\(\int \frac {(a+b x^2)^{5/4} (c+d x^2)}{x} \, dx\) [357]

Optimal result

Integrand size = 22, antiderivative size = 103 \[ \int \frac {\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{x} \, dx=2 a c \sqrt [4]{a+b x^2}+\frac {2}{5} c \left (a+b x^2\right )^{5/4}+\frac {2 d \left (a+b x^2\right )^{9/4}}{9 b}-a^{5/4} c \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )-a^{5/4} c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{x} \, dx=\frac {2 \sqrt [4]{a+b x^2} \left (54 a b c+5 a^2 d+9 b^2 c x^2+10 a b d x^2+5 b^2 d x^4\right )}{45 b}-a^{5/4} c \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )-a^{5/4} c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ) \] Input:

Integrate[((a + b*x^2)^(5/4)*(c + d*x^2))/x,x]
 

Output:

(2*(a + b*x^2)^(1/4)*(54*a*b*c + 5*a^2*d + 9*b^2*c*x^2 + 10*a*b*d*x^2 + 5* 
b^2*d*x^4))/(45*b) - a^(5/4)*c*ArcTan[(a + b*x^2)^(1/4)/a^(1/4)] - a^(5/4) 
*c*ArcTanh[(a + b*x^2)^(1/4)/a^(1/4)]
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {354, 90, 60, 60, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{2} \left (c \left (a \left (a \int \frac {1}{x^2 \left (b x^2+a\right )^{3/4}}dx^2+4 \sqrt [4]{a+b x^2}\right )+\frac {4}{5} \left (a+b x^2\right )^{5/4}\right )+\frac {4 d \left (a+b x^2\right )^{9/4}}{9 b}\right )\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} \left (c \left (a \left (\frac {4 a \left (-\frac {b \int \frac {1}{\sqrt {a}-x^4}d\sqrt [4]{b x^2+a}}{2 \sqrt {a}}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{b}+4 \sqrt [4]{a+b x^2}\right )+\frac {4}{5} \left (a+b x^2\right )^{5/4}\right )+\frac {4 d \left (a+b x^2\right )^{9/4}}{9 b}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (c \left (a \left (\frac {4 a \left (-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{b}+4 \sqrt [4]{a+b x^2}\right )+\frac {4}{5} \left (a+b x^2\right )^{5/4}\right )+\frac {4 d \left (a+b x^2\right )^{9/4}}{9 b}\right )\)

Input:

Int[((a + b*x^2)^(5/4)*(c + d*x^2))/x,x]
 

Output:

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

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 90

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

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 = 0.43 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(\frac {-9 c \left (\ln \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )\right ) b \,a^{\frac {5}{4}}+4 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (\left (d \,x^{4}+\frac {9}{5} c \,x^{2}\right ) b^{2}+\frac {54 a \left (\frac {5 x^{2} d}{27}+c \right ) b}{5}+d \,a^{2}\right )}{18 b}\)	\(108\)

Input:

int((b*x^2+a)^(5/4)*(d*x^2+c)/x,x,method=_RETURNVERBOSE)
 

Output:

1/18*(-9*c*(ln(((b*x^2+a)^(1/4)+a^(1/4))/((b*x^2+a)^(1/4)-a^(1/4)))+2*arct 
an((b*x^2+a)^(1/4)/a^(1/4)))*b*a^(5/4)+4*(b*x^2+a)^(1/4)*((d*x^4+9/5*c*x^2 
)*b^2+54/5*a*(5/27*x^2*d+c)*b+d*a^2))/b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{x} \, dx=-\frac {45 \, \left (a^{5} c^{4}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} a c + \left (a^{5} c^{4}\right )^{\frac {1}{4}}\right ) + 45 i \, \left (a^{5} c^{4}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} a c + i \, \left (a^{5} c^{4}\right )^{\frac {1}{4}}\right ) - 45 i \, \left (a^{5} c^{4}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} a c - i \, \left (a^{5} c^{4}\right )^{\frac {1}{4}}\right ) - 45 \, \left (a^{5} c^{4}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} a c - \left (a^{5} c^{4}\right )^{\frac {1}{4}}\right ) - 4 \, {\left (5 \, b^{2} d x^{4} + 54 \, a b c + 5 \, a^{2} d + {\left (9 \, b^{2} c + 10 \, a b d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}}}{90 \, b} \] Input:

integrate((b*x^2+a)^(5/4)*(d*x^2+c)/x,x, algorithm="fricas")
 

