\(\int \frac {(-b+a x^2)^{3/4} (3 b+2 a x^2)}{x} \, dx\) [2085]

Optimal result

Integrand size = 27, antiderivative size = 151 \[ \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx=\frac {2}{7} \left (-b+a x^2\right )^{3/4} \left (5 b+2 a x^2\right )+\frac {3 b^{7/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}{-\sqrt {b}+\sqrt {-b+a x^2}}\right )}{\sqrt {2}}+\frac {3 b^{7/4} \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^2}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^2}}\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.52, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {457, 81, 52, 65, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx=\frac {3 b^{7/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}\right )}{\sqrt {2}}-\frac {3 b^{7/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2}}-\frac {3 b^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{2 \sqrt {2}}+\frac {3 b^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{2 \sqrt {2}}+2 b \left (a x^2-b\right )^{3/4}+\frac {4}{7} \left (a x^2-b\right )^{7/4} \]

[In]

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

[Out]

2*b*(-b + a*x^2)^(3/4) + (4*(-b + a*x^2)^(7/4))/7 + (3*b^(7/4)*ArcTan[1 - (Sqrt[2]*(-b + a*x^2)^(1/4))/b^(1/4)
])/Sqrt[2] - (3*b^(7/4)*ArcTan[1 + (Sqrt[2]*(-b + a*x^2)^(1/4))/b^(1/4)])/Sqrt[2] - (3*b^(7/4)*Log[Sqrt[b] - S
qrt[2]*b^(1/4)*(-b + a*x^2)^(1/4) + Sqrt[-b + a*x^2]])/(2*Sqrt[2]) + (3*b^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*
(-b + a*x^2)^(1/4) + Sqrt[-b + a*x^2]])/(2*Sqrt[2])

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 210

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

Rule 303

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

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]]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx=\frac {2}{7} \left (-b+a x^2\right )^{3/4} \left (5 b+2 a x^2\right )-\frac {3 b^{7/4} \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^2}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}\right )}{\sqrt {2}}+\frac {3 b^{7/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}{\sqrt {b}+\sqrt {-b+a x^2}}\right )}{\sqrt {2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(-\frac {3 \sqrt {2}\, \left (\ln \left (\frac {\sqrt {a \,x^{2}-b}-b^{\frac {1}{4}} \left (a \,x^{2}-b \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {b}}{\sqrt {a \,x^{2}-b}+b^{\frac {1}{4}} \left (a \,x^{2}-b \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {b}}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{2}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{2}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )\right ) b^{\frac {7}{4}}}{4}+\frac {2 \left (a \,x^{2}-b \right )^{\frac {3}{4}} \left (2 a \,x^{2}+5 b \right )}{7}\)	\(158\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.11 \[ \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx=\frac {2}{7} \, {\left (2 \, a x^{2} + 5 \, b\right )} {\left (a x^{2} - b\right )}^{\frac {3}{4}} - \frac {3}{2} \, \left (-b^{7}\right )^{\frac {1}{4}} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{5} + 27 \, \left (-b^{7}\right )^{\frac {3}{4}}\right ) + \frac {3}{2} i \, \left (-b^{7}\right )^{\frac {1}{4}} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{5} + 27 i \, \left (-b^{7}\right )^{\frac {3}{4}}\right ) - \frac {3}{2} i \, \left (-b^{7}\right )^{\frac {1}{4}} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{5} - 27 i \, \left (-b^{7}\right )^{\frac {3}{4}}\right ) + \frac {3}{2} \, \left (-b^{7}\right )^{\frac {1}{4}} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{5} - 27 \, \left (-b^{7}\right )^{\frac {3}{4}}\right ) \]

[In]

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

[Out]

2/7*(2*a*x^2 + 5*b)*(a*x^2 - b)^(3/4) - 3/2*(-b^7)^(1/4)*log(27*(a*x^2 - b)^(1/4)*b^5 + 27*(-b^7)^(3/4)) + 3/2
*I*(-b^7)^(1/4)*log(27*(a*x^2 - b)^(1/4)*b^5 + 27*I*(-b^7)^(3/4)) - 3/2*I*(-b^7)^(1/4)*log(27*(a*x^2 - b)^(1/4
)*b^5 - 27*I*(-b^7)^(3/4)) + 3/2*(-b^7)^(1/4)*log(27*(a*x^2 - b)^(1/4)*b^5 - 27*(-b^7)^(3/4))

Sympy [A] (verification not implemented)

Time = 10.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.53 \[ \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx=- \frac {3 a^{\frac {3}{4}} b x^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{2}}} \right )}}{2 \Gamma \left (\frac {1}{4}\right )} + 2 a \left (\begin {cases} \frac {x^{2} \left (- b\right )^{\frac {3}{4}}}{2} & \text {for}\: a = 0 \\\frac {2 \left (a x^{2} - b\right )^{\frac {7}{4}}}{7 a} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

-3*a**(3/4)*b*x**(3/2)*gamma(-3/4)*hyper((-3/4, -3/4), (1/4,), b*exp_polar(2*I*pi)/(a*x**2))/(2*gamma(1/4)) +
2*a*Piecewise((x**2*(-b)**(3/4)/2, Eq(a, 0)), (2*(a*x**2 - b)**(7/4)/(7*a), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx=-\frac {1}{4} \, {\left (3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}}\right )} b - 8 \, {\left (a x^{2} - b\right )}^{\frac {3}{4}}\right )} b + \frac {4}{7} \, {\left (a x^{2} - b\right )}^{\frac {7}{4}} \]

[In]

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

[Out]

-1/4*(3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^2 - b)^(1/4))/b^(1/4))/b^(1/4) + 2*sqrt(2)*arc
tan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^2 - b)^(1/4))/b^(1/4))/b^(1/4) - sqrt(2)*log(sqrt(2)*(a*x^2 - b)^(1
/4)*b^(1/4) + sqrt(a*x^2 - b) + sqrt(b))/b^(1/4) + sqrt(2)*log(-sqrt(2)*(a*x^2 - b)^(1/4)*b^(1/4) + sqrt(a*x^2
- b) + sqrt(b))/b^(1/4))*b - 8*(a*x^2 - b)^(3/4))*b + 4/7*(a*x^2 - b)^(7/4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.25 \[ \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx=-\frac {3}{2} \, \sqrt {2} b^{\frac {7}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {3}{2} \, \sqrt {2} b^{\frac {7}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) + \frac {3}{4} \, \sqrt {2} b^{\frac {7}{4}} \log \left (\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right ) - \frac {3}{4} \, \sqrt {2} b^{\frac {7}{4}} \log \left (-\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right ) + \frac {4}{7} \, {\left (a x^{2} - b\right )}^{\frac {7}{4}} + 2 \, {\left (a x^{2} - b\right )}^{\frac {3}{4}} b \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.69 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.52 \[ \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx=\frac {4\,{\left (a\,x^2-b\right )}^{7/4}}{7}-3\,{\left (-b\right )}^{7/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^2-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )+3\,{\left (-b\right )}^{7/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^2-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )+2\,b\,{\left (a\,x^2-b\right )}^{3/4} \]

[In]

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

[Out]

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