3.951 \(\int \frac {(b x)^{5/2} (c+d x)^n}{(e+f x)^2} \, dx\)

Optimal. Leaf size=61 \[ \frac {2 (b x)^{7/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} F_1\left (\frac {7}{2};-n,2;\frac {9}{2};-\frac {d x}{c},-\frac {f x}{e}\right )}{7 b e^2} \]

[Out]

2/7*(b*x)^(7/2)*(d*x+c)^n*AppellF1(7/2,-n,2,9/2,-d*x/c,-f*x/e)/b/e^2/((1+d*x/c)^n)

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Rubi [A]  time = 0.08, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {130, 511, 510} \[ \frac {2 (b x)^{7/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} F_1\left (\frac {7}{2};-n,2;\frac {9}{2};-\frac {d x}{c},-\frac {f x}{e}\right )}{7 b e^2} \]

Antiderivative was successfully verified.

[In]

Int[((b*x)^(5/2)*(c + d*x)^n)/(e + f*x)^2,x]

[Out]

(2*(b*x)^(7/2)*(c + d*x)^n*AppellF1[7/2, -n, 2, 9/2, -((d*x)/c), -((f*x)/e)])/(7*b*e^2*(1 + (d*x)/c)^n)

Rule 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {(b x)^{5/2} (c+d x)^n}{(e+f x)^2} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {x^6 \left (c+\frac {d x^2}{b}\right )^n}{\left (e+\frac {f x^2}{b}\right )^2} \, dx,x,\sqrt {b x}\right )}{b}\\ &=\frac {\left (2 (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n}\right ) \operatorname {Subst}\left (\int \frac {x^6 \left (1+\frac {d x^2}{b c}\right )^n}{\left (e+\frac {f x^2}{b}\right )^2} \, dx,x,\sqrt {b x}\right )}{b}\\ &=\frac {2 (b x)^{7/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} F_1\left (\frac {7}{2};-n,2;\frac {9}{2};-\frac {d x}{c},-\frac {f x}{e}\right )}{7 b e^2}\\ \end {align*}

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Mathematica [B]  time = 0.14, size = 133, normalized size = 2.18 \[ \frac {2 b^2 \sqrt {b x} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (9 e F_1\left (\frac {1}{2};-n,1;\frac {3}{2};-\frac {d x}{c},-\frac {f x}{e}\right )-3 e F_1\left (\frac {1}{2};-n,2;\frac {3}{2};-\frac {d x}{c},-\frac {f x}{e}\right )-6 e \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};-\frac {d x}{c}\right )+f x \, _2F_1\left (\frac {3}{2},-n;\frac {5}{2};-\frac {d x}{c}\right )\right )}{3 f^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((b*x)^(5/2)*(c + d*x)^n)/(e + f*x)^2,x]

[Out]

(2*b^2*Sqrt[b*x]*(c + d*x)^n*(9*e*AppellF1[1/2, -n, 1, 3/2, -((d*x)/c), -((f*x)/e)] - 3*e*AppellF1[1/2, -n, 2,
3/2, -((d*x)/c), -((f*x)/e)] - 6*e*Hypergeometric2F1[1/2, -n, 3/2, -((d*x)/c)] + f*x*Hypergeometric2F1[3/2, -
n, 5/2, -((d*x)/c)]))/(3*f^3*(1 + (d*x)/c)^n)

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x} {\left (d x + c\right )}^{n} b^{2} x^{2}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x, algorithm="fricas")

[Out]

integral(sqrt(b*x)*(d*x + c)^n*b^2*x^2/(f^2*x^2 + 2*e*f*x + e^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((b*x)^(5/2)*(d*x + c)^n/(f*x + e)^2, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x \right )^{\frac {5}{2}} \left (d x +c \right )^{n}}{\left (f x +e \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x)

[Out]

int((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate((b*x)^(5/2)*(d*x + c)^n/(f*x + e)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^n}{{\left (e+f\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x)^(5/2)*(c + d*x)^n)/(e + f*x)^2,x)

[Out]

int(((b*x)^(5/2)*(c + d*x)^n)/(e + f*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)**(5/2)*(d*x+c)**n/(f*x+e)**2,x)

[Out]

Timed out

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