\(\int \frac {(a x+b x^2)^{5/2}}{x^4 (c+d x)} \, dx\) [120]

Optimal result

Integrand size = 24, antiderivative size = 147 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\frac {b (b c+2 a d) \sqrt {a x+b x^2}}{c d}-\frac {2 a \left (a x+b x^2\right )^{3/2}}{c x^2}-\frac {b^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{d^2}+\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a x+b x^2}}\right )}{c^{3/2} d^2} \] Output:

b*(2*a*d+b*c)*(b*x^2+a*x)^(1/2)/c/d-2*a*(b*x^2+a*x)^(3/2)/c/x^2-b^(3/2)*(- 
5*a*d+2*b*c)*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))/d^2+2*(-a*d+b*c)^(5/2)*a 
rctanh((-a*d+b*c)^(1/2)*x/c^(1/2)/(b*x^2+a*x)^(1/2))/c^(3/2)/d^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.43 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.01 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\frac {(x (a+b x))^{5/2} \left (b c^{3/2} d \sqrt {a+b x} \left (-2 a^2 d+b^2 c x\right )-2 (b c-a d)^2 \left (b c-a d-i \sqrt {a} \sqrt {d} \sqrt {b c-a d}\right ) \sqrt {-b c+2 a d-2 i \sqrt {a} \sqrt {d} \sqrt {b c-a d}} \sqrt {x} \arctan \left (\frac {\sqrt {-b c+2 a d-2 i \sqrt {a} \sqrt {d} \sqrt {b c-a d}} \sqrt {x}}{\sqrt {c} \left (-\sqrt {a}+\sqrt {a+b x}\right )}\right )-2 (b c-a d)^2 \left (b c-a d+i \sqrt {a} \sqrt {d} \sqrt {b c-a d}\right ) \sqrt {-b c+2 a d+2 i \sqrt {a} \sqrt {d} \sqrt {b c-a d}} \sqrt {x} \arctan \left (\frac {\sqrt {-b c+2 a d+2 i \sqrt {a} \sqrt {d} \sqrt {b c-a d}} \sqrt {x}}{\sqrt {c} \left (-\sqrt {a}+\sqrt {a+b x}\right )}\right )+2 b^{5/2} c^{5/2} (2 b c-5 a d) \sqrt {x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )\right )}{b c^{5/2} d^2 x^3 (a+b x)^{5/2}} \] Input:

Integrate[(a*x + b*x^2)^(5/2)/(x^4*(c + d*x)),x]
 

Output:

((x*(a + b*x))^(5/2)*(b*c^(3/2)*d*Sqrt[a + b*x]*(-2*a^2*d + b^2*c*x) - 2*( 
b*c - a*d)^2*(b*c - a*d - I*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d])*Sqrt[-(b*c) + 
 2*a*d - (2*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d]]*Sqrt[x]*ArcTan[(Sqrt[-(b*c 
) + 2*a*d - (2*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d]]*Sqrt[x])/(Sqrt[c]*(-Sqr 
t[a] + Sqrt[a + b*x]))] - 2*(b*c - a*d)^2*(b*c - a*d + I*Sqrt[a]*Sqrt[d]*S 
qrt[b*c - a*d])*Sqrt[-(b*c) + 2*a*d + (2*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d 
]]*Sqrt[x]*ArcTan[(Sqrt[-(b*c) + 2*a*d + (2*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - 
a*d]]*Sqrt[x])/(Sqrt[c]*(-Sqrt[a] + Sqrt[a + b*x]))] + 2*b^(5/2)*c^(5/2)*( 
2*b*c - 5*a*d)*Sqrt[x]*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x]) 
]))/(b*c^(5/2)*d^2*x^3*(a + b*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.26, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {1261, 109, 27, 171, 27, 175, 65, 104, 219, 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.

