\(\int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx\) [370]

Optimal result

Integrand size = 22, antiderivative size = 93 \[ \int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx=-\frac {c \left (a+b x^2\right )^{3/4}}{2 a x^2}-\frac {(b c-4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{5/4}}+\frac {(b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{5/4}} \] Output:

-1/2*c*(b*x^2+a)^(3/4)/a/x^2-1/4*(-4*a*d+b*c)*arctan((b*x^2+a)^(1/4)/a^(1/ 
4))/a^(5/4)+1/4*(-4*a*d+b*c)*arctanh((b*x^2+a)^(1/4)/a^(1/4))/a^(5/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx=-\frac {c \left (a+b x^2\right )^{3/4}}{2 a x^2}+\frac {(-b c+4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{5/4}}+\frac {(b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{5/4}} \] Input:

Integrate[(c + d*x^2)/(x^3*(a + b*x^2)^(1/4)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {354, 87, 73, 25, 27, 827, 216, 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.

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

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {d x^2+c}{x^4 \sqrt [4]{b x^2+a}}dx^2\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[(c + d*x^2)/(x^3*(a + b*x^2)^(1/4)),x]
 

Output:

(-((c*(a + b*x^2)^(3/4))/(a*x^2)) + ((b*c - 4*a*d)*(-1/2*ArcTan[(a + b*x^2 
)^(1/4)/a^(1/4)]/a^(1/4) + ArcTanh[(a + b*x^2)^(1/4)/a^(1/4)]/(2*a^(1/4))) 
)/a)/2
 

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 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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91

method	result	size
pseudoelliptic	\(-\frac {c \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{\frac {1}{4}}+\left (\ln \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )\right ) x^{2} \left (a d -\frac {b c}{4}\right )}{2 a^{\frac {5}{4}} x^{2}}\)	\(85\)

Input:

int((d*x^2+c)/x^3/(b*x^2+a)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

-1/2*(c*(b*x^2+a)^(3/4)*a^(1/4)+(ln(((b*x^2+a)^(1/4)+a^(1/4))/((b*x^2+a)^( 
1/4)-a^(1/4)))-2*arctan((b*x^2+a)^(1/4)/a^(1/4)))*x^2*(a*d-1/4*b*c))/a^(5/ 
4)/x^2
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 707, normalized size of antiderivative = 7.60 \[ \int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx =\text {Too large to display} \] Input:

integrate((d*x^2+c)/x^3/(b*x^2+a)^(1/4),x, algorithm="fricas")
 

Output:

-1/8*(a*x^2*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c* 
d^3 + 256*a^4*d^4)/a^5)^(1/4)*log(a^4*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2* 
b^2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^5)^(3/4) - (b^3*c^3 - 12*a* 
b^2*c^2*d + 48*a^2*b*c*d^2 - 64*a^3*d^3)*(b*x^2 + a)^(1/4)) - I*a*x^2*((b^ 
4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^ 
4)/a^5)^(1/4)*log(I*a^4*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 
256*a^3*b*c*d^3 + 256*a^4*d^4)/a^5)^(3/4) - (b^3*c^3 - 12*a*b^2*c^2*d + 48 
*a^2*b*c*d^2 - 64*a^3*d^3)*(b*x^2 + a)^(1/4)) + I*a*x^2*((b^4*c^4 - 16*a*b 
^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^5)^(1/4)* 
log(-I*a^4*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c*d 
^3 + 256*a^4*d^4)/a^5)^(3/4) - (b^3*c^3 - 12*a*b^2*c^2*d + 48*a^2*b*c*d^2 
- 64*a^3*d^3)*(b*x^2 + a)^(1/4)) - a*x^2*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a 
^2*b^2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^5)^(1/4)*log(-a^4*((b^4* 
c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4) 
/a^5)^(3/4) - (b^3*c^3 - 12*a*b^2*c^2*d + 48*a^2*b*c*d^2 - 64*a^3*d^3)*(b* 
x^2 + a)^(1/4)) + 4*(b*x^2 + a)^(3/4)*c)/(a*x^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91 \[ \int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx=- \frac {c \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \sqrt [4]{b} x^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} - \frac {d \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \sqrt [4]{b} \sqrt {x} \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((d*x**2+c)/x**3/(b*x**2+a)**(1/4),x)
 

Output:

-c*gamma(5/4)*hyper((1/4, 5/4), (9/4,), a*exp_polar(I*pi)/(b*x**2))/(2*b** 
(1/4)*x**(5/2)*gamma(9/4)) - d*gamma(1/4)*hyper((1/4, 1/4), (5/4,), a*exp_ 
polar(I*pi)/(b*x**2))/(2*b**(1/4)*sqrt(x)*gamma(5/4))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (73) = 146\).

Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.66 \[ \int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx=-\frac {1}{8} \, c {\left (\frac {b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{a} + \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} b}{{\left (b x^{2} + a\right )} a - a^{2}}\right )} + \frac {1}{2} \, d {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )} \] Input:

integrate((d*x^2+c)/x^3/(b*x^2+a)^(1/4),x, algorithm="maxima")
 

Output:

-1/8*c*(b*(2*arctan((b*x^2 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((b*x^2 + a)^ 
(1/4) - a^(1/4))/((b*x^2 + a)^(1/4) + a^(1/4)))/a^(1/4))/a + 4*(b*x^2 + a) 
^(3/4)*b/((b*x^2 + a)*a - a^2)) + 1/2*d*(2*arctan((b*x^2 + a)^(1/4)/a^(1/4 
))/a^(1/4) + log(((b*x^2 + a)^(1/4) - a^(1/4))/((b*x^2 + a)^(1/4) + a^(1/4 
)))/a^(1/4))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (73) = 146\).

Time = 0.13 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.72 \[ \int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx=-\frac {1}{16} \, b {\left (\frac {2 \, \sqrt {2} {\left (b c - 4 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}} a b} + \frac {2 \, \sqrt {2} {\left (b c - 4 \, a d\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}} a b} - \frac {\sqrt {2} {\left (b c - 4 \, a d\right )} \log \left (\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a b} + \frac {\sqrt {2} {\left (b c - 4 \, a d\right )} \log \left (-\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a b} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} c}{a b x^{2}}\right )} \] Input:

integrate((d*x^2+c)/x^3/(b*x^2+a)^(1/4),x, algorithm="giac")
 

Output:

-1/16*b*(2*sqrt(2)*(b*c - 4*a*d)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 
2*(b*x^2 + a)^(1/4))/(-a)^(1/4))/((-a)^(1/4)*a*b) + 2*sqrt(2)*(b*c - 4*a*d 
)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^2 + a)^(1/4))/(-a)^(1/4 
))/((-a)^(1/4)*a*b) - sqrt(2)*(b*c - 4*a*d)*log(sqrt(2)*(b*x^2 + a)^(1/4)* 
(-a)^(1/4) + sqrt(b*x^2 + a) + sqrt(-a))/((-a)^(1/4)*a*b) + sqrt(2)*(b*c - 
 4*a*d)*log(-sqrt(2)*(b*x^2 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^2 + a) + sqrt 
(-a))/((-a)^(1/4)*a*b) + 8*(b*x^2 + a)^(3/4)*c/(a*b*x^2))
 

Mupad [B] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04 \[ \int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx=\frac {d\,\left (\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )-\mathrm {atanh}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )\right )}{a^{1/4}}-\frac {c\,{\left (b\,x^2+a\right )}^{3/4}}{2\,a\,x^2}-\frac {b\,c\,\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{4\,a^{5/4}}+\frac {b\,c\,\mathrm {atanh}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{4\,a^{5/4}} \] Input:

int((c + d*x^2)/(x^3*(a + b*x^2)^(1/4)),x)
 

Output:

(d*(atan((a + b*x^2)^(1/4)/a^(1/4)) - atanh((a + b*x^2)^(1/4)/a^(1/4))))/a 
^(1/4) - (c*(a + b*x^2)^(3/4))/(2*a*x^2) - (b*c*atan((a + b*x^2)^(1/4)/a^( 
1/4)))/(4*a^(5/4)) + (b*c*atanh((a + b*x^2)^(1/4)/a^(1/4)))/(4*a^(5/4))
 

Reduce [F]

\[ \int \frac {c+d x^2}{x^3 \sqrt [4]{a+b x^2}} \, dx=\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} x^{3}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} x}d x \right ) d \] Input:

int((d*x^2+c)/x^3/(b*x^2+a)^(1/4),x)
 

Output:

int(1/((a + b*x**2)**(1/4)*x**3),x)*c + int(1/((a + b*x**2)**(1/4)*x),x)*d