\(\int \frac {(-3 b+2 a x^2) (b^2+a^2 x^2)^{3/4}}{x} \, dx\) [1369]

Optimal result

Integrand size = 29, antiderivative size = 99 \[ \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx=\frac {2 \left (b^2+a^2 x^2\right )^{3/4} \left (-7 a b+2 b^2+2 a^2 x^2\right )}{7 a}-3 b^{5/2} \arctan \left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )+3 b^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right ) \]

[Out]

2/7*(a^2*x^2+b^2)^(3/4)*(2*a^2*x^2-7*a*b+2*b^2)/a-3*b^(5/2)*arctan((a^2*x^2+b^2)^(1/4)/b^(1/2))+3*b^(5/2)*arct
anh((a^2*x^2+b^2)^(1/4)/b^(1/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {457, 81, 52, 65, 304, 209, 212} \[ \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx=-3 b^{5/2} \arctan \left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )+3 b^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )-2 b \left (a^2 x^2+b^2\right )^{3/4}+\frac {4 \left (a^2 x^2+b^2\right )^{7/4}}{7 a} \]

[In]

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

[Out]

-2*b*(b^2 + a^2*x^2)^(3/4) + (4*(b^2 + a^2*x^2)^(7/4))/(7*a) - 3*b^(5/2)*ArcTan[(b^2 + a^2*x^2)^(1/4)/Sqrt[b]]
+ 3*b^(5/2)*ArcTanh[(b^2 + a^2*x^2)^(1/4)/Sqrt[b]]

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(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 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

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

Rule 457

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(-3 b+2 a x) \left (b^2+a^2 x\right )^{3/4}}{x} \, dx,x,x^2\right ) \\ & = \frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}-\frac {1}{2} (3 b) \text {Subst}\left (\int \frac {\left (b^2+a^2 x\right )^{3/4}}{x} \, dx,x,x^2\right ) \\ & = -2 b \left (b^2+a^2 x^2\right )^{3/4}+\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}-\frac {1}{2} \left (3 b^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt [4]{b^2+a^2 x}} \, dx,x,x^2\right ) \\ & = -2 b \left (b^2+a^2 x^2\right )^{3/4}+\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}-\frac {\left (6 b^3\right ) \text {Subst}\left (\int \frac {x^2}{-\frac {b^2}{a^2}+\frac {x^4}{a^2}} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )}{a^2} \\ & = -2 b \left (b^2+a^2 x^2\right )^{3/4}+\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}+\left (3 b^3\right ) \text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )-\left (3 b^3\right ) \text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right ) \\ & = -2 b \left (b^2+a^2 x^2\right )^{3/4}+\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}-3 b^{5/2} \arctan \left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )+3 b^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx=\frac {2 \left (b^2+a^2 x^2\right )^{3/4} \left (-7 a b+2 b^2+2 a^2 x^2\right )}{7 a}-3 b^{5/2} \arctan \left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )+3 b^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right ) \]

[In]

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

[Out]

(2*(b^2 + a^2*x^2)^(3/4)*(-7*a*b + 2*b^2 + 2*a^2*x^2))/(7*a) - 3*b^(5/2)*ArcTan[(b^2 + a^2*x^2)^(1/4)/Sqrt[b]]
+ 3*b^(5/2)*ArcTanh[(b^2 + a^2*x^2)^(1/4)/Sqrt[b]]

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(\frac {21 \left (\operatorname {arctanh}\left (\frac {\left (a^{2} x^{2}+b^{2}\right )^{\frac {1}{4}}}{\sqrt {b}}\right )-\arctan \left (\frac {\left (a^{2} x^{2}+b^{2}\right )^{\frac {1}{4}}}{\sqrt {b}}\right )\right ) a \,b^{\frac {5}{2}}+2 \left (a^{2} x^{2}+b^{2}\right )^{\frac {3}{4}} \left (2 a^{2} x^{2}-7 a b +2 b^{2}\right )}{7 a}\)	\(85\)

[In]

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

[Out]

1/7*(21*(arctanh((a^2*x^2+b^2)^(1/4)/b^(1/2))-arctan((a^2*x^2+b^2)^(1/4)/b^(1/2)))*a*b^(5/2)+2*(a^2*x^2+b^2)^(
3/4)*(2*a^2*x^2-7*a*b+2*b^2))/a

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 299, normalized size of antiderivative = 3.02 \[ \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx=\left [-\frac {42 \, a b^{\frac {5}{2}} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right ) - 21 \, a b^{\frac {5}{2}} \log \left (\frac {a^{2} x^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} b^{\frac {3}{2}} + 2 \, \sqrt {a^{2} x^{2} + b^{2}} b + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} \sqrt {b}}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} x^{2} - 7 \, a b + 2 \, b^{2}\right )} {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}}}{14 \, a}, -\frac {42 \, a \sqrt {-b} b^{2} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} \sqrt {-b}}{b}\right ) - 21 \, a \sqrt {-b} b^{2} \log \left (\frac {a^{2} x^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} \sqrt {-b} b - 2 \, \sqrt {a^{2} x^{2} + b^{2}} b - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} \sqrt {-b}}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} x^{2} - 7 \, a b + 2 \, b^{2}\right )} {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}}}{14 \, a}\right ] \]

