3.5.62 \(\int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx\) [462]

3.5.62.1 Optimal result
3.5.62.2 Mathematica [A] (verified)
3.5.62.3 Rubi [A] (verified)
3.5.62.4 Maple [A] (verified)
3.5.62.5 Fricas [A] (verification not implemented)
3.5.62.6 Sympy [F]
3.5.62.7 Maxima [F(-2)]
3.5.62.8 Giac [A] (verification not implemented)
3.5.62.9 Mupad [B] (verification not implemented)

3.5.62.1 Optimal result

Integrand size = 20, antiderivative size = 207 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\frac {c (2 b c-3 a d) \sqrt {c+d x}}{4 a^2 x^2}-\frac {\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt {c+d x}}{8 a^3 x}-\frac {c (c+d x)^{3/2}}{3 a x^3}+\frac {\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 a^4 \sqrt {c}}-\frac {2 \sqrt {b} (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^4} \]

output
-1/3*c*(d*x+c)^(3/2)/a/x^3-2*(-a*d+b*c)^(5/2)*arctanh(b^(1/2)*(d*x+c)^(1/2 
)/(-a*d+b*c)^(1/2))*b^(1/2)/a^4+1/8*(-5*a^3*d^3+30*a^2*b*c*d^2-40*a*b^2*c^ 
2*d+16*b^3*c^3)*arctanh((d*x+c)^(1/2)/c^(1/2))/a^4/c^(1/2)+1/4*c*(-3*a*d+2 
*b*c)*(d*x+c)^(1/2)/a^2/x^2-1/8*(11*a^2*d^2-18*a*b*c*d+8*b^2*c^2)*(d*x+c)^ 
(1/2)/a^3/x
 
3.5.62.2 Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=-\frac {\frac {a \sqrt {c+d x} \left (24 b^2 c^2 x^2-6 a b c x (2 c+9 d x)+a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )\right )}{x^3}+48 \sqrt {b} (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )-\frac {3 \left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}}{24 a^4} \]

input
Integrate[(c + d*x)^(5/2)/(x^4*(a + b*x)),x]
 
output
-1/24*((a*Sqrt[c + d*x]*(24*b^2*c^2*x^2 - 6*a*b*c*x*(2*c + 9*d*x) + a^2*(8 
*c^2 + 26*c*d*x + 33*d^2*x^2)))/x^3 + 48*Sqrt[b]*(-(b*c) + a*d)^(5/2)*ArcT 
an[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]] - (3*(16*b^3*c^3 - 40*a*b^2 
*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/Sqrt[ 
c])/a^4
 
3.5.62.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {109, 27, 166, 27, 168, 27, 174, 73, 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.

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 166

\(\displaystyle -\frac {\frac {\int -\frac {c \left (8 b^2 c^2-18 a b d c+11 a^2 d^2\right )+d \left (6 b^2 c^2-13 a b d c+8 a^2 d^2\right ) x}{2 x^2 (a+b x) \sqrt {c+d x}}dx}{2 a}-\frac {c \sqrt {c+d x} (2 b c-3 a d)}{2 a x^2}}{2 a}-\frac {c (c+d x)^{3/2}}{3 a x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {c \left (8 b^2 c^2-18 a b d c+11 a^2 d^2\right )+d \left (6 b^2 c^2-13 a b d c+8 a^2 d^2\right ) x}{x^2 (a+b x) \sqrt {c+d x}}dx}{4 a}-\frac {c \sqrt {c+d x} (2 b c-3 a d)}{2 a x^2}}{2 a}-\frac {c (c+d x)^{3/2}}{3 a x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {c \left (16 b^3 c^3-40 a b^2 d c^2+30 a^2 b d^2 c-5 a^3 d^3+b d \left (8 b^2 c^2-18 a b d c+11 a^2 d^2\right ) x\right )}{2 x (a+b x) \sqrt {c+d x}}dx}{a c}-\frac {\sqrt {c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{a x}}{4 a}-\frac {c \sqrt {c+d x} (2 b c-3 a d)}{2 a x^2}}{2 a}-\frac {c (c+d x)^{3/2}}{3 a x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {16 b^3 c^3-40 a b^2 d c^2+30 a^2 b d^2 c-5 a^3 d^3+b d \left (8 b^2 c^2-18 a b d c+11 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}}dx}{2 a}-\frac {\sqrt {c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{a x}}{4 a}-\frac {c \sqrt {c+d x} (2 b c-3 a d)}{2 a x^2}}{2 a}-\frac {c (c+d x)^{3/2}}{3 a x^3}\)

