\(\int \frac {(a x+b x^2)^{3/2}}{x^4 (c+d x)^2} \, dx\) [106]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=-\frac {(11 b c-15 a d) \sqrt {a x+b x^2}}{3 c^3 x}-\frac {2 a \sqrt {a x+b x^2}}{3 c x^2 (c+d x)}+\frac {(3 b c-5 a d) \sqrt {a x+b x^2}}{3 c^2 x (c+d x)}+\frac {(2 b c-5 a d) \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a x+b x^2}}\right )}{c^{7/2}} \] Output:

-1/3*(-15*a*d+11*b*c)*(b*x^2+a*x)^(1/2)/c^3/x-2/3*a*(b*x^2+a*x)^(1/2)/c/x^ 
2/(d*x+c)+1/3*(-5*a*d+3*b*c)*(b*x^2+a*x)^(1/2)/c^2/x/(d*x+c)+(-5*a*d+2*b*c 
)*(-a*d+b*c)^(1/2)*arctanh((-a*d+b*c)^(1/2)*x/c^(1/2)/(b*x^2+a*x)^(1/2))/c 
^(7/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.89 (sec) , antiderivative size = 1953, normalized size of antiderivative = 11.98 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx =\text {Too large to display} \] Input:

Integrate[(a*x + b*x^2)^(3/2)/(x^4*(c + d*x)^2),x]
 

Output:

((x*(a + b*x))^(3/2)*(-((Sqrt[c]*(262144*a^10 + 1376256*a^9*b*x + 3080192* 
a^8*b^2*x^2 + 3829760*a^7*b^3*x^3 + 2888704*a^6*b^4*x^4 + 1354496*a^5*b^5* 
x^5 + 388608*a^4*b^6*x^6 + 64416*a^3*b^7*x^7 + 5460*a^2*b^8*x^8 + 181*a*b^ 
9*x^9 + b^10*x^10 - 262144*a^(19/2)*Sqrt[a + b*x] - 1245184*a^(17/2)*b*x*S 
qrt[a + b*x] - 2490368*a^(15/2)*b^2*x^2*Sqrt[a + b*x] - 2723840*a^(13/2)*b 
^3*x^3*Sqrt[a + b*x] - 1770496*a^(11/2)*b^4*x^4*Sqrt[a + b*x] - 695552*a^( 
9/2)*b^5*x^5*Sqrt[a + b*x] - 160512*a^(7/2)*b^6*x^6*Sqrt[a + b*x] - 20064* 
a^(5/2)*b^7*x^7*Sqrt[a + b*x] - 1140*a^(3/2)*b^8*x^8*Sqrt[a + b*x] - 19*Sq 
rt[a]*b^9*x^9*Sqrt[a + b*x])*(b*c*x*(8*c + 11*d*x) + a*(2*c^2 - 10*c*d*x - 
 15*d^2*x^2)))/(x^(3/2)*(c + d*x)*(-262144*a^(19/2) - 1245184*a^(17/2)*b*x 
 - 2490368*a^(15/2)*b^2*x^2 - 2723840*a^(13/2)*b^3*x^3 - 1770496*a^(11/2)* 
b^4*x^4 - 695552*a^(9/2)*b^5*x^5 - 160512*a^(7/2)*b^6*x^6 - 20064*a^(5/2)* 
b^7*x^7 - 1140*a^(3/2)*b^8*x^8 - 19*Sqrt[a]*b^9*x^9 + 262144*a^9*Sqrt[a + 
b*x] + 1114112*a^8*b*x*Sqrt[a + b*x] + 1966080*a^7*b^2*x^2*Sqrt[a + b*x] + 
 1863680*a^6*b^3*x^3*Sqrt[a + b*x] + 1025024*a^5*b^4*x^4*Sqrt[a + b*x] + 3 
29472*a^4*b^5*x^5*Sqrt[a + b*x] + 59136*a^3*b^6*x^6*Sqrt[a + b*x] + 5280*a 
^2*b^7*x^7*Sqrt[a + b*x] + 180*a*b^8*x^8*Sqrt[a + b*x] + b^9*x^9*Sqrt[a + 
b*x]))) + (21*a*b*c*d*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]))])/Sqrt[-(b*c 
) + 2*a*d - (2*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d]] + ((15*I)*a^(3/2)*d^...
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1261, 107, 105, 105, 104, 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.