Output:

-1/90*(45*(a^5*c^4)^(1/4)*b*log((b*x^2 + a)^(1/4)*a*c + (a^5*c^4)^(1/4)) + 
 45*I*(a^5*c^4)^(1/4)*b*log((b*x^2 + a)^(1/4)*a*c + I*(a^5*c^4)^(1/4)) - 4 
5*I*(a^5*c^4)^(1/4)*b*log((b*x^2 + a)^(1/4)*a*c - I*(a^5*c^4)^(1/4)) - 45* 
(a^5*c^4)^(1/4)*b*log((b*x^2 + a)^(1/4)*a*c - (a^5*c^4)^(1/4)) - 4*(5*b^2* 
d*x^4 + 54*a*b*c + 5*a^2*d + (9*b^2*c + 10*a*b*d)*x^2)*(b*x^2 + a)^(1/4))/ 
b
 

Sympy [A] (verification not implemented)

Time = 20.75 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.37 \[ \int \frac {\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{x} \, dx=- \frac {8 a^{\frac {17}{4}} b d \sqrt [4]{1 + \frac {b x^{2}}{a}}}{45 a^{2} b^{2} + 45 a b^{3} x^{2}} + \frac {8 a^{\frac {17}{4}} b d}{45 a^{2} b^{2} + 45 a b^{3} x^{2}} - \frac {6 a^{\frac {13}{4}} b^{2} d x^{2} \sqrt [4]{1 + \frac {b x^{2}}{a}}}{45 a^{2} b^{2} + 45 a b^{3} x^{2}} + \frac {8 a^{\frac {13}{4}} b^{2} d x^{2}}{45 a^{2} b^{2} + 45 a b^{3} x^{2}} + \frac {12 a^{\frac {9}{4}} b^{3} d x^{4} \sqrt [4]{1 + \frac {b x^{2}}{a}}}{45 a^{2} b^{2} + 45 a b^{3} x^{2}} + \frac {10 a^{\frac {5}{4}} b^{4} d x^{6} \sqrt [4]{1 + \frac {b x^{2}}{a}}}{45 a^{2} b^{2} + 45 a b^{3} x^{2}} - \frac {a \sqrt [4]{b} c \sqrt {x} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (\frac {3}{4}\right )} + a d \left (\begin {cases} \frac {\sqrt [4]{a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {2 \left (a + b x^{2}\right )^{\frac {5}{4}}}{5 b} & \text {otherwise} \end {cases}\right ) + b c \left (\begin {cases} \frac {\sqrt [4]{a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {2 \left (a + b x^{2}\right )^{\frac {5}{4}}}{5 b} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((b*x**2+a)**(5/4)*(d*x**2+c)/x,x)
 

Output:

-8*a**(17/4)*b*d*(1 + b*x**2/a)**(1/4)/(45*a**2*b**2 + 45*a*b**3*x**2) + 8 
*a**(17/4)*b*d/(45*a**2*b**2 + 45*a*b**3*x**2) - 6*a**(13/4)*b**2*d*x**2*( 
1 + b*x**2/a)**(1/4)/(45*a**2*b**2 + 45*a*b**3*x**2) + 8*a**(13/4)*b**2*d* 
x**2/(45*a**2*b**2 + 45*a*b**3*x**2) + 12*a**(9/4)*b**3*d*x**4*(1 + b*x**2 
/a)**(1/4)/(45*a**2*b**2 + 45*a*b**3*x**2) + 10*a**(5/4)*b**4*d*x**6*(1 + 
b*x**2/a)**(1/4)/(45*a**2*b**2 + 45*a*b**3*x**2) - a*b**(1/4)*c*sqrt(x)*ga 
mma(-1/4)*hyper((-1/4, -1/4), (3/4,), a*exp_polar(I*pi)/(b*x**2))/(2*gamma 
(3/4)) + a*d*Piecewise((a**(1/4)*x**2/2, Eq(b, 0)), (2*(a + b*x**2)**(5/4) 
/(5*b), True)) + b*c*Piecewise((a**(1/4)*x**2/2, Eq(b, 0)), (2*(a + b*x**2 
)**(5/4)/(5*b), True))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{x} \, dx=\frac {2 \, {\left (b x^{2} + a\right )}^{\frac {9}{4}} d}{9 \, b} - \frac {1}{10} \, {\left (10 \, a^{\frac {5}{4}} \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) - 5 \, a^{\frac {5}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right ) - 4 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} - 20 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} a\right )} c \] Input:

