3.425 \(\int \frac {(c x)^m}{(a x^j+b x^n)^{5/2}} \, dx\)

Optimal. Leaf size=111 \[ \frac {2 x^{1-2 n} (c x)^m \sqrt {\frac {a x^{j-n}}{b}+1} \, _2F_1\left (\frac {5}{2},\frac {m-\frac {5 n}{2}+1}{j-n};\frac {m-\frac {5 n}{2}+1}{j-n}+1;-\frac {a x^{j-n}}{b}\right )}{b^2 (2 m-5 n+2) \sqrt {a x^j+b x^n}} \]

[Out]

2*x^(1-2*n)*(c*x)^m*hypergeom([5/2, (1+m-5/2*n)/(j-n)],[1+(1+m-5/2*n)/(j-n)],-a*x^(j-n)/b)*(1+a*x^(j-n)/b)^(1/
2)/b^2/(2+2*m-5*n)/(a*x^j+b*x^n)^(1/2)

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Rubi [A]  time = 0.11, 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 = {2032, 365, 364} \[ \frac {2 x^{1-2 n} (c x)^m \sqrt {\frac {a x^{j-n}}{b}+1} \, _2F_1\left (\frac {5}{2},\frac {m-\frac {5 n}{2}+1}{j-n};\frac {m-\frac {5 n}{2}+1}{j-n}+1;-\frac {a x^{j-n}}{b}\right )}{b^2 (2 m-5 n+2) \sqrt {a x^j+b x^n}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^m/(a*x^j + b*x^n)^(5/2),x]

[Out]

(2*x^(1 - 2*n)*(c*x)^m*Sqrt[1 + (a*x^(j - n))/b]*Hypergeometric2F1[5/2, (1 + m - (5*n)/2)/(j - n), 1 + (1 + m
- (5*n)/2)/(j - n), -((a*x^(j - n))/b)])/(b^2*(2 + 2*m - 5*n)*Sqrt[a*x^j + b*x^n])

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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 )^{5/2}} \, dx &=\frac {\left (x^{-m+\frac {n}{2}} (c x)^m \sqrt {b+a x^{j-n}}\right ) \int \frac {x^{m-\frac {5 n}{2}}}{\left (b+a x^{j-n}\right )^{5/2}} \, 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 {5 n}{2}}}{\left (1+\frac {a x^{j-n}}{b}\right )^{5/2}} \, dx}{b^2 \sqrt {a x^j+b x^n}}\\ &=\frac {2 x^{1-2 n} (c x)^m \sqrt {1+\frac {a x^{j-n}}{b}} \, _2F_1\left (\frac {5}{2},\frac {1+m-\frac {5 n}{2}}{j-n};1+\frac {1+m-\frac {5 n}{2}}{j-n};-\frac {a x^{j-n}}{b}\right )}{b^2 (2+2 m-5 n) \sqrt {a x^j+b x^n}}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 166, normalized size = 1.50 \[ \frac {2 x^{1-2 j} (c x)^m \left (-(2 j-2 m+3 n-2) \sqrt {\frac {a x^{j-n}}{b}+1} \, _2F_1\left (\frac {1}{2},\frac {-4 j+2 m-n+2}{2 j-2 n};\frac {-2 j+2 m-3 n+2}{2 j-2 n};-\frac {a x^{j-n}}{b}\right )-\frac {a (j-n) x^j}{a x^j+b x^n}+2 j-2 m+3 n-2\right )}{3 a^2 (j-n)^2 \sqrt {a x^j+b x^n}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^m/(a*x^j + b*x^n)^(5/2),x]

[Out]

(2*x^(1 - 2*j)*(c*x)^m*(-2 + 2*j - 2*m + 3*n - (a*(j - n)*x^j)/(a*x^j + b*x^n) - (-2 + 2*j - 2*m + 3*n)*Sqrt[1
+ (a*x^(j - n))/b]*Hypergeometric2F1[1/2, (2 - 4*j + 2*m - n)/(2*j - 2*n), (2 - 2*j + 2*m - 3*n)/(2*j - 2*n),
-((a*x^(j - n))/b)]))/(3*a^2*(j - n)^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)^m/(a*x^j+b*x^n)^(5/2),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 \[ \int \frac {\left (c x\right )^{m}}{{\left (a x^{j} + b x^{n}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^m/(a*x^j+b*x^n)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x)^m/(a*x^j + b*x^n)^(5/2), x)

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maple [F]  time = 0.81, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{m}}{\left (a \,x^{j}+b \,x^{n}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m/(a*x^j+b*x^n)^(5/2),x)

[Out]

int((c*x)^m/(a*x^j+b*x^n)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{m}}{{\left (a x^{j} + b x^{n}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^m/(a*x^j+b*x^n)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x)^m/(a*x^j + b*x^n)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x\right )}^m}{{\left (a\,x^j+b\,x^n\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m/(a*x^j + b*x^n)^(5/2),x)

[Out]

int((c*x)^m/(a*x^j + b*x^n)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{m}}{\left (a x^{j} + b x^{n}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**m/(a*x**j+b*x**n)**(5/2),x)

[Out]

Integral((c*x)**m/(a*x**j + b*x**n)**(5/2), x)

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