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

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

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.159572, antiderivative size = 94, normalized size of antiderivative = 1., 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 = {2030, 2031, 2029, 206} \[ \frac{2 c^3 \sqrt{c x} \tanh ^{-1}\left (\frac{\sqrt{a} x^{3/2}}{\sqrt{a x^3+b x^n}}\right )}{a^{3/2} (3-n) \sqrt{x}}-\frac{2 c^2 (c x)^{3/2}}{a (3-n) \sqrt{a x^3+b x^n}} \]

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 2030

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 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]

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 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])

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 ) \operatorname{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*}

Mathematica [A]  time = 0.187117, size = 109, normalized size = 1.16 \[ \frac{2 c^3 \sqrt{c x} \left (\sqrt{a} x^{3/2}-\sqrt{b} x^{n/2} \sqrt{\frac{a x^{3-n}}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{a} x^{\frac{3}{2}-\frac{n}{2}}}{\sqrt{b}}\right )\right )}{a^{3/2} (n-3) \sqrt{x} \sqrt{a x^3+b x^n}} \]

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])

________________________________________________________________________________________

Maple [F]  time = 0.322, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{{\frac{7}{2}}} \left ( a{x}^{3}+b{x}^{n} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{7}{2}}}{{\left (a x^{3} + b x^{n}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)

________________________________________________________________________________________

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

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: UnboundLocalError

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{7}{2}}}{{\left (a x^{3} + b x^{n}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)