\(\int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx\) [496]

Optimal result

Integrand size = 20, antiderivative size = 113 \[ \int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx=-\frac {a \sqrt {c+d x}}{3 c x^3}+\frac {5 a d \sqrt {c+d x}}{12 c^2 x^2}-\frac {\left (8 b c^2+5 a d^2\right ) \sqrt {c+d x}}{8 c^3 x}+\frac {d \left (8 b c^2+5 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 c^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.80 \[ \int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx=-\frac {\sqrt {c+d x} \left (24 b c^2 x^2+a \left (8 c^2-10 c d x+15 d^2 x^2\right )\right )}{24 c^3 x^3}+\frac {d \left (8 b c^2+5 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 c^{7/2}} \] Input:

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

Output:

-1/24*(Sqrt[c + d*x]*(24*b*c^2*x^2 + a*(8*c^2 - 10*c*d*x + 15*d^2*x^2)))/( 
c^3*x^3) + (d*(8*b*c^2 + 5*a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(8*c^(7/ 
2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {517, 1471, 25, 298, 215, 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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 1471

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 298

\(\displaystyle 2 d \left (\frac {\frac {3 \left (5 a d^2+8 b c^2\right ) \int \frac {1}{d^2 x^2}d\sqrt {c+d x}}{4 c}+\frac {5 a \sqrt {c+d x}}{4 c x^2}}{6 c}-\frac {a \sqrt {c+d x}}{6 c d x^3}\right )\)

\(\Big \downarrow \) 215

\(\displaystyle 2 d \left (\frac {\frac {3 \left (5 a d^2+8 b c^2\right ) \left (\frac {\int -\frac {1}{d x}d\sqrt {c+d x}}{2 c}-\frac {\sqrt {c+d x}}{2 c d x}\right )}{4 c}+\frac {5 a \sqrt {c+d x}}{4 c x^2}}{6 c}-\frac {a \sqrt {c+d x}}{6 c d x^3}\right )\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

2*d*(-1/6*(a*Sqrt[c + d*x])/(c*d*x^3) + ((5*a*Sqrt[c + d*x])/(4*c*x^2) + ( 
3*(8*b*c^2 + 5*a*d^2)*(-1/2*Sqrt[c + d*x]/(c*d*x) + ArcTanh[Sqrt[c + d*x]/ 
Sqrt[c]]/(2*c^(3/2))))/(4*c))/(6*c))
 

Defintions of rubi rules used

rule 25

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

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 298

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(-\frac {\left (-\frac {15}{8} a \,d^{3}-3 b \,c^{2} d \right ) x^{3} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+\left (\left (3 b \,x^{2}+a \right ) c^{\frac {5}{2}}+\frac {15 d x \left (d x \sqrt {c}-\frac {2 c^{\frac {3}{2}}}{3}\right ) a}{8}\right ) \sqrt {d x +c}}{3 c^{\frac {7}{2}} x^{3}}\)	\(78\)
risch	\(-\frac {\sqrt {d x +c}\, \left (15 a \,d^{2} x^{2}+24 b \,c^{2} x^{2}-10 a d x c +8 a \,c^{2}\right )}{24 c^{3} x^{3}}+\frac {d \left (5 a \,d^{2}+8 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 c^{\frac {7}{2}}}\)	\(79\)
derivativedivides	\(2 d \left (-\frac {\frac {\left (5 a \,d^{2}+8 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{16 c^{3}}-\frac {\left (5 a \,d^{2}+6 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{6 c^{2}}+\frac {\left (11 a \,d^{2}+8 b \,c^{2}\right ) \sqrt {d x +c}}{16 c}}{d^{3} x^{3}}+\frac {\left (5 a \,d^{2}+8 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 c^{\frac {7}{2}}}\right )\)	\(119\)
default	\(2 d \left (-\frac {\frac {\left (5 a \,d^{2}+8 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{16 c^{3}}-\frac {\left (5 a \,d^{2}+6 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{6 c^{2}}+\frac {\left (11 a \,d^{2}+8 b \,c^{2}\right ) \sqrt {d x +c}}{16 c}}{d^{3} x^{3}}+\frac {\left (5 a \,d^{2}+8 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 c^{\frac {7}{2}}}\right )\)	\(119\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.67 \[ \int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (8 \, b c^{2} d + 5 \, a d^{3}\right )} \sqrt {c} x^{3} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (10 \, a c^{2} d x - 8 \, a c^{3} - 3 \, {\left (8 \, b c^{3} + 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{48 \, c^{4} x^{3}}, -\frac {3 \, {\left (8 \, b c^{2} d + 5 \, a d^{3}\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (10 \, a c^{2} d x - 8 \, a c^{3} - 3 \, {\left (8 \, b c^{3} + 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{24 \, c^{4} x^{3}}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 24.56 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.64 \[ \int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx=- \frac {a}{3 \sqrt {d} x^{\frac {7}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {a \sqrt {d}}{12 c x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {5 a d^{\frac {3}{2}}}{24 c^{2} x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {5 a d^{\frac {5}{2}}}{8 c^{3} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} + \frac {5 a d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{8 c^{\frac {7}{2}}} - \frac {b \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{c \sqrt {x}} + \frac {b d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{c^{\frac {3}{2}}} \] Input:

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

Output:

-a/(3*sqrt(d)*x**(7/2)*sqrt(c/(d*x) + 1)) + a*sqrt(d)/(12*c*x**(5/2)*sqrt( 
c/(d*x) + 1)) - 5*a*d**(3/2)/(24*c**2*x**(3/2)*sqrt(c/(d*x) + 1)) - 5*a*d* 
*(5/2)/(8*c**3*sqrt(x)*sqrt(c/(d*x) + 1)) + 5*a*d**3*asinh(sqrt(c)/(sqrt(d 
)*sqrt(x)))/(8*c**(7/2)) - b*sqrt(d)*sqrt(c/(d*x) + 1)/(c*sqrt(x)) + b*d*a 
sinh(sqrt(c)/(sqrt(d)*sqrt(x)))/c**(3/2)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.60 \[ \int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx=-\frac {1}{48} \, d^{3} {\left (\frac {2 \, {\left (3 \, {\left (8 \, b c^{2} + 5 \, a d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} - 8 \, {\left (6 \, b c^{3} + 5 \, a c d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 3 \, {\left (8 \, b c^{4} + 11 \, a c^{2} d^{2}\right )} \sqrt {d x + c}\right )}}{{\left (d x + c\right )}^{3} c^{3} d^{2} - 3 \, {\left (d x + c\right )}^{2} c^{4} d^{2} + 3 \, {\left (d x + c\right )} c^{5} d^{2} - c^{6} d^{2}} + \frac {3 \, {\left (8 \, b c^{2} + 5 \, a d^{2}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {7}{2}} d^{2}}\right )} \] Input:

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

Output:

-1/48*d^3*(2*(3*(8*b*c^2 + 5*a*d^2)*(d*x + c)^(5/2) - 8*(6*b*c^3 + 5*a*c*d 
^2)*(d*x + c)^(3/2) + 3*(8*b*c^4 + 11*a*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) + 3*(8 
*b*c^2 + 5*a*d^2)*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 = 139, normalized size of antiderivative = 1.23 \[ \int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx=-\frac {1}{24} \, d^{3} {\left (\frac {3 \, {\left (8 \, b c^{2} + 5 \, a d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{3} d^{2}} + \frac {24 \, {\left (d x + c\right )}^{\frac {5}{2}} b c^{2} - 48 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{3} + 24 \, \sqrt {d x + c} b c^{4} + 15 \, {\left (d x + c\right )}^{\frac {5}{2}} a d^{2} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{2} + 33 \, \sqrt {d x + c} a c^{2} d^{2}}{c^{3} d^{5} x^{3}}\right )} \] Input:

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

Output:

-1/24*d^3*(3*(8*b*c^2 + 5*a*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)* 
c^3*d^2) + (24*(d*x + c)^(5/2)*b*c^2 - 48*(d*x + c)^(3/2)*b*c^3 + 24*sqrt( 
d*x + c)*b*c^4 + 15*(d*x + c)^(5/2)*a*d^2 - 40*(d*x + c)^(3/2)*a*c*d^2 + 3 
3*sqrt(d*x + c)*a*c^2*d^2)/(c^3*d^5*x^3))
 

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.30 \[ \int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx=\frac {\frac {\left (8\,b\,c^2\,d+11\,a\,d^3\right )\,\sqrt {c+d\,x}}{8\,c}-\frac {\left (6\,b\,c^2\,d+5\,a\,d^3\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,c^2}+\frac {\left (8\,b\,c^2\,d+5\,a\,d^3\right )\,{\left (c+d\,x\right )}^{5/2}}{8\,c^3}}{3\,c\,{\left (c+d\,x\right )}^2-3\,c^2\,\left (c+d\,x\right )-{\left (c+d\,x\right )}^3+c^3}+\frac {d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (8\,b\,c^2+5\,a\,d^2\right )}{8\,c^{7/2}} \] Input:

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

Output:

(((11*a*d^3 + 8*b*c^2*d)*(c + d*x)^(1/2))/(8*c) - ((5*a*d^3 + 6*b*c^2*d)*( 
c + d*x)^(3/2))/(3*c^2) + ((5*a*d^3 + 8*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) + (d*atanh((c + d 
*x)^(1/2)/c^(1/2))*(5*a*d^2 + 8*b*c^2))/(8*c^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.38 \[ \int \frac {a+b x^2}{x^4 \sqrt {c+d x}} \, dx=\frac {-16 \sqrt {d x +c}\, a \,c^{3}+20 \sqrt {d x +c}\, a \,c^{2} d x -30 \sqrt {d x +c}\, a c \,d^{2} x^{2}-48 \sqrt {d x +c}\, b \,c^{3} x^{2}-15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{3} x^{3}-24 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{2} d \,x^{3}+15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{3} x^{3}+24 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{2} d \,x^{3}}{48 c^{4} x^{3}} \] Input:

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

Output:

( - 16*sqrt(c + d*x)*a*c**3 + 20*sqrt(c + d*x)*a*c**2*d*x - 30*sqrt(c + d* 
x)*a*c*d**2*x**2 - 48*sqrt(c + d*x)*b*c**3*x**2 - 15*sqrt(c)*log(sqrt(c + 
d*x) - sqrt(c))*a*d**3*x**3 - 24*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*b*c* 
*2*d*x**3 + 15*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*d**3*x**3 + 24*sqrt( 
c)*log(sqrt(c + d*x) + sqrt(c))*b*c**2*d*x**3)/(48*c**4*x**3)