\(\int \frac {c+d x^2}{x^3 (a+b x^2)^{3/4}} \, dx\) [381]

Optimal result

Integrand size = 22, antiderivative size = 95 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{3/4}} \, dx=-\frac {c \sqrt [4]{a+b x^2}}{2 a x^2}+\frac {(3 b c-4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{7/4}}+\frac {(3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{7/4}} \] Output:

-1/2*c*(b*x^2+a)^(1/4)/a/x^2+1/4*(-4*a*d+3*b*c)*arctan((b*x^2+a)^(1/4)/a^( 
1/4))/a^(7/4)+1/4*(-4*a*d+3*b*c)*arctanh((b*x^2+a)^(1/4)/a^(1/4))/a^(7/4)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{3/4}} \, dx=-\frac {c \sqrt [4]{a+b x^2}}{2 a x^2}+\frac {(3 b c-4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{7/4}}+\frac {(3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{7/4}} \] Input:

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

Output:

-1/2*(c*(a + b*x^2)^(1/4))/(a*x^2) + ((3*b*c - 4*a*d)*ArcTan[(a + b*x^2)^( 
1/4)/a^(1/4)])/(4*a^(7/4)) + ((3*b*c - 4*a*d)*ArcTanh[(a + b*x^2)^(1/4)/a^ 
(1/4)])/(4*a^(7/4))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {354, 87, 73, 756, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 756

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

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09

method	result	size
pseudoelliptic	\(\frac {x^{2} \left (-4 d \,a^{2}+3 a b c \right ) \ln \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right )+2 x^{2} \left (-4 d \,a^{2}+3 a b c \right ) \arctan \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )-4 a^{\frac {7}{4}} c \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{8 a^{\frac {11}{4}} x^{2}}\)	\(104\)

Input:

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

Output:

1/8*(x^2*(-4*a^2*d+3*a*b*c)*ln(((b*x^2+a)^(1/4)+a^(1/4))/((b*x^2+a)^(1/4)- 
a^(1/4)))+2*x^2*(-4*a^2*d+3*a*b*c)*arctan((b*x^2+a)^(1/4)/a^(1/4))-4*a^(7/ 
4)*c*(b*x^2+a)^(1/4))/a^(11/4)/x^2
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 607, normalized size of antiderivative = 6.39 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{3/4}} \, dx=-\frac {a x^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (a^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (3 \, b c - 4 \, a d\right )}\right ) + i \, a x^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (i \, a^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (3 \, b c - 4 \, a d\right )}\right ) - i \, a x^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-i \, a^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (3 \, b c - 4 \, a d\right )}\right ) - a x^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-a^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (3 \, b c - 4 \, a d\right )}\right ) + 4 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} c}{8 \, a x^{2}} \] Input:

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

Output:

-1/8*(a*x^2*((81*b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 - 768*a^3 
*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4)*log(a^2*((81*b^4*c^4 - 432*a*b^3*c^3*d 
+ 864*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4) - (b*x^2 
 + a)^(1/4)*(3*b*c - 4*a*d)) + I*a*x^2*((81*b^4*c^4 - 432*a*b^3*c^3*d + 86 
4*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4)*log(I*a^2*(( 
81*b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256 
*a^4*d^4)/a^7)^(1/4) - (b*x^2 + a)^(1/4)*(3*b*c - 4*a*d)) - I*a*x^2*((81*b 
^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4 
*d^4)/a^7)^(1/4)*log(-I*a^2*((81*b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c 
^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4) - (b*x^2 + a)^(1/4)*(3* 
b*c - 4*a*d)) - a*x^2*((81*b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 
 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4)*log(-a^2*((81*b^4*c^4 - 432*a 
*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/ 
4) - (b*x^2 + a)^(1/4)*(3*b*c - 4*a*d)) + 4*(b*x^2 + a)^(1/4)*c)/(a*x^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.59 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{3/4}} \, dx=- \frac {c \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {3}{4}} x^{\frac {7}{2}} \Gamma \left (\frac {11}{4}\right )} - \frac {d \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {3}{4}} x^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

-c*gamma(7/4)*hyper((3/4, 7/4), (11/4,), a*exp_polar(I*pi)/(b*x**2))/(2*b* 
*(3/4)*x**(7/2)*gamma(11/4)) - d*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*ex 
p_polar(I*pi)/(b*x**2))/(2*b**(3/4)*x**(3/2)*gamma(7/4))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (75) = 150\).

Time = 0.13 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.66 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{3/4}} \, dx=-\frac {1}{8} \, c {\left (\frac {4 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} b}{{\left (b x^{2} + a\right )} a - a^{2}} - \frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )}}{a}\right )} - \frac {1}{2} \, d {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {\log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (75) = 150\).

Time = 0.13 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.71 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{3/4}} \, dx=-\frac {1}{16} \, b {\left (\frac {2 \, \sqrt {2} {\left (3 \, b c - 4 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a b} + \frac {2 \, \sqrt {2} {\left (3 \, b c - 4 \, a d\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a b} + \frac {\sqrt {2} {\left (3 \, b c - 4 \, a d\right )} \log \left (\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a b} - \frac {\sqrt {2} {\left (3 \, b c - 4 \, a d\right )} \log \left (-\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a b} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} c}{a b x^{2}}\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{3/4}} \, dx=\frac {3\,b\,c\,\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{4\,a^{7/4}}-\frac {d\,\mathrm {atanh}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{a^{3/4}}-\frac {c\,{\left (b\,x^2+a\right )}^{1/4}}{2\,a\,x^2}-\frac {d\,\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{a^{3/4}}+\frac {3\,b\,c\,\mathrm {atanh}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{4\,a^{7/4}} \] Input:

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

Output:

(3*b*c*atan((a + b*x^2)^(1/4)/a^(1/4)))/(4*a^(7/4)) - (d*atanh((a + b*x^2) 
^(1/4)/a^(1/4)))/a^(3/4) - (c*(a + b*x^2)^(1/4))/(2*a*x^2) - (d*atan((a + 
b*x^2)^(1/4)/a^(1/4)))/a^(3/4) + (3*b*c*atanh((a + b*x^2)^(1/4)/a^(1/4)))/ 
(4*a^(7/4))
 

Reduce [F]

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

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

Output:

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