3.7.59 \(\int \frac {(a+b (c x)^n)^{3/2}}{x} \, dx\) [659]

Optimal. Leaf size=70 \[ \frac {2 a \sqrt {a+b (c x)^n}}{n}+\frac {2 \left (a+b (c x)^n\right )^{3/2}}{3 n}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{n} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {374, 12, 272, 52, 65, 214} \begin {gather*} -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\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} \end {gather*}

Antiderivative was successfully verified.

[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)*ArcTanh[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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[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*} \int \frac {\left (a+b (c x)^n\right )^{3/2}}{x} \, dx &=\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} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{n}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 61, normalized size = 0.87 \begin {gather*} \frac {2 \sqrt {a+b (c x)^n} \left (4 a+b (c x)^n\right )-6 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{3 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c*x)^n)^(3/2)/x,x]

[Out]

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

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Maple [A]
time = 0.62, size = 54, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +b \left (c x \right )^{n}\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +b \left (c x \right )^{n}}-2 a^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{n}\)	\(54\)
default	\(\frac {\frac {2 \left (a +b \left (c x \right )^{n}\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +b \left (c x \right )^{n}}-2 a^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{n}\)	\(54\)
risch	\(\frac {2 \left (b \,{\mathrm e}^{n \ln \left (c x \right )}+4 a \right ) \sqrt {a +b \,{\mathrm e}^{n \ln \left (c x \right )}}}{3 n}-\frac {2 a^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {a +b \,{\mathrm e}^{n \ln \left (c x \right )}}}{\sqrt {a}}\right )}{n}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[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)*arctanh((a+b*(c*x)^n)^(1/2)/a^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*x)^n*b + a)^(3/2)/x, x)

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Fricas [A]
time = 0.40, size = 130, normalized size = 1.86 \begin {gather*} \left [\frac {3 \, a^{\frac {3}{2}} \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 \, {\left (\left (c x\right )^{n} b + 4 \, a\right )} \sqrt {\left (c x\right )^{n} b + a}}{3 \, n}, \frac {2 \, {\left (3 \, \sqrt {-a} a \arctan \left (\frac {\sqrt {\left (c x\right )^{n} b + a} \sqrt {-a}}{a}\right ) + {\left (\left (c x\right )^{n} b + 4 \, a\right )} \sqrt {\left (c x\right )^{n} b + a}\right )}}{3 \, n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 36.80, size = 102, normalized size = 1.46 \begin {gather*} \begin {cases} \frac {- a \left (- \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b \left (c x\right )^{n}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - 2 \sqrt {a + b \left (c x\right )^{n}}\right ) - b \left (\begin {cases} - \sqrt {a} \left (c x\right )^{n} & \text {for}\: b = 0 \\- \frac {2 \left (a + b \left (c x\right )^{n}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right )}{n} & \text {for}\: n \neq 0 \\\left (a \sqrt {a + b} + b \sqrt {a + b}\right ) \log {\left (x \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*x)^n*b + a)^(3/2)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,{\left (c\,x\right )}^n\right )}^{3/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c*x)^n)^(3/2)/x,x)

[Out]

int((a + b*(c*x)^n)^(3/2)/x, x)

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