integrate((b*x^2+a)^(5/4)*(d*x^2+c)/x,x, algorithm="maxima")
 

Output:

2/9*(b*x^2 + a)^(9/4)*d/b - 1/10*(10*a^(5/4)*arctan((b*x^2 + a)^(1/4)/a^(1 
/4)) - 5*a^(5/4)*log(((b*x^2 + a)^(1/4) - a^(1/4))/((b*x^2 + a)^(1/4) + a^ 
(1/4))) - 4*(b*x^2 + a)^(5/4) - 20*(b*x^2 + a)^(1/4)*a)*c
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (81) = 162\).

Time = 0.13 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.27 \[ \int \frac {\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{x} \, dx=-\frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a c \log \left (\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a c \log \left (-\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right ) + \frac {2 \, {\left (9 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{9} c + 45 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} a b^{9} c + 5 \, {\left (b x^{2} + a\right )}^{\frac {9}{4}} b^{8} d\right )}}{45 \, b^{9}} \] Input:

integrate((b*x^2+a)^(5/4)*(d*x^2+c)/x,x, algorithm="giac")
 

Output:

-1/2*sqrt(2)*(-a)^(1/4)*a*c*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b* 
x^2 + a)^(1/4))/(-a)^(1/4)) - 1/2*sqrt(2)*(-a)^(1/4)*a*c*arctan(-1/2*sqrt( 
2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^2 + a)^(1/4))/(-a)^(1/4)) - 1/4*sqrt(2)*(- 
a)^(1/4)*a*c*log(sqrt(2)*(b*x^2 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^2 + a) + 
sqrt(-a)) + 1/4*sqrt(2)*(-a)^(1/4)*a*c*log(-sqrt(2)*(b*x^2 + a)^(1/4)*(-a) 
^(1/4) + sqrt(b*x^2 + a) + sqrt(-a)) + 2/45*(9*(b*x^2 + a)^(5/4)*b^9*c + 4 
5*(b*x^2 + a)^(1/4)*a*b^9*c + 5*(b*x^2 + a)^(9/4)*b^8*d)/b^9
 

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{x} \, dx=\frac {2\,c\,{\left (b\,x^2+a\right )}^{5/4}}{5}+{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2\,a^2\,d}{9\,b}+\frac {4\,a\,d\,x^2}{9}+\frac {2\,b\,d\,x^4}{9}\right )-a^{5/4}\,c\,\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )+2\,a\,c\,{\left (b\,x^2+a\right )}^{1/4}+a^{5/4}\,c\,\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,1{}\mathrm {i} \] Input:

int(((a + b*x^2)^(5/4)*(c + d*x^2))/x,x)
 

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{x} \, dx=\frac {10 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} d +108 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b c +20 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b d \,x^{2}+18 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} c \,x^{2}+10 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} d \,x^{4}+45 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{3}+a x}d x \right ) a^{2} b c}{45 b} \] Input:

int((b*x^2+a)^(5/4)*(d*x^2+c)/x,x)
 

Output:

(10*(a + b*x**2)**(1/4)*a**2*d + 108*(a + b*x**2)**(1/4)*a*b*c + 20*(a + b 
*x**2)**(1/4)*a*b*d*x**2 + 18*(a + b*x**2)**(1/4)*b**2*c*x**2 + 10*(a + b* 
x**2)**(1/4)*b**2*d*x**4 + 45*int((a + b*x**2)**(1/4)/(a*x + b*x**3),x)*a* 
*2*b*c)/(45*b)