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

\(\Big \downarrow \) 1261

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2} \left (\frac {\frac {b \sqrt {x} \sqrt {a+b x} (2 a d+b c)}{d}-\frac {\frac {b^2 c (2 b c-5 a d) \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx}{d}-\frac {2 (b c-a d)^3 \int \frac {1}{\sqrt {x} \sqrt {a+b x} (c+d x)}dx}{d}}{2 d}}{c}-\frac {2 a (a+b x)^{3/2}}{c \sqrt {x}}\right )}{x^{5/2} (a+b x)^{5/2}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2} \left (\frac {\frac {b \sqrt {x} \sqrt {a+b x} (2 a d+b c)}{d}-\frac {\frac {2 b^2 c (2 b c-5 a d) \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{d}-\frac {2 (b c-a d)^3 \int \frac {1}{\sqrt {x} \sqrt {a+b x} (c+d x)}dx}{d}}{2 d}}{c}-\frac {2 a (a+b x)^{3/2}}{c \sqrt {x}}\right )}{x^{5/2} (a+b x)^{5/2}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2} \left (\frac {\frac {b \sqrt {x} \sqrt {a+b x} (2 a d+b c)}{d}-\frac {\frac {2 b^2 c (2 b c-5 a d) \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{d}-\frac {4 (b c-a d)^3 \int \frac {1}{c-\frac {(b c-a d) x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{d}}{2 d}}{c}-\frac {2 a (a+b x)^{3/2}}{c \sqrt {x}}\right )}{x^{5/2} (a+b x)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2} \left (\frac {\frac {b \sqrt {x} \sqrt {a+b x} (2 a d+b c)}{d}-\frac {\frac {2 b^{3/2} c \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) (2 b c-5 a d)}{d}-\frac {4 (b c-a d)^3 \int \frac {1}{c-\frac {(b c-a d) x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{d}}{2 d}}{c}-\frac {2 a (a+b x)^{3/2}}{c \sqrt {x}}\right )}{x^{5/2} (a+b x)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2} \left (\frac {\frac {b \sqrt {x} \sqrt {a+b x} (2 a d+b c)}{d}-\frac {\frac {2 b^{3/2} c \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) (2 b c-5 a d)}{d}-\frac {4 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {x} \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x}}\right )}{\sqrt {c} d}}{2 d}}{c}-\frac {2 a (a+b x)^{3/2}}{c \sqrt {x}}\right )}{x^{5/2} (a+b x)^{5/2}}\)

Input:

Int[(a*x + b*x^2)^(5/2)/(x^4*(c + d*x)),x]
 

Output:

((a*x + b*x^2)^(5/2)*((-2*a*(a + b*x)^(3/2))/(c*Sqrt[x]) + ((b*(b*c + 2*a* 
d)*Sqrt[x]*Sqrt[a + b*x])/d - ((2*b^(3/2)*c*(2*b*c - 5*a*d)*ArcTanh[(Sqrt[ 
b]*Sqrt[x])/Sqrt[a + b*x]])/d - (4*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b*c - a 
*d]*Sqrt[x])/(Sqrt[c]*Sqrt[a + b*x])])/(Sqrt[c]*d))/(2*d))/c))/(x^(5/2)*(a 
 + b*x)^(5/2))
 

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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*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 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegersQ[2*m, 2*n, 2*p]
 

rule 175

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

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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]
 

rule 1261

Int[((e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((b_.)*(x_) + (c_.)*(x_)^2) 
^(p_), x_Symbol] :> Simp[(e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p)) 
  Int[x^(m + p)*(f + g*x)^n*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m, 
 n}, x] &&  !IGtQ[n, 0]
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(-\frac {2 \left (x \left (-a d +b c \right )^{3} \sqrt {b}\, \arctan \left (\frac {\sqrt {x \left (b x +a \right )}\, c}{x \sqrt {c \left (a d -b c \right )}}\right )-\frac {\sqrt {c \left (a d -b c \right )}\, \left (x \left (5 c d a \,b^{2}-2 b^{3} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )+d \left (b^{2} c x -2 a^{2} d \right ) \sqrt {b}\, \sqrt {x \left (b x +a \right )}\right )}{2}\right )}{\sqrt {b}\, \sqrt {c \left (a d -b c \right )}\, c x \,d^{2}}\)	\(148\)
risch	\(-\frac {\left (b x +a \right ) \left (-b^{2} c x +2 a^{2} d \right )}{c \sqrt {x \left (b x +a \right )}\, d}+\frac {\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}+\frac {b^{\frac {3}{2}} c \left (5 a d -2 b c \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{d}}{2 c d}\)	\(255\)
default	\(\text {Expression too large to display}\)	\(1523\)

Input:

int((b*x^2+a*x)^(5/2)/x^4/(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

-2/b^(1/2)/(c*(a*d-b*c))^(1/2)*(x*(-a*d+b*c)^3*b^(1/2)*arctan((x*(b*x+a))^ 
(1/2)/x*c/(c*(a*d-b*c))^(1/2))-1/2*(c*(a*d-b*c))^(1/2)*(x*(5*a*b^2*c*d-2*b 
^3*c^2)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))+d*(b^2*c*x-2*a^2*d)*b^(1/2)*( 
x*(b*x+a))^(1/2)))/c/x/d^2
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 685, normalized size of antiderivative = 4.66 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\left [-\frac {{\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {b} x \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x + 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right ) - 2 \, {\left (b^{2} c d x - 2 \, a^{2} d^{2}\right )} \sqrt {b x^{2} + a x}}{2 \, c d^{2} x}, \frac {4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right ) - {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {b} x \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (b^{2} c d x - 2 \, a^{2} d^{2}\right )} \sqrt {b x^{2} + a x}}{2 \, c d^{2} x}, \frac {{\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x + 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right ) + {\left (b^{2} c d x - 2 \, a^{2} d^{2}\right )} \sqrt {b x^{2} + a x}}{c d^{2} x}, \frac {{\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right ) + {\left (b^{2} c d x - 2 \, a^{2} d^{2}\right )} \sqrt {b x^{2} + a x}}{c d^{2} x}\right ] \] Input:

integrate((b*x^2+a*x)^(5/2)/x^4/(d*x+c),x, algorithm="fricas")
 

