3.4.99 \(\int \frac {(c x)^{7/2}}{(a x^3+b x^n)^{3/2}} \, dx\) [399]

Optimal. Leaf size=94 \[ -\frac {2 c^2 (c x)^{3/2}}{a (3-n) \sqrt {a x^3+b x^n}}+\frac {2 c^3 \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{a^{3/2} (3-n) \sqrt {x}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 94, 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 = {2055, 2056, 2054, 212} \begin {gather*} \frac {2 c^3 \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{a^{3/2} (3-n) \sqrt {x}}-\frac {2 c^2 (c x)^{3/2}}{a (3-n) \sqrt {a x^3+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^(7/2)/(a*x^3 + b*x^n)^(3/2),x]

[Out]

(-2*c^2*(c*x)^(3/2))/(a*(3 - n)*Sqrt[a*x^3 + b*x^n]) + (2*c^3*Sqrt[c*x]*ArcTanh[(Sqrt[a]*x^(3/2))/Sqrt[a*x^3 +
b*x^n]])/(a^(3/2)*(3 - n)*Sqrt[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 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 2055

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

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 {(c x)^{7/2}}{\left (a x^3+b x^n\right )^{3/2}} \, dx &=-\frac {2 c^2 (c x)^{3/2}}{a (3-n) \sqrt {a x^3+b x^n}}+\frac {c^3 \int \frac {\sqrt {c x}}{\sqrt {a x^3+b x^n}} \, dx}{a}\\ &=-\frac {2 c^2 (c x)^{3/2}}{a (3-n) \sqrt {a x^3+b x^n}}+\frac {\left (c^3 \sqrt {c x}\right ) \int \frac {\sqrt {x}}{\sqrt {a x^3+b x^n}} \, dx}{a \sqrt {x}}\\ &=-\frac {2 c^2 (c x)^{3/2}}{a (3-n) \sqrt {a x^3+b x^n}}+\frac {\left (2 c^3 \sqrt {c x}\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{a (3-n) \sqrt {x}}\\ &=-\frac {2 c^2 (c x)^{3/2}}{a (3-n) \sqrt {a x^3+b x^n}}+\frac {2 c^3 \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{a^{3/2} (3-n) \sqrt {x}}\\ \end {align*}

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Mathematica [A]
time = 1.64, size = 109, normalized size = 1.16 \begin {gather*} \frac {2 c^3 \sqrt {c x} \left (\sqrt {a} x^{3/2}-\sqrt {b} x^{n/2} \sqrt {1+\frac {a x^{3-n}}{b}} \sinh ^{-1}\left (\frac {\sqrt {a} x^{\frac {3}{2}-\frac {n}{2}}}{\sqrt {b}}\right )\right )}{a^{3/2} (-3+n) \sqrt {x} \sqrt {a x^3+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(7/2)/(a*x^3 + b*x^n)^(3/2),x]

[Out]

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

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (c x \right )^{\frac {7}{2}}}{\left (a \,x^{3}+b \,x^{n}\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)/(a*x^3+b*x^n)^(3/2),x)

[Out]

int((c*x)^(7/2)/(a*x^3+b*x^n)^(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((c*x)^(7/2)/(a*x^3+b*x^n)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x)^(7/2)/(a*x^3 + b*x^n)^(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((c*x)^(7/2)/(a*x^3+b*x^n)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(7/2)/(a*x**3+b*x**n)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3878 deep

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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((c*x)^(7/2)/(a*x^3+b*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x)^(7/2)/(a*x^3 + b*x^n)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x\right )}^{7/2}}{{\left (b\,x^n+a\,x^3\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)/(b*x^n + a*x^3)^(3/2),x)

[Out]

int((c*x)^(7/2)/(b*x^n + a*x^3)^(3/2), x)

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