\(\int \frac {c+d x^2}{x^3 (a+b x^2)^{9/4}} \, dx\) [417]

Optimal result

Integrand size = 22, antiderivative size = 146 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 (b c-a d)}{5 a^2 \left (a+b x^2\right )^{5/4}}-\frac {2 (2 b c-a d)}{a^3 \sqrt [4]{a+b x^2}}-\frac {c \left (a+b x^2\right )^{3/4}}{2 a^3 x^2}-\frac {(9 b c-4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{13/4}}+\frac {(9 b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 a^{13/4}} \] Output:

1/5*(2*a*d-2*b*c)/a^2/(b*x^2+a)^(5/4)-2*(-a*d+2*b*c)/a^3/(b*x^2+a)^(1/4)-1 
/2*c*(b*x^2+a)^(3/4)/a^3/x^2-1/4*(-4*a*d+9*b*c)*arctan((b*x^2+a)^(1/4)/a^( 
1/4))/a^(13/4)+1/4*(-4*a*d+9*b*c)*arctanh((b*x^2+a)^(1/4)/a^(1/4))/a^(13/4 
)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.88 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx=\frac {-\frac {2 \sqrt [4]{a} \left (45 b^2 c x^4+a^2 \left (5 c-24 d x^2\right )+a b \left (54 c x^2-20 d x^4\right )\right )}{x^2 \left (a+b x^2\right )^{5/4}}+5 (-9 b c+4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )+5 (9 b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{20 a^{13/4}} \] Input:

Integrate[(c + d*x^2)/(x^3*(a + b*x^2)^(9/4)),x]
 

Output:

((-2*a^(1/4)*(45*b^2*c*x^4 + a^2*(5*c - 24*d*x^2) + a*b*(54*c*x^2 - 20*d*x 
^4)))/(x^2*(a + b*x^2)^(5/4)) + 5*(-9*b*c + 4*a*d)*ArcTan[(a + b*x^2)^(1/4 
)/a^(1/4)] + 5*(9*b*c - 4*a*d)*ArcTanh[(a + b*x^2)^(1/4)/a^(1/4)])/(20*a^( 
13/4))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {354, 87, 61, 61, 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.

\(\displaystyle \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {d x^2+c}{x^4 \left (b x^2+a\right )^{9/4}}dx^2\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \int \frac {1}{x^2 \left (b x^2+a\right )^{9/4}}dx^2}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \left (\frac {\int \frac {1}{x^2 \left (b x^2+a\right )^{5/4}}dx^2}{a}+\frac {4}{5 a \left (a+b x^2\right )^{5/4}}\right )}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \left (\frac {\frac {\int \frac {1}{x^2 \sqrt [4]{b x^2+a}}dx^2}{a}+\frac {4}{a \sqrt [4]{a+b x^2}}}{a}+\frac {4}{5 a \left (a+b x^2\right )^{5/4}}\right )}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \left (\frac {\frac {4 \int -\frac {b x^4}{a-x^8}d\sqrt [4]{b x^2+a}}{a b}+\frac {4}{a \sqrt [4]{a+b x^2}}}{a}+\frac {4}{5 a \left (a+b x^2\right )^{5/4}}\right )}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \left (\frac {\frac {4}{a \sqrt [4]{a+b x^2}}-\frac {4 \int \frac {b x^4}{a-x^8}d\sqrt [4]{b x^2+a}}{a b}}{a}+\frac {4}{5 a \left (a+b x^2\right )^{5/4}}\right )}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \left (\frac {\frac {4}{a \sqrt [4]{a+b x^2}}-\frac {4 \int \frac {x^4}{a-x^8}d\sqrt [4]{b x^2+a}}{a}}{a}+\frac {4}{5 a \left (a+b x^2\right )^{5/4}}\right )}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \left (\frac {\frac {4}{a \sqrt [4]{a+b x^2}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^4}d\sqrt [4]{b x^2+a}-\frac {1}{2} \int \frac {1}{x^4+\sqrt {a}}d\sqrt [4]{b x^2+a}\right )}{a}}{a}+\frac {4}{5 a \left (a+b x^2\right )^{5/4}}\right )}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \left (\frac {\frac {4}{a \sqrt [4]{a+b x^2}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^4}d\sqrt [4]{b x^2+a}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}}{a}+\frac {4}{5 a \left (a+b x^2\right )^{5/4}}\right )}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {(9 b c-4 a d) \left (\frac {\frac {4}{a \sqrt [4]{a+b x^2}}-\frac {4 \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}}{a}+\frac {4}{5 a \left (a+b x^2\right )^{5/4}}\right )}{4 a}-\frac {c}{a x^2 \left (a+b x^2\right )^{5/4}}\right )\)

Input:

Int[(c + d*x^2)/(x^3*(a + b*x^2)^(9/4)),x]
 

Output:

