3.4.74 \(\int \frac {(a x^3+b x^n)^{3/2}}{(c x)^{11/2}} \, dx\) [374]

Optimal. Leaf size=128 \[ -\frac {2 a \sqrt {a x^3+b x^n}}{c^4 (3-n) (c x)^{3/2}}-\frac {2 \left (a x^3+b x^n\right )^{3/2}}{3 c (3-n) (c x)^{9/2}}+\frac {2 a^{3/2} \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{c^6 (3-n) \sqrt {x}} \]

[Out]

-2/3*(a*x^3+b*x^n)^(3/2)/c/(3-n)/(c*x)^(9/2)+2*a^(3/2)*arctanh(x^(3/2)*a^(1/2)/(a*x^3+b*x^n)^(1/2))*(c*x)^(1/2
)/c^6/(3-n)/x^(1/2)-2*a*(a*x^3+b*x^n)^(1/2)/c^4/(3-n)/(c*x)^(3/2)

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Rubi [A]
time = 0.15, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2053, 2056, 2054, 212} \begin {gather*} \frac {2 a^{3/2} \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{c^6 (3-n) \sqrt {x}}-\frac {2 a \sqrt {a x^3+b x^n}}{c^4 (3-n) (c x)^{3/2}}-\frac {2 \left (a x^3+b x^n\right )^{3/2}}{3 c (3-n) (c x)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x^3 + b*x^n)^(3/2)/(c*x)^(11/2),x]

[Out]

(-2*a*Sqrt[a*x^3 + b*x^n])/(c^4*(3 - n)*(c*x)^(3/2)) - (2*(a*x^3 + b*x^n)^(3/2))/(3*c*(3 - n)*(c*x)^(9/2)) + (
2*a^(3/2)*Sqrt[c*x]*ArcTanh[(Sqrt[a]*x^(3/2))/Sqrt[a*x^3 + b*x^n]])/(c^6*(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 2053

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 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 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 {\left (a x^3+b x^n\right )^{3/2}}{(c x)^{11/2}} \, dx &=-\frac {2 \left (a x^3+b x^n\right )^{3/2}}{3 c (3-n) (c x)^{9/2}}+\frac {a \int \frac {\sqrt {a x^3+b x^n}}{(c x)^{5/2}} \, dx}{c^3}\\ &=-\frac {2 a \sqrt {a x^3+b x^n}}{c^4 (3-n) (c x)^{3/2}}-\frac {2 \left (a x^3+b x^n\right )^{3/2}}{3 c (3-n) (c x)^{9/2}}+\frac {a^2 \int \frac {\sqrt {c x}}{\sqrt {a x^3+b x^n}} \, dx}{c^6}\\ &=-\frac {2 a \sqrt {a x^3+b x^n}}{c^4 (3-n) (c x)^{3/2}}-\frac {2 \left (a x^3+b x^n\right )^{3/2}}{3 c (3-n) (c x)^{9/2}}+\frac {\left (a^2 \sqrt {c x}\right ) \int \frac {\sqrt {x}}{\sqrt {a x^3+b x^n}} \, dx}{c^6 \sqrt {x}}\\ &=-\frac {2 a \sqrt {a x^3+b x^n}}{c^4 (3-n) (c x)^{3/2}}-\frac {2 \left (a x^3+b x^n\right )^{3/2}}{3 c (3-n) (c x)^{9/2}}+\frac {\left (2 a^2 \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 )}{c^6 (3-n) \sqrt {x}}\\ &=-\frac {2 a \sqrt {a x^3+b x^n}}{c^4 (3-n) (c x)^{3/2}}-\frac {2 \left (a x^3+b x^n\right )^{3/2}}{3 c (3-n) (c x)^{9/2}}+\frac {2 a^{3/2} \sqrt {c x} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b x^n}}\right )}{c^6 (3-n) \sqrt {x}}\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 126, normalized size = 0.98 \begin {gather*} \frac {2 \sqrt {c x} \left (4 a^2 x^6+b^2 x^{2 n}+5 a b x^{3+n}-3 a^{3/2} \sqrt {b} x^{\frac {9+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 )}{3 c^6 (-3+n) x^5 \sqrt {a x^3+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x^3 + b*x^n)^(3/2)/(c*x)^(11/2),x]

[Out]

(2*Sqrt[c*x]*(4*a^2*x^6 + b^2*x^(2*n) + 5*a*b*x^(3 + n) - 3*a^(3/2)*Sqrt[b]*x^((9 + n)/2)*Sqrt[1 + (a*x^(3 - n
))/b]*ArcSinh[(Sqrt[a]*x^(3/2 - n/2))/Sqrt[b]]))/(3*c^6*(-3 + n)*x^5*Sqrt[a*x^3 + b*x^n])

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{3}+b \,x^{n}\right )^{\frac {3}{2}}}{\left (c x \right )^{\frac {11}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^3+b*x^n)^(3/2)/(c*x)^(11/2),x)

[Out]

int((a*x^3+b*x^n)^(3/2)/(c*x)^(11/2),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((a*x^3+b*x^n)^(3/2)/(c*x)^(11/2),x, algorithm="maxima")

[Out]

integrate((a*x^3 + b*x^n)^(3/2)/(c*x)^(11/2), 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)^(3/2)/(c*x)^(11/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
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((a*x**3+b*x**n)**(3/2)/(c*x)**(11/2),x)

[Out]

Timed out

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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((a*x^3+b*x^n)^(3/2)/(c*x)^(11/2),x, algorithm="giac")

[Out]

integrate((a*x^3 + b*x^n)^(3/2)/(c*x)^(11/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^n+a\,x^3\right )}^{3/2}}{{\left (c\,x\right )}^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^n + a*x^3)^(3/2)/(c*x)^(11/2),x)

[Out]

int((b*x^n + a*x^3)^(3/2)/(c*x)^(11/2), x)

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