3.426 \(\int (a x^j+b x^n)^{3/2} \, dx\)

Optimal. Leaf size=97 \[ \frac {2 b x^{n+1} \sqrt {a x^j+b x^n} \, _2F_1\left (-\frac {3}{2},\frac {\frac {3 n}{2}+1}{j-n};\frac {2 j+n+2}{2 (j-n)};-\frac {a x^{j-n}}{b}\right )}{(3 n+2) \sqrt {\frac {a x^{j-n}}{b}+1}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2011, 365, 364} \[ \frac {2 b x^{n+1} \sqrt {a x^j+b x^n} \, _2F_1\left (-\frac {3}{2},\frac {\frac {3 n}{2}+1}{j-n};\frac {2 j+n+2}{2 (j-n)};-\frac {a x^{j-n}}{b}\right )}{(3 n+2) \sqrt {\frac {a x^{j-n}}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^j + b*x^n)^(3/2),x]

[Out]

(2*b*x^(1 + n)*Sqrt[a*x^j + b*x^n]*Hypergeometric2F1[-3/2, (1 + (3*n)/2)/(j - n), (2 + 2*j + n)/(2*(j - n)), -
((a*x^(j - n))/b)])/((2 + 3*n)*Sqrt[1 + (a*x^(j - n))/b])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rubi steps

\begin {align*} \int \left (a x^j+b x^n\right )^{3/2} \, dx &=\frac {\left (x^{-n/2} \sqrt {a x^j+b x^n}\right ) \int x^{3 n/2} \left (b+a x^{j-n}\right )^{3/2} \, dx}{\sqrt {b+a x^{j-n}}}\\ &=\frac {\left (b x^{-n/2} \sqrt {a x^j+b x^n}\right ) \int x^{3 n/2} \left (1+\frac {a x^{j-n}}{b}\right )^{3/2} \, dx}{\sqrt {1+\frac {a x^{j-n}}{b}}}\\ &=\frac {2 b x^{1+n} \sqrt {a x^j+b x^n} \, _2F_1\left (-\frac {3}{2},\frac {1+\frac {3 n}{2}}{j-n};\frac {2+2 j+n}{2 (j-n)};-\frac {a x^{j-n}}{b}\right )}{(2+3 n) \sqrt {1+\frac {a x^{j-n}}{b}}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 177, normalized size = 1.82 \[ \frac {2 x \left (3 a^2 (j-n)^2 x^{2 j} \sqrt {\frac {a x^{j-n}}{b}+1} \, _2F_1\left (\frac {1}{2},\frac {4 j-n+2}{2 j-2 n};\frac {6 j-3 n+2}{2 j-2 n};-\frac {a x^{j-n}}{b}\right )+(4 j-n+2) \left (a x^j+b x^n\right ) \left (a (-j+4 n+2) x^j+b (2 j+n+2) x^n\right )\right )}{(3 n+2) (4 j-n+2) (2 j+n+2) \sqrt {a x^j+b x^n}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^j + b*x^n)^(3/2),x]

[Out]

(2*x*((2 + 4*j - n)*(a*x^j + b*x^n)*(a*(2 - j + 4*n)*x^j + b*(2 + 2*j + n)*x^n) + 3*a^2*(j - n)^2*x^(2*j)*Sqrt
[1 + (a*x^(j - n))/b]*Hypergeometric2F1[1/2, (2 + 4*j - n)/(2*j - 2*n), (2 + 6*j - 3*n)/(2*j - 2*n), -((a*x^(j
- n))/b)]))/((2 + 4*j - n)*(2 + 2*j + n)*(2 + 3*n)*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((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 \[ \int {\left (a x^{j} + b x^{n}\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^j + b*x^n)^(3/2), x)

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maple [F]  time = 0.81, size = 0, normalized size = 0.00 \[ \int \left (a \,x^{j}+b \,x^{n}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^j + b*x^n)^(3/2), x)

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mupad [B]  time = 5.25, size = 82, normalized size = 0.85 \[ \frac {x\,{\left (a\,x^j+b\,x^n\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {\frac {3\,n}{2}+1}{j-n};\ \frac {\frac {3\,n}{2}+1}{j-n}+1;\ -\frac {a\,x^{j-n}}{b}\right )}{\left (\frac {3\,n}{2}+1\right )\,{\left (\frac {a\,x^{j-n}}{b}+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^j + b*x^n)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x**j + b*x**n)**(3/2), x)

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