(-(c/(a*x^2*(a + b*x^2)^(5/4))) - ((9*b*c - 4*a*d)*(4/(5*a*(a + b*x^2)^(5/ 
4)) + (4/(a*(a + b*x^2)^(1/4)) - (4*(-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)/a))/(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 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, 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 = 0.62 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(-\frac {\frac {54 \left (-\frac {10 x^{2} d}{27}+c \right ) b \,x^{2} a^{\frac {5}{4}}}{5}+\left (-\frac {24 x^{2} d}{5}+c \right ) a^{\frac {9}{4}}+\left (9 b^{2} c \,x^{2} a^{\frac {1}{4}}+\left (a d -\frac {9 b c}{4}\right ) \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (\ln \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )\right )\right ) x^{2}}{2 \left (b \,x^{2}+a \right )^{\frac {5}{4}} a^{\frac {13}{4}} x^{2}}\)	\(132\)

Input:

int((d*x^2+c)/x^3/(b*x^2+a)^(9/4),x,method=_RETURNVERBOSE)
 

Output:

-1/2*(54/5*(-10/27*x^2*d+c)*b*x^2*a^(5/4)+(-24/5*x^2*d+c)*a^(9/4)+(9*b^2*c 
*x^2*a^(1/4)+(a*d-9/4*b*c)*(b*x^2+a)^(5/4)*(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)/(b*x^2+ 
a)^(5/4)/a^(13/4)/x^2
                                                                                    
                                                                                    
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 879, normalized size of antiderivative = 6.02 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx =\text {Too large to display} \] Input:

integrate((d*x^2+c)/x^3/(b*x^2+a)^(9/4),x, algorithm="fricas")
 

Output:

-1/40*(5*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*((6561*b^4*c^4 - 11664*a*b^ 
3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 2304*a^3*b*c*d^3 + 256*a^4*d^4)/a^13)^(1/ 
4)*log(a^10*((6561*b^4*c^4 - 11664*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 23 
04*a^3*b*c*d^3 + 256*a^4*d^4)/a^13)^(3/4) - (729*b^3*c^3 - 972*a*b^2*c^2*d 
 + 432*a^2*b*c*d^2 - 64*a^3*d^3)*(b*x^2 + a)^(1/4)) + 5*(-I*a^3*b^2*x^6 - 
2*I*a^4*b*x^4 - I*a^5*x^2)*((6561*b^4*c^4 - 11664*a*b^3*c^3*d + 7776*a^2*b 
^2*c^2*d^2 - 2304*a^3*b*c*d^3 + 256*a^4*d^4)/a^13)^(1/4)*log(I*a^10*((6561 
*b^4*c^4 - 11664*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 2304*a^3*b*c*d^3 + 2 
56*a^4*d^4)/a^13)^(3/4) - (729*b^3*c^3 - 972*a*b^2*c^2*d + 432*a^2*b*c*d^2 
 - 64*a^3*d^3)*(b*x^2 + a)^(1/4)) + 5*(I*a^3*b^2*x^6 + 2*I*a^4*b*x^4 + I*a 
^5*x^2)*((6561*b^4*c^4 - 11664*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 2304*a 
^3*b*c*d^3 + 256*a^4*d^4)/a^13)^(1/4)*log(-I*a^10*((6561*b^4*c^4 - 11664*a 
*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 2304*a^3*b*c*d^3 + 256*a^4*d^4)/a^13)^ 
(3/4) - (729*b^3*c^3 - 972*a*b^2*c^2*d + 432*a^2*b*c*d^2 - 64*a^3*d^3)*(b* 
x^2 + a)^(1/4)) - 5*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*((6561*b^4*c^4 - 
 11664*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 2304*a^3*b*c*d^3 + 256*a^4*d^4 
)/a^13)^(1/4)*log(-a^10*((6561*b^4*c^4 - 11664*a*b^3*c^3*d + 7776*a^2*b^2* 
c^2*d^2 - 2304*a^3*b*c*d^3 + 256*a^4*d^4)/a^13)^(3/4) - (729*b^3*c^3 - 972 
*a*b^2*c^2*d + 432*a^2*b*c*d^2 - 64*a^3*d^3)*(b*x^2 + a)^(1/4)) + 4*(5*(9* 
b^2*c - 4*a*b*d)*x^4 + 5*a^2*c + 6*(9*a*b*c - 4*a^2*d)*x^2)*(b*x^2 + a)...
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 22.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.58 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx=- \frac {c \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {9}{4}} x^{\frac {13}{2}} \Gamma \left (\frac {17}{4}\right )} - \frac {d \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {9}{4}} x^{\frac {9}{2}} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((d*x**2+c)/x**3/(b*x**2+a)**(9/4),x)
 

Output:

-c*gamma(13/4)*hyper((9/4, 13/4), (17/4,), a*exp_polar(I*pi)/(b*x**2))/(2* 
b**(9/4)*x**(13/2)*gamma(17/4)) - d*gamma(9/4)*hyper((9/4, 9/4), (13/4,), 
a*exp_polar(I*pi)/(b*x**2))/(2*b**(9/4)*x**(9/2)*gamma(13/4))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.49 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx=-\frac {1}{40} \, c {\left (\frac {4 \, {\left (45 \, {\left (b x^{2} + a\right )}^{2} b - 36 \, {\left (b x^{2} + a\right )} a b - 4 \, a^{2} b\right )}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} a^{3} - {\left (b x^{2} + a\right )}^{\frac {5}{4}} a^{4}} + \frac {45 \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{a^{3}}\right )} + \frac {1}{10} \, d {\left (\frac {5 \, {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{a^{2}} + \frac {4 \, {\left (5 \, b x^{2} + 6 \, a\right )}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} a^{2}}\right )} \] Input:

integrate((d*x^2+c)/x^3/(b*x^2+a)^(9/4),x, algorithm="maxima")
 

