\(\int \frac {(a+b x^2)^2}{x^4 \sqrt {c+d x}} \, dx\) [545]

Optimal result

Integrand size = 22, antiderivative size = 136 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \sqrt {c+d x}} \, dx=\frac {2 b^2 \sqrt {c+d x}}{d}-\frac {a^2 \sqrt {c+d x}}{3 c x^3}+\frac {5 a^2 d \sqrt {c+d x}}{12 c^2 x^2}-\frac {a \left (16 b c^2+5 a d^2\right ) \sqrt {c+d x}}{8 c^3 x}+\frac {a d \left (16 b c^2+5 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 c^{7/2}} \] Output:

2*b^2*(d*x+c)^(1/2)/d-1/3*a^2*(d*x+c)^(1/2)/c/x^3+5/12*a^2*d*(d*x+c)^(1/2) 
/c^2/x^2-1/8*a*(5*a*d^2+16*b*c^2)*(d*x+c)^(1/2)/c^3/x+1/8*a*d*(5*a*d^2+16* 
b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \sqrt {c+d x}} \, dx=\frac {\sqrt {c+d x} \left (-48 a b c^2 d x^2+48 b^2 c^3 x^3+a^2 d \left (-8 c^2+10 c d x-15 d^2 x^2\right )\right )}{24 c^3 d x^3}+\frac {a d \left (16 b c^2+5 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 c^{7/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*(-48*a*b*c^2*d*x^2 + 48*b^2*c^3*x^3 + a^2*d*(-8*c^2 + 10*c* 
d*x - 15*d^2*x^2)))/(24*c^3*d*x^3) + (a*d*(16*b*c^2 + 5*a*d^2)*ArcTanh[Sqr 
t[c + d*x]/Sqrt[c]])/(8*c^(7/2))
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {517, 1471, 25, 2345, 27, 1471, 25, 299, 219}

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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 1471

\(\displaystyle \frac {2 \left (-\frac {\int \frac {6 b^2 c^4+12 a b d^2 c^2+18 b^2 (c+d x)^2 c^2-6 b^2 (c+d x)^3 c-6 b \left (3 b c^2+2 a d^2\right ) (c+d x) c+5 a^2 d^4}{d^3 x^3}d\sqrt {c+d x}}{6 c}-\frac {a^2 d \sqrt {c+d x}}{6 c x^3}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {\int -\frac {6 b^2 c^4+12 a b d^2 c^2+18 b^2 (c+d x)^2 c^2-6 b^2 (c+d x)^3 c-6 b \left (3 b c^2+2 a d^2\right ) (c+d x) c+5 a^2 d^4}{d^3 x^3}d\sqrt {c+d x}}{6 c}-\frac {a^2 d \sqrt {c+d x}}{6 c x^3}\right )}{d}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {2 \left (\frac {\frac {5 a^2 d^2 \sqrt {c+d x}}{4 c x^2}-\frac {\int -\frac {3 \left (8 b^2 c^4-16 b^2 (c+d x) c^3+16 a b d^2 c^2+8 b^2 (c+d x)^2 c^2+5 a^2 d^4\right )}{d^2 x^2}d\sqrt {c+d x}}{4 c}}{6 c}-\frac {a^2 d \sqrt {c+d x}}{6 c x^3}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {\frac {3 \int \frac {8 b^2 c^4-16 b^2 (c+d x) c^3+16 a b d^2 c^2+8 b^2 (c+d x)^2 c^2+5 a^2 d^4}{d^2 x^2}d\sqrt {c+d x}}{4 c}+\frac {5 a^2 d^2 \sqrt {c+d x}}{4 c x^2}}{6 c}-\frac {a^2 d \sqrt {c+d x}}{6 c x^3}\right )}{d}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {2 \left (\frac {\frac {3 \left (-\frac {\int \frac {16 b^2 c^4-16 b^2 (c+d x) c^3+16 a b d^2 c^2+5 a^2 d^4}{d x}d\sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x} \left (5 a d^2+16 b c^2\right )}{2 c x}\right )}{4 c}+\frac {5 a^2 d^2 \sqrt {c+d x}}{4 c x^2}}{6 c}-\frac {a^2 d \sqrt {c+d x}}{6 c x^3}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {\frac {3 \left (\frac {\int -\frac {16 b^2 c^4-16 b^2 (c+d x) c^3+16 a b d^2 c^2+5 a^2 d^4}{d x}d\sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x} \left (5 a d^2+16 b c^2\right )}{2 c x}\right )}{4 c}+\frac {5 a^2 d^2 \sqrt {c+d x}}{4 c x^2}}{6 c}-\frac {a^2 d \sqrt {c+d x}}{6 c x^3}\right )}{d}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {2 \left (\frac {\frac {3 \left (\frac {a d^2 \left (5 a d^2+16 b c^2\right ) \int -\frac {1}{d x}d\sqrt {c+d x}+16 b^2 c^3 \sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x} \left (5 a d^2+16 b c^2\right )}{2 c x}\right )}{4 c}+\frac {5 a^2 d^2 \sqrt {c+d x}}{4 c x^2}}{6 c}-\frac {a^2 d \sqrt {c+d x}}{6 c x^3}\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \left (\frac {\frac {5 a^2 d^2 \sqrt {c+d x}}{4 c x^2}+\frac {3 \left (\frac {\frac {a d^2 \left (5 a d^2+16 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}+16 b^2 c^3 \sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x} \left (5 a d^2+16 b c^2\right )}{2 c x}\right )}{4 c}}{6 c}-\frac {a^2 d \sqrt {c+d x}}{6 c x^3}\right )}{d}\)

