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

Optimal. Leaf size=141 \[ \frac {2 a^{3/2} x^{3 j/2} (c x)^{-3 j/2} \tanh ^{-1}\left (\frac {\sqrt {a} x^{j/2}}{\sqrt {a x^j+b x^n}}\right )}{c (j-n)}-\frac {2 (c x)^{-3 j/2} \left (a x^j+b x^n\right )^{3/2}}{3 c (j-n)}-\frac {2 a x^j (c x)^{-3 j/2} \sqrt {a x^j+b x^n}}{c (j-n)} \]

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Rubi [A]  time = 0.23, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2031, 2028, 2029, 206} \begin {gather*} \frac {2 a^{3/2} 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 )}{c (j-n)}-\frac {2 (c x)^{-3 j/2} \left (a x^j+b x^n\right )^{3/2}}{3 c (j-n)}-\frac {2 a x^j (c x)^{-3 j/2} \sqrt {a x^j+b x^n}}{c (j-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*x^j*Sqrt[a*x^j + b*x^n])/(c*(j - n)*(c*x)^((3*j)/2)) - (2*(a*x^j + b*x^n)^(3/2))/(3*c*(j - n)*(c*x)^((3*
j)/2)) + (2*a^(3/2)*x^((3*j)/2)*ArcTanh[(Sqrt[a]*x^(j/2))/Sqrt[a*x^j + b*x^n]])/(c*(j - n)*(c*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 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 (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 x^{-1-\frac {3 j}{2}} \left (a x^j+b x^n\right )^{3/2} \, dx}{c}\\ &=-\frac {2 (c x)^{-3 j/2} \left (a x^j+b x^n\right )^{3/2}}{3 c (j-n)}+\frac {\left (a x^{3 j/2} (c x)^{-3 j/2}\right ) \int x^{-1-\frac {j}{2}} \sqrt {a x^j+b x^n} \, dx}{c}\\ &=-\frac {2 a x^j (c x)^{-3 j/2} \sqrt {a x^j+b x^n}}{c (j-n)}-\frac {2 (c x)^{-3 j/2} \left (a x^j+b x^n\right )^{3/2}}{3 c (j-n)}+\frac {\left (a^2 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}{c}\\ &=-\frac {2 a x^j (c x)^{-3 j/2} \sqrt {a x^j+b x^n}}{c (j-n)}-\frac {2 (c x)^{-3 j/2} \left (a x^j+b x^n\right )^{3/2}}{3 c (j-n)}+\frac {\left (2 a^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 )}{c (j-n)}\\ &=-\frac {2 a x^j (c x)^{-3 j/2} \sqrt {a x^j+b x^n}}{c (j-n)}-\frac {2 (c x)^{-3 j/2} \left (a x^j+b x^n\right )^{3/2}}{3 c (j-n)}+\frac {2 a^{3/2} 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 )}{c (j-n)}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 131, normalized size = 0.93 \begin {gather*} -\frac {2 (c x)^{-3 j/2} \left (-3 a^{3/2} \sqrt {b} x^{\frac {1}{2} (3 j+n)} \sqrt {\frac {a x^{j-n}}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} x^{\frac {j-n}{2}}}{\sqrt {b}}\right )+4 a^2 x^{2 j}+5 a b x^{j+n}+b^2 x^{2 n}\right )}{3 c (j-n) \sqrt {a x^j+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(4*a^2*x^(2*j) + b^2*x^(2*n) + 5*a*b*x^(j + n) - 3*a^(3/2)*Sqrt[b]*x^((3*j + n)/2)*Sqrt[1 + (a*x^(j - n))/
b]*ArcSinh[(Sqrt[a]*x^((j - n)/2))/Sqrt[b]]))/(3*c*(j - n)*(c*x)^((3*j)/2)*Sqrt[a*x^j + b*x^n])

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IntegrateAlgebraic [F]  time = 0.14, size = 0, normalized size = 0.00 \begin {gather*} \int (c x)^{-1-\frac {3 j}{2}} \left (a x^j+b x^n\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[IntegrateAlgebraic][(c*x)^(-1 - (3*j)/2)*(a*x^j + 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)^(-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 (ha
s polynomial part)

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

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

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maple [F]  time = 0.87, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \,x^{j}+b \,x^{n}\right )^{\frac {3}{2}} \left (c x \right )^{-\frac {3 j}{2}-1}\, dx \end {gather*}

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 \begin {gather*} \int {\left (a x^{j} + b x^{n}\right )}^{\frac {3}{2}} \left (c x\right )^{-\frac {3}{2} \, j - 1}\,{d x} \end {gather*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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