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

\(\Big \downarrow \) 1261

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

Int[(a*x + b*x^2)^(3/2)/(x^4*(c + d*x)^2),x]
 

Output:

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

Defintions of rubi rules used

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 105

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 + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(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] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

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 = 143, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {-2 \sqrt {c \left (a d -b c \right )}\, \sqrt {x \left (b x +a \right )}\, \left (\left (4 b x +a \right ) c^{2}-5 d x \left (-\frac {11 b x}{10}+a \right ) c -\frac {15 a \,d^{2} x^{2}}{2}\right )-3 x^{2} \arctan \left (\frac {\sqrt {x \left (b x +a \right )}\, c}{x \sqrt {c \left (a d -b c \right )}}\right ) \left (d x +c \right ) \left (5 a^{2} d^{2}-7 a b c d +2 b^{2} c^{2}\right )}{3 \sqrt {c \left (a d -b c \right )}\, c^{3} x^{2} \left (d x +c \right )}\)	\(143\)
risch	\(-\frac {2 \left (b x +a \right ) \left (-6 a d x +4 c b x +a c \right )}{3 c^{3} \sqrt {x \left (b x +a \right )}\, x}+\frac {\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (\frac {d^{2} \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{c \left (a d -b c \right ) \left (x +\frac {c}{d}\right )}-\frac {\left (a d -2 b c \right ) d \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{2 c \left (a d -b c \right ) \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}\right )}{d^{2}}-\frac {2 a \left (a d -b c \right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{\sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}}{c^{3}}\)	\(430\)
default	\(\text {Expression too large to display}\)	\(1992\)

Input:

int((b*x^2+a*x)^(3/2)/x^4/(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

1/3*(-2*(c*(a*d-b*c))^(1/2)*(x*(b*x+a))^(1/2)*((4*b*x+a)*c^2-5*d*x*(-11/10 
*b*x+a)*c-15/2*a*d^2*x^2)-3*x^2*arctan((x*(b*x+a))^(1/2)/x*c/(c*(a*d-b*c)) 
^(1/2))*(d*x+c)*(5*a^2*d^2-7*a*b*c*d+2*b^2*c^2))/(c*(a*d-b*c))^(1/2)/c^3/x 
^2/(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.08 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, b c d - 5 \, a d^{2}\right )} x^{3} + {\left (2 \, b c^{2} - 5 \, a c d\right )} x^{2}\right )} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x - 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right ) + 2 \, {\left (2 \, a c^{2} + {\left (11 \, b c d - 15 \, a d^{2}\right )} x^{2} + 2 \, {\left (4 \, b c^{2} - 5 \, a c d\right )} x\right )} \sqrt {b x^{2} + a x}}{6 \, {\left (c^{3} d x^{3} + c^{4} x^{2}\right )}}, \frac {3 \, {\left ({\left (2 \, b c d - 5 \, a d^{2}\right )} x^{3} + {\left (2 \, b c^{2} - 5 \, a c d\right )} x^{2}\right )} \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right ) - {\left (2 \, a c^{2} + {\left (11 \, b c d - 15 \, a d^{2}\right )} x^{2} + 2 \, {\left (4 \, b c^{2} - 5 \, a c d\right )} x\right )} \sqrt {b x^{2} + a x}}{3 \, {\left (c^{3} d x^{3} + c^{4} x^{2}\right )}}\right ] \] Input:

integrate((b*x^2+a*x)^(3/2)/x^4/(d*x+c)^2,x, algorithm="fricas")
 

Output:

[-1/6*(3*((2*b*c*d - 5*a*d^2)*x^3 + (2*b*c^2 - 5*a*c*d)*x^2)*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*(2*a*c^2 + (11*b*c*d - 15*a*d^2)*x^2 + 2*(4*b*c^2 - 5* 
a*c*d)*x)*sqrt(b*x^2 + a*x))/(c^3*d*x^3 + c^4*x^2), 1/3*(3*((2*b*c*d - 5*a 
*d^2)*x^3 + (2*b*c^2 - 5*a*c*d)*x^2)*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*a*c^2 + (11*b*c*d - 
 15*a*d^2)*x^2 + 2*(4*b*c^2 - 5*a*c*d)*x)*sqrt(b*x^2 + a*x))/(c^3*d*x^3 + 
c^4*x^2)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{4} \left (c + d x\right )^{2}}\, dx \] Input:

integrate((b*x**2+a*x)**(3/2)/x**4/(d*x+c)**2,x)
 

Output:

Integral((x*(a + b*x))**(3/2)/(x**4*(c + d*x)**2), x)
 

Maxima [F]

\[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\int { \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{2} x^{4}} \,d x } \] Input:

integrate((b*x^2+a*x)^(3/2)/x^4/(d*x+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a*x)^(3/2)/((d*x + c)^2*x^4), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1345 vs. \(2 (141) = 282\).

