\(\int \frac {(-a+b (c x)^n)^{3/2}}{x} \, dx\) [665]

Optimal result

Integrand size = 19, antiderivative size = 76 \[ \int \frac {\left (-a+b (c x)^n\right )^{3/2}}{x} \, dx=-\frac {2 a \sqrt {-a+b (c x)^n}}{n}+\frac {2 \left (-a+b (c x)^n\right )^{3/2}}{3 n}+\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {-a+b (c x)^n}}{\sqrt {a}}\right )}{n} \]

[Out]

2/3*(-a+b*(c*x)^n)^(3/2)/n+2*a^(3/2)*arctan((-a+b*(c*x)^n)^(1/2)/a^(1/2))/n-2*a*(-a+b*(c*x)^n)^(1/2)/n

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {374, 12, 272, 52, 65, 211} \[ \int \frac {\left (-a+b (c x)^n\right )^{3/2}}{x} \, dx=\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {b (c x)^n-a}}{\sqrt {a}}\right )}{n}-\frac {2 a \sqrt {b (c x)^n-a}}{n}+\frac {2 \left (b (c x)^n-a\right )^{3/2}}{3 n} \]

[In]

Int[(-a + b*(c*x)^n)^(3/2)/x,x]

[Out]

(-2*a*Sqrt[-a + b*(c*x)^n])/n + (2*(-a + b*(c*x)^n)^(3/2))/(3*n) + (2*a^(3/2)*ArcTan[Sqrt[-a + b*(c*x)^n]/Sqrt
[a]])/n

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 211

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

Rule 272

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

Rule 374

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {c \left (-a+b x^n\right )^{3/2}}{x} \, dx,x,c x\right )}{c} \\ & = \text {Subst}\left (\int \frac {\left (-a+b x^n\right )^{3/2}}{x} \, dx,x,c x\right ) \\ & = \frac {\text {Subst}\left (\int \frac {(-a+b x)^{3/2}}{x} \, dx,x,(c x)^n\right )}{n} \\ & = \frac {2 \left (-a+b (c x)^n\right )^{3/2}}{3 n}-\frac {a \text {Subst}\left (\int \frac {\sqrt {-a+b x}}{x} \, dx,x,(c x)^n\right )}{n} \\ & = -\frac {2 a \sqrt {-a+b (c x)^n}}{n}+\frac {2 \left (-a+b (c x)^n\right )^{3/2}}{3 n}+\frac {a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {-a+b x}} \, dx,x,(c x)^n\right )}{n} \\ & = -\frac {2 a \sqrt {-a+b (c x)^n}}{n}+\frac {2 \left (-a+b (c x)^n\right )^{3/2}}{3 n}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b (c x)^n}\right )}{b n} \\ & = -\frac {2 a \sqrt {-a+b (c x)^n}}{n}+\frac {2 \left (-a+b (c x)^n\right )^{3/2}}{3 n}+\frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {-a+b (c x)^n}}{\sqrt {a}}\right )}{n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-a+b (c x)^n\right )^{3/2}}{x} \, dx=\frac {-2 \left (4 a-b (c x)^n\right ) \sqrt {-a+b (c x)^n}+6 a^{3/2} \arctan \left (\frac {\sqrt {-a+b (c x)^n}}{\sqrt {a}}\right )}{3 n} \]

[In]

Integrate[(-a + b*(c*x)^n)^(3/2)/x,x]

[Out]

(-2*(4*a - b*(c*x)^n)*Sqrt[-a + b*(c*x)^n] + 6*a^(3/2)*ArcTan[Sqrt[-a + b*(c*x)^n]/Sqrt[a]])/(3*n)

Maple [A] (verified)

Time = 1.41 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {2 \left (-a +b \left (c x \right )^{n}\right )^{\frac {3}{2}}}{3}-2 a \sqrt {-a +b \left (c x \right )^{n}}+2 a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{n}\)	\(60\)
default	\(\frac {\frac {2 \left (-a +b \left (c x \right )^{n}\right )^{\frac {3}{2}}}{3}-2 a \sqrt {-a +b \left (c x \right )^{n}}+2 a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{n}\)	\(60\)
risch	\(\frac {2 \left (-b \,{\mathrm e}^{n \ln \left (c x \right )}+4 a \right ) \left (a -b \,{\mathrm e}^{n \ln \left (c x \right )}\right )}{3 n \sqrt {-a +b \,{\mathrm e}^{n \ln \left (c x \right )}}}+\frac {2 a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-a +b \,{\mathrm e}^{n \ln \left (c x \right )}}}{\sqrt {a}}\right )}{n}\)	\(76\)

