3.3.69 \(\int \frac {\sqrt {a x^3+b x^n}}{(c x)^{5/2}} \, dx\)

Optimal. Leaf size=91 \[ \frac {2 \sqrt {a} \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{c^3 (3-n) \sqrt {x}}-\frac {2 \sqrt {a x^3+b x^n}}{c (3-n) (c x)^{3/2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2028, 2031, 2029, 206} \begin {gather*} \frac {2 \sqrt {a} \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{c^3 (3-n) \sqrt {x}}-\frac {2 \sqrt {a x^3+b x^n}}{c (3-n) (c x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^3 + b*x^n]/(c*x)^(5/2),x]

[Out]

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

Rule 206

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

Rule 2028

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 2029

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 2031

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^3+b x^n}}{(c x)^{5/2}} \, dx &=-\frac {2 \sqrt {a x^3+b x^n}}{c (3-n) (c x)^{3/2}}+\frac {a \int \frac {\sqrt {c x}}{\sqrt {a x^3+b x^n}} \, dx}{c^3}\\ &=-\frac {2 \sqrt {a x^3+b x^n}}{c (3-n) (c x)^{3/2}}+\frac {\left (a \sqrt {c x}\right ) \int \frac {\sqrt {x}}{\sqrt {a x^3+b x^n}} \, dx}{c^3 \sqrt {x}}\\ &=-\frac {2 \sqrt {a x^3+b x^n}}{c (3-n) (c x)^{3/2}}+\frac {\left (2 a \sqrt {c x}\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{c^3 (3-n) \sqrt {x}}\\ &=-\frac {2 \sqrt {a x^3+b x^n}}{c (3-n) (c x)^{3/2}}+\frac {2 \sqrt {a} \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{c^3 (3-n) \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 103, normalized size = 1.13 \begin {gather*} \frac {2 x \left (-\sqrt {a} \sqrt {b} x^{\frac {n+3}{2}} \sqrt {\frac {a x^{3-n}}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} x^{\frac {3}{2}-\frac {n}{2}}}{\sqrt {b}}\right )+a x^3+b x^n\right )}{(n-3) (c x)^{5/2} \sqrt {a x^3+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^3 + b*x^n]/(c*x)^(5/2),x]

[Out]

(2*x*(a*x^3 + b*x^n - Sqrt[a]*Sqrt[b]*x^((3 + n)/2)*Sqrt[1 + (a*x^(3 - n))/b]*ArcSinh[(Sqrt[a]*x^(3/2 - n/2))/
Sqrt[b]]))/((-3 + n)*(c*x)^(5/2)*Sqrt[a*x^3 + b*x^n])

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IntegrateAlgebraic [F]  time = 3.30, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a x^3+b x^n}}{(c x)^{5/2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[Sqrt[a*x^3 + b*x^n]/(c*x)^(5/2),x]

[Out]

Defer[IntegrateAlgebraic][Sqrt[a*x^3 + b*x^n]/(c*x)^(5/2), 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^3+b*x^n)^(1/2)/(c*x)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a x^{3} + b x^{n}}}{\left (c x\right )^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^3+b*x^n)^(1/2)/(c*x)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*x^3 + b*x^n)/(c*x)^(5/2), x)

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maple [F]  time = 0.76, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a \,x^{3}+b \,x^{n}}}{\left (c x \right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^3+b*x^n)^(1/2)/(c*x)^(5/2),x)

[Out]

int((a*x^3+b*x^n)^(1/2)/(c*x)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a x^{3} + b x^{n}}}{\left (c x\right )^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^3+b*x^n)^(1/2)/(c*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*x^3 + b*x^n)/(c*x)^(5/2), 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^3}}{{\left (c\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^n + a*x^3)^(1/2)/(c*x)^(5/2),x)

[Out]

int((b*x^n + a*x^3)^(1/2)/(c*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a x^{3} + b x^{n}}}{\left (c x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**3+b*x**n)**(1/2)/(c*x)**(5/2),x)

[Out]

Integral(sqrt(a*x**3 + b*x**n)/(c*x)**(5/2), x)

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