Output:

[-1/2*((2*b^2*c^2 - 5*a*b*c*d)*sqrt(b)*x*log(2*b*x + a + 2*sqrt(b*x^2 + a* 
x)*sqrt(b)) - 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*sqrt((b*c - a*d)/c)*log( 
(a*c + (2*b*c - a*d)*x + 2*sqrt(b*x^2 + a*x)*c*sqrt((b*c - a*d)/c))/(d*x + 
 c)) - 2*(b^2*c*d*x - 2*a^2*d^2)*sqrt(b*x^2 + a*x))/(c*d^2*x), 1/2*(4*(b^2 
*c^2 - 2*a*b*c*d + a^2*d^2)*x*sqrt(-(b*c - a*d)/c)*arctan(-sqrt(b*x^2 + a* 
x)*c*sqrt(-(b*c - a*d)/c)/((b*c - a*d)*x)) - (2*b^2*c^2 - 5*a*b*c*d)*sqrt( 
b)*x*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(b^2*c*d*x - 2*a^2*d 
^2)*sqrt(b*x^2 + a*x))/(c*d^2*x), ((2*b^2*c^2 - 5*a*b*c*d)*sqrt(-b)*x*arct 
an(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x + a)) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2) 
*x*sqrt((b*c - a*d)/c)*log((a*c + (2*b*c - a*d)*x + 2*sqrt(b*x^2 + a*x)*c* 
sqrt((b*c - a*d)/c))/(d*x + c)) + (b^2*c*d*x - 2*a^2*d^2)*sqrt(b*x^2 + a*x 
))/(c*d^2*x), ((2*b^2*c^2 - 5*a*b*c*d)*sqrt(-b)*x*arctan(sqrt(b*x^2 + a*x) 
*sqrt(-b)/(b*x + a)) + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*sqrt(-(b*c - a* 
d)/c)*arctan(-sqrt(b*x^2 + a*x)*c*sqrt(-(b*c - a*d)/c)/((b*c - a*d)*x)) + 
(b^2*c*d*x - 2*a^2*d^2)*sqrt(b*x^2 + a*x))/(c*d^2*x)]
 

Sympy [F]

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

integrate((b*x**2+a*x)**(5/2)/x**4/(d*x+c),x)
 

Output:

Integral((x*(a + b*x))**(5/2)/(x**4*(c + d*x)), x)
 

Maxima [F]

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

integrate((b*x^2+a*x)^(5/2)/x^4/(d*x+c),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a*x)^(5/2)/((d*x + c)*x^4), x)
 

Giac [F(-2)]

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

integrate((b*x^2+a*x)^(5/2)/x^4/(d*x+c),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

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

int((a*x + b*x^2)^(5/2)/(x^4*(c + d*x)),x)
 

Output:

int((a*x + b*x^2)^(5/2)/(x^4*(c + d*x)), x)
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.17 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4 (c+d x)} \, dx=\frac {8 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a^{2} d^{2} x -16 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a b c d x +8 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x +8 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a^{2} d^{2} x -16 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a b c d x +8 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x -8 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c \,d^{2}+4 \sqrt {x}\, \sqrt {b x +a}\, b^{2} c^{2} d x +20 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a b \,c^{2} d x -8 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) b^{2} c^{3} x -8 \sqrt {b}\, a^{2} c \,d^{2} x -\sqrt {b}\, a b \,c^{2} d x}{4 c^{2} d^{2} x} \] Input:

int((b*x^2+a*x)^(5/2)/x^4/(d*x+c),x)
 

Output:

(8*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - 
 sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a**2*d**2*x - 16*sqrt(c)*sqrt 
(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d 
)*sqrt(b))/(sqrt(c)*sqrt(b)))*a*b*c*d*x + 8*sqrt(c)*sqrt(a*d - b*c)*atan(( 
sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c 
)*sqrt(b)))*b**2*c**2*x + 8*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) 
+ sqrt(d)*sqrt(a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a**2 
*d**2*x - 16*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt( 
a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a*b*c*d*x + 8*sqrt( 
c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(x) 
*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*b**2*c**2*x - 8*sqrt(x)*sqrt(a + b*x) 
*a**2*c*d**2 + 4*sqrt(x)*sqrt(a + b*x)*b**2*c**2*d*x + 20*sqrt(b)*log((sqr 
t(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a*b*c**2*d*x - 8*sqrt(b)*log((sqrt( 
a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*b**2*c**3*x - 8*sqrt(b)*a**2*c*d**2*x 
 - sqrt(b)*a*b*c**2*d*x)/(4*c**2*d**2*x)