Input:

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

Output:

(2*(-1/6*(a^2*d*Sqrt[c + d*x])/(c*x^3) + ((5*a^2*d^2*Sqrt[c + d*x])/(4*c*x 
^2) + (3*(-1/2*(a*d*(16*b*c^2 + 5*a*d^2)*Sqrt[c + d*x])/(c*x) + (16*b^2*c^ 
3*Sqrt[c + d*x] + (a*d^2*(16*b*c^2 + 5*a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c 
]])/Sqrt[c])/(2*c)))/(4*c))/(6*c)))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 517

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1))   Subst[Int[x^(2*n + 1)*(-c + x^ 
2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {\sqrt {d x +c}\, \left (-48 b^{2} c^{3} x^{3}+15 a^{2} d^{3} x^{2}+48 a b \,c^{2} d \,x^{2}-10 a^{2} c \,d^{2} x +8 c^{2} d \,a^{2}\right )}{24 d \,c^{3} x^{3}}+\frac {a d \left (5 a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 c^{\frac {7}{2}}}\)	\(105\)
pseudoelliptic	\(-\frac {3 x^{3} \left (-\frac {5}{8} a^{2} d^{4}-2 b \,c^{2} d^{2} a \right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+\left (a d \left (6 b \,x^{2}+a \right ) c^{\frac {5}{2}}+\frac {15 \sqrt {c}\, a^{2} d^{3} x^{2}}{8}-\frac {5 c^{\frac {3}{2}} a^{2} d^{2} x}{4}-6 c^{\frac {7}{2}} b^{2} x^{3}\right ) \sqrt {d x +c}}{3 c^{\frac {7}{2}} d \,x^{3}}\)	\(109\)
derivativedivides	\(\frac {2 b^{2} \sqrt {d x +c}+2 a \,d^{2} \left (-\frac {\frac {\left (5 a \,d^{2}+16 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{16 c^{3}}-\frac {\left (5 a \,d^{2}+12 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{6 c^{2}}+\frac {\left (11 a \,d^{2}+16 b \,c^{2}\right ) \sqrt {d x +c}}{16 c}}{d^{3} x^{3}}+\frac {\left (5 a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 c^{\frac {7}{2}}}\right )}{d}\)	\(138\)
default	\(\frac {2 b^{2} \sqrt {d x +c}+2 a \,d^{2} \left (-\frac {\frac {\left (5 a \,d^{2}+16 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{16 c^{3}}-\frac {\left (5 a \,d^{2}+12 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{6 c^{2}}+\frac {\left (11 a \,d^{2}+16 b \,c^{2}\right ) \sqrt {d x +c}}{16 c}}{d^{3} x^{3}}+\frac {\left (5 a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 c^{\frac {7}{2}}}\right )}{d}\)	\(138\)

Input:

int((b*x^2+a)^2/x^4/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(d*x+c)^(1/2)*(-48*b^2*c^3*x^3+15*a^2*d^3*x^2+48*a*b*c^2*d*x^2-10*a^ 
2*c*d^2*x+8*a^2*c^2*d)/d/c^3/x^3+1/8*a*d*(5*a*d^2+16*b*c^2)*arctanh((d*x+c 
)^(1/2)/c^(1/2))/c^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.83 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (16 \, a b c^{2} d^{2} + 5 \, a^{2} d^{4}\right )} \sqrt {c} x^{3} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (48 \, b^{2} c^{4} x^{3} + 10 \, a^{2} c^{2} d^{2} x - 8 \, a^{2} c^{3} d - 3 \, {\left (16 \, a b c^{3} d + 5 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x + c}}{48 \, c^{4} d x^{3}}, -\frac {3 \, {\left (16 \, a b c^{2} d^{2} + 5 \, a^{2} d^{4}\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (48 \, b^{2} c^{4} x^{3} + 10 \, a^{2} c^{2} d^{2} x - 8 \, a^{2} c^{3} d - 3 \, {\left (16 \, a b c^{3} d + 5 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x + c}}{24 \, c^{4} d x^{3}}\right ] \] Input:

integrate((b*x^2+a)^2/x^4/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/48*(3*(16*a*b*c^2*d^2 + 5*a^2*d^4)*sqrt(c)*x^3*log((d*x + 2*sqrt(d*x + 
c)*sqrt(c) + 2*c)/x) + 2*(48*b^2*c^4*x^3 + 10*a^2*c^2*d^2*x - 8*a^2*c^3*d 
- 3*(16*a*b*c^3*d + 5*a^2*c*d^3)*x^2)*sqrt(d*x + c))/(c^4*d*x^3), -1/24*(3 
*(16*a*b*c^2*d^2 + 5*a^2*d^4)*sqrt(-c)*x^3*arctan(sqrt(-c)/sqrt(d*x + c)) 
- (48*b^2*c^4*x^3 + 10*a^2*c^2*d^2*x - 8*a^2*c^3*d - 3*(16*a*b*c^3*d + 5*a 
^2*c*d^3)*x^2)*sqrt(d*x + c))/(c^4*d*x^3)]
 

Sympy [A] (verification not implemented)

Time = 63.01 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \sqrt {c+d x}} \, dx=- \frac {a^{2}}{3 \sqrt {d} x^{\frac {7}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {a^{2} \sqrt {d}}{12 c x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {5 a^{2} d^{\frac {3}{2}}}{24 c^{2} x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {5 a^{2} d^{\frac {5}{2}}}{8 c^{3} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} + \frac {5 a^{2} d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{8 c^{\frac {7}{2}}} - \frac {2 a b \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{c \sqrt {x}} + \frac {2 a b d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{c^{\frac {3}{2}}} + b^{2} \left (\begin {cases} \frac {2 \sqrt {c + d x}}{d} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {c}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((b*x**2+a)**2/x**4/(d*x+c)**(1/2),x)
 

Output:

-a**2/(3*sqrt(d)*x**(7/2)*sqrt(c/(d*x) + 1)) + a**2*sqrt(d)/(12*c*x**(5/2) 
*sqrt(c/(d*x) + 1)) - 5*a**2*d**(3/2)/(24*c**2*x**(3/2)*sqrt(c/(d*x) + 1)) 
 - 5*a**2*d**(5/2)/(8*c**3*sqrt(x)*sqrt(c/(d*x) + 1)) + 5*a**2*d**3*asinh( 
sqrt(c)/(sqrt(d)*sqrt(x)))/(8*c**(7/2)) - 2*a*b*sqrt(d)*sqrt(c/(d*x) + 1)/ 
(c*sqrt(x)) + 2*a*b*d*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/c**(3/2) + b**2*Pie 
cewise((2*sqrt(c + d*x)/d, Ne(d, 0)), (x/sqrt(c), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.51 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \sqrt {c+d x}} \, dx=-\frac {1}{48} \, d^{3} {\left (\frac {2 \, {\left (3 \, {\left (16 \, a b c^{2} + 5 \, a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} - 8 \, {\left (12 \, a b c^{3} + 5 \, a^{2} c d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 3 \, {\left (16 \, a b c^{4} + 11 \, a^{2} c^{2} d^{2}\right )} \sqrt {d x + c}\right )}}{{\left (d x + c\right )}^{3} c^{3} d^{2} - 3 \, {\left (d x + c\right )}^{2} c^{4} d^{2} + 3 \, {\left (d x + c\right )} c^{5} d^{2} - c^{6} d^{2}} - \frac {96 \, \sqrt {d x + c} b^{2}}{d^{4}} + \frac {3 \, {\left (16 \, b c^{2} + 5 \, a d^{2}\right )} a \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {7}{2}} d^{2}}\right )} \] Input:

integrate((b*x^2+a)^2/x^4/(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

-1/48*d^3*(2*(3*(16*a*b*c^2 + 5*a^2*d^2)*(d*x + c)^(5/2) - 8*(12*a*b*c^3 + 
 5*a^2*c*d^2)*(d*x + c)^(3/2) + 3*(16*a*b*c^4 + 11*a^2*c^2*d^2)*sqrt(d*x + 
 c))/((d*x + c)^3*c^3*d^2 - 3*(d*x + c)^2*c^4*d^2 + 3*(d*x + c)*c^5*d^2 - 
c^6*d^2) - 96*sqrt(d*x + c)*b^2/d^4 + 3*(16*b*c^2 + 5*a*d^2)*a*log((sqrt(d 
*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/(c^(7/2)*d^2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \sqrt {c+d x}} \, dx=\frac {1}{24} \, d^{3} {\left (\frac {48 \, \sqrt {d x + c} b^{2}}{d^{4}} - \frac {3 \, {\left (16 \, a b c^{2} + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{3} d^{2}} - \frac {48 \, {\left (d x + c\right )}^{\frac {5}{2}} a b c^{2} - 96 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c^{3} + 48 \, \sqrt {d x + c} a b c^{4} + 15 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} d^{2} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} c d^{2} + 33 \, \sqrt {d x + c} a^{2} c^{2} d^{2}}{c^{3} d^{5} x^{3}}\right )} \] Input:

integrate((b*x^2+a)^2/x^4/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

1/24*d^3*(48*sqrt(d*x + c)*b^2/d^4 - 3*(16*a*b*c^2 + 5*a^2*d^2)*arctan(sqr 
t(d*x + c)/sqrt(-c))/(sqrt(-c)*c^3*d^2) - (48*(d*x + c)^(5/2)*a*b*c^2 - 96 
*(d*x + c)^(3/2)*a*b*c^3 + 48*sqrt(d*x + c)*a*b*c^4 + 15*(d*x + c)^(5/2)*a 
^2*d^2 - 40*(d*x + c)^(3/2)*a^2*c*d^2 + 33*sqrt(d*x + c)*a^2*c^2*d^2)/(c^3 
*d^5*x^3))
 

Mupad [B] (verification not implemented)

Time = 8.73 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \sqrt {c+d x}} \, dx=\frac {\frac {\left (11\,a^2\,d^3+16\,b\,a\,c^2\,d\right )\,\sqrt {c+d\,x}}{8\,c}-\frac {\left (5\,a^2\,d^3+12\,b\,a\,c^2\,d\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,c^2}+\frac {\left (5\,a^2\,d^3+16\,b\,a\,c^2\,d\right )\,{\left (c+d\,x\right )}^{5/2}}{8\,c^3}}{3\,c\,{\left (c+d\,x\right )}^2-3\,c^2\,\left (c+d\,x\right )-{\left (c+d\,x\right )}^3+c^3}+\frac {2\,b^2\,\sqrt {c+d\,x}}{d}+\frac {a\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (16\,b\,c^2+5\,a\,d^2\right )}{8\,c^{7/2}} \] Input:

int((a + b*x^2)^2/(x^4*(c + d*x)^(1/2)),x)
 

Output:

(((11*a^2*d^3 + 16*a*b*c^2*d)*(c + d*x)^(1/2))/(8*c) - ((5*a^2*d^3 + 12*a* 
b*c^2*d)*(c + d*x)^(3/2))/(3*c^2) + ((5*a^2*d^3 + 16*a*b*c^2*d)*(c + d*x)^ 
(5/2))/(8*c^3))/(3*c*(c + d*x)^2 - 3*c^2*(c + d*x) - (c + d*x)^3 + c^3) + 
(2*b^2*(c + d*x)^(1/2))/d + (a*d*atanh((c + d*x)^(1/2)/c^(1/2))*(5*a*d^2 + 
 16*b*c^2))/(8*c^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \sqrt {c+d x}} \, dx=\frac {-16 \sqrt {d x +c}\, a^{2} c^{3} d +20 \sqrt {d x +c}\, a^{2} c^{2} d^{2} x -30 \sqrt {d x +c}\, a^{2} c \,d^{3} x^{2}-96 \sqrt {d x +c}\, a b \,c^{3} d \,x^{2}+96 \sqrt {d x +c}\, b^{2} c^{4} x^{3}-15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} d^{4} x^{3}-48 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a b \,c^{2} d^{2} x^{3}+15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} d^{4} x^{3}+48 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a b \,c^{2} d^{2} x^{3}}{48 c^{4} d \,x^{3}} \] Input:

int((b*x^2+a)^2/x^4/(d*x+c)^(1/2),x)
 

Output:

( - 16*sqrt(c + d*x)*a**2*c**3*d + 20*sqrt(c + d*x)*a**2*c**2*d**2*x - 30* 
sqrt(c + d*x)*a**2*c*d**3*x**2 - 96*sqrt(c + d*x)*a*b*c**3*d*x**2 + 96*sqr 
t(c + d*x)*b**2*c**4*x**3 - 15*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a**2*d 
**4*x**3 - 48*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a*b*c**2*d**2*x**3 + 15 
*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**2*d**4*x**3 + 48*sqrt(c)*log(sqrt 
(c + d*x) + sqrt(c))*a*b*c**2*d**2*x**3)/(48*c**4*d*x**3)