Output:

-1/40*c*(4*(45*(b*x^2 + a)^2*b - 36*(b*x^2 + a)*a*b - 4*a^2*b)/((b*x^2 + a 
)^(9/4)*a^3 - (b*x^2 + a)^(5/4)*a^4) + 45*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^3) + 1/10*d*(5*(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^2 + 4*(5*b*x^2 + 6*a)/((b*x^2 + a)^(5/4)*a^2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (120) = 240\).

Time = 0.13 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.97 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx=-\frac {\sqrt {2} {\left (9 \, b c - 4 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{8 \, \left (-a\right )^{\frac {1}{4}} a^{3}} - \frac {\sqrt {2} {\left (9 \, b c - 4 \, a d\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{8 \, \left (-a\right )^{\frac {1}{4}} a^{3}} + \frac {\sqrt {2} {\left (9 \, b c - 4 \, a d\right )} \log \left (\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{16 \, \left (-a\right )^{\frac {1}{4}} a^{3}} - \frac {\sqrt {2} {\left (9 \, b c - 4 \, a d\right )} \log \left (-\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{16 \, \left (-a\right )^{\frac {1}{4}} a^{3}} - \frac {2 \, {\left (10 \, {\left (b x^{2} + a\right )} b c + a b c - 5 \, {\left (b x^{2} + a\right )} a d - a^{2} d\right )}}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} a^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} c}{2 \, a^{3} x^{2}} \] Input:

integrate((d*x^2+c)/x^3/(b*x^2+a)^(9/4),x, algorithm="giac")
 

Output:

-1/8*sqrt(2)*(9*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^3) - 1/8*sqrt(2)*(9*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^3) + 1/16*sqrt(2)*(9*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^3) - 1/16*sq 
rt(2)*(9*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^3) - 2/5*(10*(b*x^2 + a)*b*c + a*b*c - 5 
*(b*x^2 + a)*a*d - a^2*d)/((b*x^2 + a)^(5/4)*a^3) - 1/2*(b*x^2 + a)^(3/4)* 
c/(a^3*x^2)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx=\frac {\frac {2\,d}{5\,a}+\frac {2\,d\,\left (b\,x^2+a\right )}{a^2}}{{\left (b\,x^2+a\right )}^{5/4}}-\frac {\frac {2\,b\,c}{5\,a}+\frac {18\,b\,c\,\left (b\,x^2+a\right )}{5\,a^2}-\frac {9\,b\,c\,{\left (b\,x^2+a\right )}^2}{2\,a^3}}{a\,{\left (b\,x^2+a\right )}^{5/4}-{\left (b\,x^2+a\right )}^{9/4}}+\frac {d\,\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{a^{9/4}}-\frac {d\,\mathrm {atanh}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{a^{9/4}}-\frac {9\,b\,c\,\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{4\,a^{13/4}}+\frac {9\,b\,c\,\mathrm {atanh}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )}{4\,a^{13/4}} \] Input:

int((c + d*x^2)/(x^3*(a + b*x^2)^(9/4)),x)
 

Output:

((2*d)/(5*a) + (2*d*(a + b*x^2))/a^2)/(a + b*x^2)^(5/4) - ((2*b*c)/(5*a) + 
 (18*b*c*(a + b*x^2))/(5*a^2) - (9*b*c*(a + b*x^2)^2)/(2*a^3))/(a*(a + b*x 
^2)^(5/4) - (a + b*x^2)^(9/4)) + (d*atan((a + b*x^2)^(1/4)/a^(1/4)))/a^(9/ 
4) - (d*atanh((a + b*x^2)^(1/4)/a^(1/4)))/a^(9/4) - (9*b*c*atan((a + b*x^2 
)^(1/4)/a^(1/4)))/(4*a^(13/4)) + (9*b*c*atanh((a + b*x^2)^(1/4)/a^(1/4)))/ 
(4*a^(13/4))
 

Reduce [F]

\[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^{9/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} x^{3}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{5}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{7}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} x +2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{3}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{5}}d x \right ) d \] Input:

int((d*x^2+c)/x^3/(b*x^2+a)^(9/4),x)
 

Output:

int(1/((a + b*x**2)**(1/4)*a**2*x**3 + 2*(a + b*x**2)**(1/4)*a*b*x**5 + (a 
 + b*x**2)**(1/4)*b**2*x**7),x)*c + int(1/((a + b*x**2)**(1/4)*a**2*x + 2* 
(a + b*x**2)**(1/4)*a*b*x**3 + (a + b*x**2)**(1/4)*b**2*x**5),x)*d