3.4.68 \(\int \frac {\sqrt {a x+b x^n}}{(c x)^{3/2}} \, dx\) [368]

Optimal. Leaf size=87 \[ -\frac {2 \sqrt {a x+b x^n}}{c (1-n) \sqrt {c x}}+\frac {2 \sqrt {a} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^n}}\right )}{c (1-n) \sqrt {c x}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2053, 2056, 2054, 212} \begin {gather*} \frac {2 \sqrt {a} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^n}}\right )}{c (1-n) \sqrt {c x}}-\frac {2 \sqrt {a x+b x^n}}{c (1-n) \sqrt {c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x + b*x^n]/(c*x)^(3/2),x]

[Out]

(-2*Sqrt[a*x + b*x^n])/(c*(1 - n)*Sqrt[c*x]) + (2*Sqrt[a]*Sqrt[x]*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[a*x + b*x^n]]
)/(c*(1 - n)*Sqrt[c*x])

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 2053

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*p*(n - j))), x] + Dist[a/c^j, Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c,
j, m, n}, x] && IGtQ[p + 1/2, 0] && NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rule 2056

Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[c^IntPart[m]*((c*x)^FracPar
t[m]/x^FracPart[m]), Int[x^m*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2]
&& NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0]

Rubi steps

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

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Mathematica [A]
time = 0.36, size = 100, normalized size = 1.15 \begin {gather*} \frac {x \left (2 a x+2 b x^n-2 \sqrt {a} \sqrt {b} x^{\frac {1+n}{2}} \sqrt {1+\frac {a x^{1-n}}{b}} \sinh ^{-1}\left (\frac {\sqrt {a} x^{\frac {1}{2}-\frac {n}{2}}}{\sqrt {b}}\right )\right )}{(-1+n) (c x)^{3/2} \sqrt {a x+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2*a*x + 2*b*x^n - 2*Sqrt[a]*Sqrt[b]*x^((1 + n)/2)*Sqrt[1 + (a*x^(1 - n))/b]*ArcSinh[(Sqrt[a]*x^(1/2 - n/2)
)/Sqrt[b]]))/((-1 + n)*(c*x)^(3/2)*Sqrt[a*x + b*x^n])

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {a x +b \,x^{n}}}{\left (c x \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+b*x^n)^(1/2)/(c*x)^(3/2),x)

[Out]

int((a*x+b*x^n)^(1/2)/(c*x)^(3/2),x)

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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*x+b*x^n)^(1/2)/(c*x)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+b*x^n)^(1/2)/(c*x)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a x + b x^{n}}}{\left (c x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+b*x**n)**(1/2)/(c*x)**(3/2),x)

[Out]

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

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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*x+b*x^n)^(1/2)/(c*x)^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^n + a*x)^(1/2)/(c*x)^(3/2),x)

[Out]

int((b*x^n + a*x)^(1/2)/(c*x)^(3/2), x)

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