Time = 8.72 (sec) , antiderivative size = 1345, normalized size of antiderivative = 8.25 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\text {Too large to display} \] Input:

integrate((b*x^2+a*x)^(3/2)/x^4/(d*x+c)^2,x, algorithm="giac")
 

Output:

-1/10*d^2*(2*(2*sqrt(b*c^2 - a*c*d)*b^2*c^7*d^2*abs(d)*sgn(1/(d*x + c))*sg 
n(d) - 7*sqrt(b*c^2 - a*c*d)*a*b*c^6*d^3*abs(d)*sgn(1/(d*x + c))*sgn(d) + 
5*sqrt(b*c^2 - a*c*d)*a^2*c^5*d^4*abs(d)*sgn(1/(d*x + c))*sgn(d))*log(abs( 
b*c^2 - a*c*d))/(b*c^10*d^5 - a*c^9*d^6) + sqrt(b*c^2 - a*c*d)*(2*b*c - 5* 
a*d)*log(abs(-2*(b*c^4 - a*c^3*d)*(sqrt(b - 2*b*c/(d*x + c) + b*c^2/(d*x + 
 c)^2 + a*d/(d*x + c) - a*c*d/(d*x + c)^2) + sqrt(b*c^2*d^2 - a*c*d^3)/((d 
*x + c)*d))^5*abs(d) + (10*b*c^3*d - 9*a*c^2*d^2)*sqrt(b*c^2 - a*c*d)*(sqr 
t(b - 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a*d/(d*x + c) - a*c*d/(d*x + c 
)^2) + sqrt(b*c^2*d^2 - a*c*d^3)/((d*x + c)*d))^4 - 4*(5*b^2*c^4 - 9*a*b*c 
^3*d + 4*a^2*c^2*d^2)*(sqrt(b - 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a*d/ 
(d*x + c) - a*c*d/(d*x + c)^2) + sqrt(b*c^2*d^2 - a*c*d^3)/((d*x + c)*d))^ 
3*abs(d) + 2*(10*b^2*c^3*d - 17*a*b*c^2*d^2 + 7*a^2*c*d^3)*sqrt(b*c^2 - a* 
c*d)*(sqrt(b - 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a*d/(d*x + c) - a*c*d 
/(d*x + c)^2) + sqrt(b*c^2*d^2 - a*c*d^3)/((d*x + c)*d))^2 - 2*(5*b^3*c^4 
- 13*a*b^2*c^3*d + 11*a^2*b*c^2*d^2 - 3*a^3*c*d^3)*(sqrt(b - 2*b*c/(d*x + 
c) + b*c^2/(d*x + c)^2 + a*d/(d*x + c) - a*c*d/(d*x + c)^2) + sqrt(b*c^2*d 
^2 - a*c*d^3)/((d*x + c)*d))*abs(d) + (2*b^3*c^3*d - 5*a*b^2*c^2*d^2 + 4*a 
^2*b*c*d^3 - a^3*d^4)*sqrt(b*c^2 - a*c*d)))*sgn(1/(d*x + c))*sgn(d)/(c^4*d 
*abs(d)) - (10*b^(3/2)*c^2*abs(d) - 10*a*sqrt(b)*c*d*abs(d) + 2*sqrt(b*c^2 
 - a*c*d)*b*c*d*log(abs(-32*b^(7/2)*c^4*abs(d) + 64*a*b^(5/2)*c^3*d*abs...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{3/2}}{x^4\,{\left (c+d\,x\right )}^2} \,d x \] Input:

int((a*x + b*x^2)^(3/2)/(x^4*(c + d*x)^2),x)
 

Output:

int((a*x + b*x^2)^(3/2)/(x^4*(c + d*x)^2), x)
 

Reduce [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 980, normalized size of antiderivative = 6.01 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx =\text {Too large to display} \] Input:

int((b*x^2+a*x)^(3/2)/x^4/(d*x+c)^2,x)
 

Output:

( - 75*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*c*d**2*x**2 - 75*sqr 
t(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**3*x**3 + 90*sqrt(c)*sqrt(a* 
d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*s 
qrt(b))/(sqrt(c)*sqrt(b)))*a*b*c**2*d*x**2 + 90*sqrt(c)*sqrt(a*d - b*c)*at 
an((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*sqrt(b))/(sq 
rt(c)*sqrt(b)))*a*b*c*d**2*x**3 - 24*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**3*x**2 - 24*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - s 
qrt(d)*sqrt(a + b*x) - sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*b**2*c* 
*2*d*x**3 - 75*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqr 
t(a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a**2*c*d**2*x**2 
- 75*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**3*x**3 + 90*sqrt(c) 
*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(x)*s 
qrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a*b*c**2*d*x**2 + 90*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**2*x**3 - 24*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))/(sq...