a^3*b^3 
*c^2*d^2 + 420*a^4*b^2*c*d^3 - 231*a^5*b*d^4)*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^6*d^4*x^5 + 15*a^2*b^4*c^3*d - 
1743*a^3*b^3*c^2*d^2 + 5145*a^4*b^2*c*d^3 - 3465*a^5*b*d^4 + 8*(17*b^6*c*d 
^3 - 11*a*b^5*d^4)*x^4 + 2*(59*b^6*c^2*d^2 - 158*a*b^5*c*d^3 + 99*a^2*b...
 

Sympy [F]

\[ \int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\int \frac {x^{3} \left (c + d x\right )^{\frac {5}{2}}}{\left (a + b x\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^3*(d*x+c)^(5/2)/(b*x+a)^(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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 943 vs. \(2 (324) = 648\).

Time = 0.47 (sec) , antiderivative size = 943, normalized size of antiderivative = 2.52 \[ \int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^3*(d*x+c)^(5/2)/(b*x+a)^(5/2),x, algorithm="giac")
 

Output:

1/192*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b* 
x + a)*d^2*abs(b)/b^8 + (17*b^32*c*d^7*abs(b) - 41*a*b^31*d^8*abs(b))/(b^3 
9*d^6)) + (59*b^33*c^2*d^6*abs(b) - 430*a*b^32*c*d^7*abs(b) + 515*a^2*b^31 
*d^8*abs(b))/(b^39*d^6)) + 3*(5*b^34*c^3*d^5*abs(b) - 279*a*b^33*c^2*d^6*a 
bs(b) + 975*a^2*b^32*c*d^7*abs(b) - 765*a^3*b^31*d^8*abs(b))/(b^39*d^6))*s 
qrt(b*x + a) + 5/128*(b^4*c^4*abs(b) + 20*a*b^3*c^3*d*abs(b) - 210*a^2*b^2 
*c^2*d^2*abs(b) + 420*a^3*b*c*d^3*abs(b) - 231*a^4*d^4*abs(b))*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^7 
*d) - 4/3*(9*a^2*b^7*c^5*d*abs(b) - 52*a^3*b^6*c^4*d^2*abs(b) + 118*a^4*b^ 
5*c^3*d^3*abs(b) - 132*a^5*b^4*c^2*d^4*abs(b) + 73*a^6*b^3*c*d^5*abs(b) - 
16*a^7*b^2*d^6*abs(b) - 18*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + 
a)*b*d - a*b*d))^2*a^2*b^5*c^4*d*abs(b) + 84*(sqrt(b*d)*sqrt(b*x + a) - sq 
rt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^4*c^3*d^2*abs(b) - 144*(sqrt(b* 
d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^3*c^2*d^3* 
abs(b) + 108*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d 
))^2*a^5*b^2*c*d^4*abs(b) - 30*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b* 
x + a)*b*d - a*b*d))^2*a^6*b*d^5*abs(b) + 9*(sqrt(b*d)*sqrt(b*x + a) - sqr 
t(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^3*c^3*d*abs(b) - 36*(sqrt(b*d)*s 
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^2*c^2*d^2*abs( 
b) + 45*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\int \frac {x^3\,{\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,x\right )}^{5/2}} \,d x \] Input:

int((x^3*(c + d*x)^(5/2))/(a + b*x)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int((x^3*(c + d*x)^(5/2))/(a + b*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^3 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\int \frac {x^{3} \left (d x +c \right )^{\frac {5}{2}}}{\left (b x +a \right )^{\frac {5}{2}}}d x \] Input:

int(x^3*(d*x+c)^(5/2)/(b*x+a)^(5/2),x)
 

Output:

int(x^3*(d*x+c)^(5/2)/(b*x+a)^(5/2),x)