[In]

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

[Out]

1/n*(2/3*(-a+b*(c*x)^n)^(3/2)-2*a*(-a+b*(c*x)^n)^(1/2)+2*a^(3/2)*arctan((-a+b*(c*x)^n)^(1/2)/a^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.78 \[ \int \frac {\left (-a+b (c x)^n\right )^{3/2}}{x} \, dx=\left [\frac {3 \, \sqrt {-a} a \log \left (\frac {\left (c x\right )^{n} b + 2 \, \sqrt {\left (c x\right )^{n} b - a} \sqrt {-a} - 2 \, a}{\left (c x\right )^{n}}\right ) + 2 \, \sqrt {\left (c x\right )^{n} b - a} {\left (\left (c x\right )^{n} b - 4 \, a\right )}}{3 \, n}, \frac {2 \, {\left (3 \, a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\left (c x\right )^{n} b - a}}{\sqrt {a}}\right ) + \sqrt {\left (c x\right )^{n} b - a} {\left (\left (c x\right )^{n} b - 4 \, a\right )}\right )}}{3 \, n}\right ] \]

[In]

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

[Out]

[1/3*(3*sqrt(-a)*a*log(((c*x)^n*b + 2*sqrt((c*x)^n*b - a)*sqrt(-a) - 2*a)/(c*x)^n) + 2*sqrt((c*x)^n*b - a)*((c
*x)^n*b - 4*a))/n, 2/3*(3*a^(3/2)*arctan(sqrt((c*x)^n*b - a)/sqrt(a)) + sqrt((c*x)^n*b - a)*((c*x)^n*b - 4*a))
/n]

Sympy [A] (verification not implemented)

Time = 11.99 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.24 \[ \int \frac {\left (-a+b (c x)^n\right )^{3/2}}{x} \, dx=\begin {cases} \frac {\begin {cases} 2 a^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt {- a + b \left (c x\right )^{n}}}{\sqrt {a}} \right )} - 2 a \sqrt {- a + b \left (c x\right )^{n}} + \frac {2 \left (- a + b \left (c x\right )^{n}\right )^{\frac {3}{2}}}{3} & \text {for}\: b \neq 0 \\- a \sqrt {- a} \log {\left (\left (c x\right )^{n} \right )} & \text {otherwise} \end {cases}}{n} & \text {for}\: n \neq 0 \\\left (- a \sqrt {- a + b} + b \sqrt {- a + b}\right ) \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((Piecewise((2*a**(3/2)*atan(sqrt(-a + b*(c*x)**n)/sqrt(a)) - 2*a*sqrt(-a + b*(c*x)**n) + 2*(-a + b*(
c*x)**n)**(3/2)/3, Ne(b, 0)), (-a*sqrt(-a)*log((c*x)**n), True))/n, Ne(n, 0)), ((-a*sqrt(-a + b) + b*sqrt(-a +
b))*log(x), True))

Maxima [F]

\[ \int \frac {\left (-a+b (c x)^n\right )^{3/2}}{x} \, dx=\int { \frac {{\left (\left (c x\right )^{n} b - a\right )}^{\frac {3}{2}}}{x} \,d x } \]

[In]

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

[Out]

integrate(((c*x)^n*b - a)^(3/2)/x, x)

Giac [F]

\[ \int \frac {\left (-a+b (c x)^n\right )^{3/2}}{x} \, dx=\int { \frac {{\left (\left (c x\right )^{n} b - a\right )}^{\frac {3}{2}}}{x} \,d x } \]

[In]

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

[Out]

integrate(((c*x)^n*b - a)^(3/2)/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-a+b (c x)^n\right )^{3/2}}{x} \, dx=\int \frac {{\left (b\,{\left (c\,x\right )}^n-a\right )}^{3/2}}{x} \,d x \]

[In]

int((b*(c*x)^n - a)^(3/2)/x,x)

[Out]

int((b*(c*x)^n - a)^(3/2)/x, x)