\(\Big \downarrow \) 174

\(\displaystyle -\frac {-\frac {-\frac {\frac {\left (-5 a^3 d^3+30 a^2 b c d^2-40 a b^2 c^2 d+16 b^3 c^3\right ) \int \frac {1}{x \sqrt {c+d x}}dx}{a}-\frac {16 b (b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{a}}{2 a}-\frac {\sqrt {c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{a x}}{4 a}-\frac {c \sqrt {c+d x} (2 b c-3 a d)}{2 a x^2}}{2 a}-\frac {c (c+d x)^{3/2}}{3 a x^3}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {-\frac {-\frac {\frac {2 \left (-5 a^3 d^3+30 a^2 b c d^2-40 a b^2 c^2 d+16 b^3 c^3\right ) \int \frac {1}{\frac {c+d x}{d}-\frac {c}{d}}d\sqrt {c+d x}}{a d}-\frac {32 b (b c-a d)^3 \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{a d}}{2 a}-\frac {\sqrt {c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{a x}}{4 a}-\frac {c \sqrt {c+d x} (2 b c-3 a d)}{2 a x^2}}{2 a}-\frac {c (c+d x)^{3/2}}{3 a x^3}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {-\frac {-\frac {\sqrt {c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{a x}-\frac {\frac {32 \sqrt {b} (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a}-\frac {2 \left (-5 a^3 d^3+30 a^2 b c d^2-40 a b^2 c^2 d+16 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a \sqrt {c}}}{2 a}}{4 a}-\frac {c \sqrt {c+d x} (2 b c-3 a d)}{2 a x^2}}{2 a}-\frac {c (c+d x)^{3/2}}{3 a x^3}\)

input
Int[(c + d*x)^(5/2)/(x^4*(a + b*x)),x]
 
output
-1/3*(c*(c + d*x)^(3/2))/(a*x^3) - (-1/2*(c*(2*b*c - 3*a*d)*Sqrt[c + d*x]) 
/(a*x^2) - (-(((8*b^2*c^2 - 18*a*b*c*d + 11*a^2*d^2)*Sqrt[c + d*x])/(a*x)) 
 - ((-2*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh 
[Sqrt[c + d*x]/Sqrt[c]])/(a*Sqrt[c]) + (32*Sqrt[b]*(b*c - a*d)^(5/2)*ArcTa 
nh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/a)/(2*a))/(4*a))/(2*a)
 

3.5.62.3.1 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 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 109
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])
 

rule 166
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] && ILtQ[m, -1] && GtQ[n, 0]
 

rule 168
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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 174
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

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]
 
3.5.62.4 Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.85

method result size
pseudoelliptic \(\frac {-\frac {2 \left (a d -b c \right )^{3} b \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {\sqrt {d x +c}\, a \left (33 a^{2} d^{2} x^{2}-54 a b c d \,x^{2}+24 b^{2} c^{2} x^{2}+26 a^{2} c d x -12 a b \,c^{2} x +8 a^{2} c^{2}\right )}{24 x^{3}}-\frac {\left (5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}}{a^{4}}\) \(176\)
risch \(-\frac {\sqrt {d x +c}\, \left (33 a^{2} d^{2} x^{2}-54 a b c d \,x^{2}+24 b^{2} c^{2} x^{2}+26 a^{2} c d x -12 a b \,c^{2} x +8 a^{2} c^{2}\right )}{24 a^{3} x^{3}}-\frac {d \left (\frac {16 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) b \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a d \sqrt {\left (a d -b c \right ) b}}-\frac {\left (-5 a^{3} d^{3}+30 a^{2} b c \,d^{2}-40 a \,b^{2} c^{2} d +16 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d \sqrt {c}}\right )}{8 a^{3}}\) \(219\)
derivativedivides \(2 d^{4} \left (\frac {-\frac {\left (\frac {11}{16} a^{3} d^{3}-\frac {9}{8} a^{2} b c \,d^{2}+\frac {1}{2} a \,b^{2} c^{2} d \right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a^{3} c \,d^{3}+2 a^{2} d^{2} b \,c^{2}-a \,b^{2} c^{3} d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} b \,c^{3} d^{2}+\frac {1}{2} a \,b^{2} c^{4} d +\frac {5}{16} a^{3} c^{2} d^{3}\right ) \sqrt {d x +c}}{d^{3} x^{3}}-\frac {\left (5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{a^{4} d^{4}}-\frac {\left (a d -b c \right )^{3} b \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a^{4} d^{4} \sqrt {\left (a d -b c \right ) b}}\right )\) \(250\)
default \(2 d^{4} \left (\frac {-\frac {\left (\frac {11}{16} a^{3} d^{3}-\frac {9}{8} a^{2} b c \,d^{2}+\frac {1}{2} a \,b^{2} c^{2} d \right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a^{3} c \,d^{3}+2 a^{2} d^{2} b \,c^{2}-a \,b^{2} c^{3} d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} b \,c^{3} d^{2}+\frac {1}{2} a \,b^{2} c^{4} d +\frac {5}{16} a^{3} c^{2} d^{3}\right ) \sqrt {d x +c}}{d^{3} x^{3}}-\frac {\left (5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{a^{4} d^{4}}-\frac {\left (a d -b c \right )^{3} b \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a^{4} d^{4} \sqrt {\left (a d -b c \right ) b}}\right )\) \(250\)