[In]

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

[Out]

[-1/14*(42*a*b^(5/2)*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b)) - 21*a*b^(5/2)*log((a^2*x^2 + 2*b^2 + 2*(a^2*x^2 +
b^2)^(1/4)*b^(3/2) + 2*sqrt(a^2*x^2 + b^2)*b + 2*(a^2*x^2 + b^2)^(3/4)*sqrt(b))/x^2) - 4*(2*a^2*x^2 - 7*a*b +
2*b^2)*(a^2*x^2 + b^2)^(3/4))/a, -1/14*(42*a*sqrt(-b)*b^2*arctan((a^2*x^2 + b^2)^(1/4)*sqrt(-b)/b) - 21*a*sqrt
(-b)*b^2*log((a^2*x^2 + 2*b^2 + 2*(a^2*x^2 + b^2)^(1/4)*sqrt(-b)*b - 2*sqrt(a^2*x^2 + b^2)*b - 2*(a^2*x^2 + b^
2)^(3/4)*sqrt(-b))/x^2) - 4*(2*a^2*x^2 - 7*a*b + 2*b^2)*(a^2*x^2 + b^2)^(3/4))/a]

Sympy [A] (verification not implemented)

Time = 8.42 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.39 \[ \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx=- a \left (\begin {cases} - x^{2} \left (b^{2}\right )^{\frac {3}{4}} & \text {for}\: a^{2} = 0 \\- \frac {4 \left (a^{2} x^{2} + b^{2}\right )^{\frac {7}{4}}}{7 a^{2}} & \text {otherwise} \end {cases}\right ) - \frac {3 b \left (\begin {cases} 2 b^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt [4]{a^{2} x^{2} + b^{2}}}{\sqrt {b}} \right )} + \frac {2 b^{2} \operatorname {atan}{\left (\frac {\sqrt [4]{a^{2} x^{2} + b^{2}}}{\sqrt {- b}} \right )}}{\sqrt {- b}} + \frac {4 \left (a^{2} x^{2} + b^{2}\right )^{\frac {3}{4}}}{3} & \text {for}\: a^{2} \neq 0 \\- \left (b^{2}\right )^{\frac {3}{4}} \log {\left (\frac {1}{x^{2}} \right )} & \text {otherwise} \end {cases}\right )}{2} \]

[In]

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

[Out]

-a*Piecewise((-x**2*(b**2)**(3/4), Eq(a**2, 0)), (-4*(a**2*x**2 + b**2)**(7/4)/(7*a**2), True)) - 3*b*Piecewis
e((2*b**(3/2)*atan((a**2*x**2 + b**2)**(1/4)/sqrt(b)) + 2*b**2*atan((a**2*x**2 + b**2)**(1/4)/sqrt(-b))/sqrt(-
b) + 4*(a**2*x**2 + b**2)**(3/4)/3, Ne(a**2, 0)), (-(b**2)**(3/4)*log(x**(-2)), True))/2

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx=-\frac {1}{2} \, {\left (6 \, b^{\frac {3}{2}} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right ) + 3 \, b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b} + {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}\right ) + 4 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}}\right )} b + \frac {4 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {7}{4}}}{7 \, a} \]

[In]

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

[Out]

-1/2*(6*b^(3/2)*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b)) + 3*b^(3/2)*log(-(sqrt(b) - (a^2*x^2 + b^2)^(1/4))/(sqrt
(b) + (a^2*x^2 + b^2)^(1/4))) + 4*(a^2*x^2 + b^2)^(3/4))*b + 4/7*(a^2*x^2 + b^2)^(7/4)/a

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98 \[ \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx=-\frac {3 \, b^{3} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {-b}}\right )}{\sqrt {-b}} - 3 \, b^{\frac {5}{2}} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right ) - \frac {2 \, {\left (7 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} a^{7} b - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {7}{4}} a^{6}\right )}}{7 \, a^{7}} \]

[In]

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

[Out]

-3*b^3*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(-b))/sqrt(-b) - 3*b^(5/2)*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b)) - 2/7
*(7*(a^2*x^2 + b^2)^(3/4)*a^7*b - 2*(a^2*x^2 + b^2)^(7/4)*a^6)/a^7

Mupad [B] (verification not implemented)

Time = 6.39 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.82 \[ \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx=3\,b^{5/2}\,\mathrm {atanh}\left (\frac {{\left (a^2\,x^2+b^2\right )}^{1/4}}{\sqrt {b}}\right )-3\,b^{5/2}\,\mathrm {atan}\left (\frac {{\left (a^2\,x^2+b^2\right )}^{1/4}}{\sqrt {b}}\right )-2\,b\,{\left (a^2\,x^2+b^2\right )}^{3/4}+\frac {4\,{\left (a^2\,x^2+b^2\right )}^{7/4}}{7\,a} \]

[In]

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

[Out]

3*b^(5/2)*atanh((b^2 + a^2*x^2)^(1/4)/b^(1/2)) - 3*b^(5/2)*atan((b^2 + a^2*x^2)^(1/4)/b^(1/2)) - 2*b*(b^2 + a^
2*x^2)^(3/4) + (4*(b^2 + a^2*x^2)^(7/4))/(7*a)