3.5.24 \(\int \frac {(c x)^m}{(a x^j+b x^n)^{3/2}} \, dx\) [424]

Optimal. Leaf size=111 \[ \frac {2 x^{1-n} (c x)^m \sqrt {1+\frac {a x^{j-n}}{b}} \, _2F_1\left (\frac {3}{2},\frac {1+m-\frac {3 n}{2}}{j-n};1+\frac {1+m-\frac {3 n}{2}}{j-n};-\frac {a x^{j-n}}{b}\right )}{b (2+2 m-3 n) \sqrt {a x^j+b x^n}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2057, 372, 371} \begin {gather*} \frac {2 x^{1-n} (c x)^m \sqrt {\frac {a x^{j-n}}{b}+1} \, _2F_1\left (\frac {3}{2},\frac {m-\frac {3 n}{2}+1}{j-n};\frac {m-\frac {3 n}{2}+1}{j-n}+1;-\frac {a x^{j-n}}{b}\right )}{b (2 m-3 n+2) \sqrt {a x^j+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^m/(a*x^j + b*x^n)^(3/2),x]

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 116, normalized size = 1.05 \begin {gather*} \frac {2 x^{1-j} (c x)^m \left (-1+\sqrt {1+\frac {a x^{j-n}}{b}} \, _2F_1\left (\frac {1}{2},\frac {2-2 j+2 m-n}{2 j-2 n};\frac {2+2 m-3 n}{2 j-2 n};-\frac {a x^{j-n}}{b}\right )\right )}{a (j-n) \sqrt {a x^j+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (c x \right )^{m}}{\left (a \,x^{j}+b \,x^{n}\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((c*x)^m/(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*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m/(a*x^j + b*x^n)^(3/2),x)

[Out]

int((c*x)^m/(a*x^j + b*x^n)^(3/2), x)

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