input
int((d*x+c)^(5/2)/x^4/(b*x+a),x,method=_RETURNVERBOSE)
 
output
1/a^4*(-2*(a*d-b*c)^3*b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b 
*c)*b)^(1/2))-1/24*(d*x+c)^(1/2)*a*(33*a^2*d^2*x^2-54*a*b*c*d*x^2+24*b^2*c 
^2*x^2+26*a^2*c*d*x-12*a*b*c^2*x+8*a^2*c^2)/x^3-1/8*(5*a^3*d^3-30*a^2*b*c* 
d^2+40*a*b^2*c^2*d-16*b^3*c^3)/c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1/2)))
 
3.5.62.5 Fricas [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 930, normalized size of antiderivative = 4.49 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\left [\frac {48 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {b^{2} c - a b d} x^{3} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) - 3 \, {\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {c} x^{3} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (8 \, a^{3} c^{3} + 3 \, {\left (8 \, a b^{2} c^{3} - 18 \, a^{2} b c^{2} d + 11 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (6 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt {d x + c}}{48 \, a^{4} c x^{3}}, \frac {96 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {-b^{2} c + a b d} x^{3} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - 3 \, {\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {c} x^{3} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (8 \, a^{3} c^{3} + 3 \, {\left (8 \, a b^{2} c^{3} - 18 \, a^{2} b c^{2} d + 11 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (6 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt {d x + c}}{48 \, a^{4} c x^{3}}, -\frac {3 \, {\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - 24 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {b^{2} c - a b d} x^{3} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + {\left (8 \, a^{3} c^{3} + 3 \, {\left (8 \, a b^{2} c^{3} - 18 \, a^{2} b c^{2} d + 11 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (6 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt {d x + c}}{24 \, a^{4} c x^{3}}, \frac {48 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {-b^{2} c + a b d} x^{3} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - 3 \, {\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (8 \, a^{3} c^{3} + 3 \, {\left (8 \, a b^{2} c^{3} - 18 \, a^{2} b c^{2} d + 11 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (6 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt {d x + c}}{24 \, a^{4} c x^{3}}\right ] \]

input
integrate((d*x+c)^(5/2)/x^4/(b*x+a),x, algorithm="fricas")
 
output
[1/48*(48*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(b^2*c - a*b*d)*x^3*log( 
(b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) - 3 
*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(c)*x^3*lo 
g((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a^3*c^3 + 3*(8*a*b^2*c^3 
 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 - 2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)* 
sqrt(d*x + c))/(a^4*c*x^3), 1/48*(96*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*s 
qrt(-b^2*c + a*b*d)*x^3*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + 
 b*c)) - 3*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt 
(c)*x^3*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a^3*c^3 + 3*(8 
*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 - 2*(6*a^2*b*c^3 - 13*a^3* 
c^2*d)*x)*sqrt(d*x + c))/(a^4*c*x^3), -1/24*(3*(16*b^3*c^3 - 40*a*b^2*c^2* 
d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-c)*x^3*arctan(sqrt(d*x + c)*sqrt(-c) 
/c) - 24*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(b^2*c - a*b*d)*x^3*log(( 
b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) + (8 
*a^3*c^3 + 3*(8*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 - 2*(6*a^2* 
b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*c*x^3), 1/24*(48*(b^2*c^3 - 2 
*a*b*c^2*d + a^2*c*d^2)*sqrt(-b^2*c + a*b*d)*x^3*arctan(sqrt(-b^2*c + a*b* 
d)*sqrt(d*x + c)/(b*d*x + b*c)) - 3*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2* 
b*c*d^2 - 5*a^3*d^3)*sqrt(-c)*x^3*arctan(sqrt(d*x + c)*sqrt(-c)/c) - (8*a^ 
3*c^3 + 3*(8*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 - 2*(6*a^2*...
 
