Integrand size = 22, antiderivative size = 86 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx=2 c \sqrt [4]{a+b x^2}+\frac {2 d \left (a+b x^2\right )^{5/4}}{5 b}-\sqrt [4]{a} c \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )-\sqrt [4]{a} c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ) \] Output:
2*c*(b*x^2+a)^(1/4)+2/5*d*(b*x^2+a)^(5/4)/b-a^(1/4)*c*arctan((b*x^2+a)^(1/ 4)/a^(1/4))-a^(1/4)*c*arctanh((b*x^2+a)^(1/4)/a^(1/4))
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx=\frac {2 \sqrt [4]{a+b x^2} \left (5 b c+a d+b d x^2\right )}{5 b}-\sqrt [4]{a} c \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )-\sqrt [4]{a} c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ) \] Input:
Integrate[((a + b*x^2)^(1/4)*(c + d*x^2))/x,x]
Output:
(2*(a + b*x^2)^(1/4)*(5*b*c + a*d + b*d*x^2))/(5*b) - a^(1/4)*c*ArcTan[(a + b*x^2)^(1/4)/a^(1/4)] - a^(1/4)*c*ArcTanh[(a + b*x^2)^(1/4)/a^(1/4)]
Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.20, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {354, 90, 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.
\(\displaystyle \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt [4]{b x^2+a} \left (d x^2+c\right )}{x^2}dx^2\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} \left (c \int \frac {\sqrt [4]{b x^2+a}}{x^2}dx^2+\frac {4 d \left (a+b x^2\right )^{5/4}}{5 b}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (c \left (a \int \frac {1}{x^2 \left (b x^2+a\right )^{3/4}}dx^2+4 \sqrt [4]{a+b x^2}\right )+\frac {4 d \left (a+b x^2\right )^{5/4}}{5 b}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (c \left (\frac {4 a \int \frac {1}{\frac {x^8}{b}-\frac {a}{b}}d\sqrt [4]{b x^2+a}}{b}+4 \sqrt [4]{a+b x^2}\right )+\frac {4 d \left (a+b x^2\right )^{5/4}}{5 b}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{2} \left (c \left (\frac {4 a \left (-\frac {b \int \frac {1}{\sqrt {a}-x^4}d\sqrt [4]{b x^2+a}}{2 \sqrt {a}}-\frac {b \int \frac {1}{x^4+\sqrt {a}}d\sqrt [4]{b x^2+a}}{2 \sqrt {a}}\right )}{b}+4 \sqrt [4]{a+b x^2}\right )+\frac {4 d \left (a+b x^2\right )^{5/4}}{5 b}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (c \left (\frac {4 a \left (-\frac {b \int \frac {1}{\sqrt {a}-x^4}d\sqrt [4]{b x^2+a}}{2 \sqrt {a}}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{b}+4 \sqrt [4]{a+b x^2}\right )+\frac {4 d \left (a+b x^2\right )^{5/4}}{5 b}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (c \left (\frac {4 a \left (-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{b}+4 \sqrt [4]{a+b x^2}\right )+\frac {4 d \left (a+b x^2\right )^{5/4}}{5 b}\right )\) |
Input:
Int[((a + b*x^2)^(1/4)*(c + d*x^2))/x,x]
Output:
((4*d*(a + b*x^2)^(5/4))/(5*b) + c*(4*(a + b*x^2)^(1/4) + (4*a*(-1/2*(b*Ar cTan[(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
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]
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]
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]
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])
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])
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]
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]
Time = 1.00 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03
method | result | size |
pseudoelliptic | \(\frac {-5 c b \left (\ln \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )\right ) a^{\frac {1}{4}}+4 \left (\left (x^{2} d +5 c \right ) b +a d \right ) \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{10 b}\) | \(89\) |
Input:
int((b*x^2+a)^(1/4)*(d*x^2+c)/x,x,method=_RETURNVERBOSE)
Output:
1/10*(-5*c*b*(ln(((b*x^2+a)^(1/4)+a^(1/4))/((b*x^2+a)^(1/4)-a^(1/4)))+2*ar ctan((b*x^2+a)^(1/4)/a^(1/4)))*a^(1/4)+4*((d*x^2+5*c)*b+a*d)*(b*x^2+a)^(1/ 4))/b
