3.398 \(\int \frac {(c x)^{-1+\frac {3 j}{2}}}{(a x^j+b x^n)^{3/2}} \, dx\)

Optimal. Leaf size=107 \[ \frac {2 x^{-3 j/2} (c x)^{3 j/2} \tanh ^{-1}\left (\frac {\sqrt {a} x^{j/2}}{\sqrt {a x^j+b x^n}}\right )}{a^{3/2} c (j-n)}-\frac {2 x^{-j} (c x)^{3 j/2}}{a c (j-n) \sqrt {a x^j+b x^n}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2031, 2030, 2029, 206} \[ \frac {2 x^{-3 j/2} (c x)^{3 j/2} \tanh ^{-1}\left (\frac {\sqrt {a} x^{j/2}}{\sqrt {a x^j+b x^n}}\right )}{a^{3/2} c (j-n)}-\frac {2 x^{-j} (c x)^{3 j/2}}{a c (j-n) \sqrt {a x^j+b x^n}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(-1 + (3*j)/2)/(a*x^j + b*x^n)^(3/2),x]

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 117, normalized size = 1.09 \[ -\frac {2 x^{-3 j/2} (c x)^{3 j/2} \left (\sqrt {a} x^{j/2}-\sqrt {b} x^{n/2} \sqrt {\frac {a x^{j-n}}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} x^{\frac {j-n}{2}}}{\sqrt {b}}\right )\right )}{a^{3/2} c (j-n) \sqrt {a x^j+b x^n}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(-1 + (3*j)/2)/(a*x^j + b*x^n)^(3/2),x]

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-1+3/2*j)/(a*x^j+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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {3}{2} \, j - 1}}{{\left (a x^{j} + b x^{n}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.83, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{\frac {3 j}{2}-1}}{\left (a \,x^{j}+b \,x^{n}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {3}{2} \, j - 1}}{{\left (a x^{j} + b x^{n}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {3 j}{2} - 1}}{\left (a x^{j} + b x^{n}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*x)**(3*j/2 - 1)/(a*x**j + b*x**n)**(3/2), x)

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