3.5.62.6 Sympy [F]

\[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{4} \left (a + b x\right )}\, dx \]

input
integrate((d*x+c)**(5/2)/x**4/(b*x+a),x)
 
output
Integral((c + d*x)**(5/2)/(x**4*(a + b*x)), x)
 
3.5.62.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\text {Exception raised: ValueError} \]

input
integrate((d*x+c)^(5/2)/x^4/(b*x+a),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
 
3.5.62.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.45 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\frac {2 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{4}} - \frac {{\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{8 \, a^{4} \sqrt {-c}} - \frac {24 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} c^{2} d - 48 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c^{3} d + 24 \, \sqrt {d x + c} b^{2} c^{4} d - 54 \, {\left (d x + c\right )}^{\frac {5}{2}} a b c d^{2} + 96 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c^{2} d^{2} - 42 \, \sqrt {d x + c} a b c^{3} d^{2} + 33 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} d^{3} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} c d^{3} + 15 \, \sqrt {d x + c} a^{2} c^{2} d^{3}}{24 \, a^{3} d^{3} x^{3}} \]

input
integrate((d*x+c)^(5/2)/x^4/(b*x+a),x, algorithm="giac")
 
output
2*(b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*arctan(sqrt(d*x 
+ c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^4) - 1/8*(16*b^3*c^3 
- 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*arctan(sqrt(d*x + c)/sqrt(- 
c))/(a^4*sqrt(-c)) - 1/24*(24*(d*x + c)^(5/2)*b^2*c^2*d - 48*(d*x + c)^(3/ 
2)*b^2*c^3*d + 24*sqrt(d*x + c)*b^2*c^4*d - 54*(d*x + c)^(5/2)*a*b*c*d^2 + 
 96*(d*x + c)^(3/2)*a*b*c^2*d^2 - 42*sqrt(d*x + c)*a*b*c^3*d^2 + 33*(d*x + 
 c)^(5/2)*a^2*d^3 - 40*(d*x + c)^(3/2)*a^2*c*d^3 + 15*sqrt(d*x + c)*a^2*c^ 
2*d^3)/(a^3*d^3*x^3)
 
3.5.62.9 Mupad [B] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 2147, normalized size of antiderivative = 10.37 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\text {Too large to display} \]

input
int((c + d*x)^(5/2)/(x^4*(a + b*x)),x)
 
output
(((c + d*x)^(1/2)*(8*b^2*c^4*d + 5*a^2*c^2*d^3 - 14*a*b*c^3*d^2))/(8*a^3) 
- ((c + d*x)^(3/2)*(5*a^2*c*d^3 + 6*b^2*c^3*d - 12*a*b*c^2*d^2))/(3*a^3) + 
 (d*(c + d*x)^(5/2)*(11*a^2*d^2 + 8*b^2*c^2 - 18*a*b*c*d))/(8*a^3))/(3*c*( 
c + d*x)^2 - 3*c^2*(c + d*x) - (c + d*x)^3 + c^3) + (2*atanh((25*b^3*d^8*( 
c + d*x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c^3*d^2 
 - 10*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/(16*((25*a^3*b^3*d^11)/16 - 
(217*b^6*c^3*d^8)/16 + (227*a*b^5*c^2*d^9)/16 - (119*a^2*b^4*c*d^10)/16 + 
(13*b^7*c^4*d^7)/(2*a) - (5*b^8*c^5*d^6)/(4*a^2))) + (5*b^5*c^2*d^6*(c + d 
*x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c^3*d^2 - 10 
*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/(4*((25*a^5*b^3*d^11)/16 - (5*b^8 
*c^5*d^6)/4 + (13*a*b^7*c^4*d^7)/2 - (119*a^4*b^4*c*d^10)/16 - (217*a^2*b^ 
6*c^3*d^8)/16 + (227*a^3*b^5*c^2*d^9)/16)) - (11*b^4*c*d^7*(c + d*x)^(1/2) 
*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c^3*d^2 - 10*a^3*b^3* 
c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/(4*((25*a^4*b^3*d^11)/16 + (13*b^7*c^4*d^7 
)/2 - (217*a*b^6*c^3*d^8)/16 - (119*a^3*b^4*c*d^10)/16 + (227*a^2*b^5*c^2* 
d^9)/16 - (5*b^8*c^5*d^6)/(4*a))))*(-b*(a*d - b*c)^5)^(1/2))/a^4 + (atan(( 
((((c + d*x)^(1/2)*(281*a^6*b^3*d^8 + 512*b^9*c^6*d^2 - 2816*a*b^8*c^5*d^3 
 - 1836*a^5*b^4*c*d^7 + 6400*a^2*b^7*c^4*d^4 - 7680*a^3*b^6*c^3*d^5 + 5140 
*a^4*b^5*c^2*d^6))/(32*a^6) - (((80*a^11*b^2*d^6 - 304*a^10*b^3*c*d^5 - 12 
8*a^8*b^5*c^3*d^3 + 352*a^9*b^4*c^2*d^4)/(32*a^9) - ((256*a^9*b^2*d^3 -...