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.83 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx=-\frac {5 \, \left (a c^{4}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} c + \left (a c^{4}\right )^{\frac {1}{4}}\right ) + 5 i \, \left (a c^{4}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} c + i \, \left (a c^{4}\right )^{\frac {1}{4}}\right ) - 5 i \, \left (a c^{4}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} c - i \, \left (a c^{4}\right )^{\frac {1}{4}}\right ) - 5 \, \left (a c^{4}\right )^{\frac {1}{4}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} c - \left (a c^{4}\right )^{\frac {1}{4}}\right ) - 4 \, {\left (b d x^{2} + 5 \, b c + a d\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}}}{10 \, b} \] Input:
integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x,x, algorithm="fricas")
Output:
-1/10*(5*(a*c^4)^(1/4)*b*log((b*x^2 + a)^(1/4)*c + (a*c^4)^(1/4)) + 5*I*(a *c^4)^(1/4)*b*log((b*x^2 + a)^(1/4)*c + I*(a*c^4)^(1/4)) - 5*I*(a*c^4)^(1/ 4)*b*log((b*x^2 + a)^(1/4)*c - I*(a*c^4)^(1/4)) - 5*(a*c^4)^(1/4)*b*log((b *x^2 + a)^(1/4)*c - (a*c^4)^(1/4)) - 4*(b*d*x^2 + 5*b*c + a*d)*(b*x^2 + a) ^(1/4))/b
Time = 10.37 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx=- \frac {\sqrt [4]{b} c \sqrt {x} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (\frac {3}{4}\right )} + d \left (\begin {cases} \frac {\sqrt [4]{a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {2 \left (a + b x^{2}\right )^{\frac {5}{4}}}{5 b} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x**2+a)**(1/4)*(d*x**2+c)/x,x)
Output:
-b**(1/4)*c*sqrt(x)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), a*exp_polar(I* pi)/(b*x**2))/(2*gamma(3/4)) + d*Piecewise((a**(1/4)*x**2/2, Eq(b, 0)), (2 *(a + b*x**2)**(5/4)/(5*b), True))
Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx=-\frac {1}{2} \, {\left (2 \, a^{\frac {1}{4}} \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) - a^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right ) - 4 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )} c + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} d}{5 \, b} \] Input:
integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x,x, algorithm="maxima")
Output:
-1/2*(2*a^(1/4)*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))) - 4*(b*x^2 + a)^(1/4))* c + 2/5*(b*x^2 + a)^(5/4)*d/b
Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (68) = 136\).
Time = 0.13 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.48 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx=-\frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} c \log \left (\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} c \log \left (-\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right ) + \frac {2 \, {\left (5 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} b^{5} c + {\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{4} d\right )}}{5 \, b^{5}} \] Input:
integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x,x, algorithm="giac")
Output:
-1/2*sqrt(2)*(-a)^(1/4)*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)*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)*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)*c*log(-sqrt(2)*(b*x^2 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^2 + a) + sqrt(-a)) + 2/5*(5*(b*x^2 + a)^(1/4)*b^5*c + (b*x^2 + a )^(5/4)*b^4*d)/b^5
Time = 0.70 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx=2\,c\,{\left (b\,x^2+a\right )}^{1/4}+\frac {2\,d\,{\left (b\,x^2+a\right )}^{5/4}}{5\,b}-a^{1/4}\,c\,\mathrm {atan}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right )-a^{1/4}\,c\,\mathrm {atanh}\left (\frac {{\left (b\,x^2+a\right )}^{1/4}}{a^{1/4}}\right ) \] Input:
int(((a + b*x^2)^(1/4)*(c + d*x^2))/x,x)
Output:
2*c*(a + b*x^2)^(1/4) + (2*d*(a + b*x^2)^(5/4))/(5*b) - a^(1/4)*c*atan((a + b*x^2)^(1/4)/a^(1/4)) - a^(1/4)*c*atanh((a + b*x^2)^(1/4)/a^(1/4))
\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x} \, dx=\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d +10 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c +2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{2}+5 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{3}+a x}d x \right ) a b c}{5 b} \] Input:
int((b*x^2+a)^(1/4)*(d*x^2+c)/x,x)
Output:
(2*(a + b*x**2)**(1/4)*a*d + 10*(a + b*x**2)**(1/4)*b*c + 2*(a + b*x**2)** (1/4)*b*d*x**2 + 5*int((a + b*x**2)**(1/4)/(a*x + b*x**3),x)*a*b*c)/(5*b)