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

Optimal. Leaf size=107 \[ \frac{2 f (b x)^{7/2} (c+d x)^{n+1}}{b d (2 n+9)}-\frac{2 (b x)^{7/2} (c+d x)^n \left (\frac{d x}{c}+1\right )^{-n} (7 c f-d e (2 n+9)) \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};-\frac{d x}{c}\right )}{7 b d (2 n+9)} \]

[Out]

(2*f*(b*x)^(7/2)*(c + d*x)^(1 + n))/(b*d*(9 + 2*n)) - (2*(7*c*f - d*e*(9 + 2*n))*(b*x)^(7/2)*(c + d*x)^n*Hyper
geometric2F1[7/2, -n, 9/2, -((d*x)/c)])/(7*b*d*(9 + 2*n)*(1 + (d*x)/c)^n)

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Rubi [A]  time = 0.0438067, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {80, 66, 64} \[ \frac{2 f (b x)^{7/2} (c+d x)^{n+1}}{b d (2 n+9)}-\frac{2 (b x)^{7/2} (c+d x)^n \left (\frac{d x}{c}+1\right )^{-n} (7 c f-d e (2 n+9)) \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};-\frac{d x}{c}\right )}{7 b d (2 n+9)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*f*(b*x)^(7/2)*(c + d*x)^(1 + n))/(b*d*(9 + 2*n)) - (2*(7*c*f - d*e*(9 + 2*n))*(b*x)^(7/2)*(c + d*x)^n*Hyper
geometric2F1[7/2, -n, 9/2, -((d*x)/c)])/(7*b*d*(9 + 2*n)*(1 + (d*x)/c)^n)

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c^IntPart[n]*(c + d*x)^FracPart[n])/(1 + (d
*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-(d/(b*c)), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0]))
||  !RationalQ[n])

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx &=\frac{2 f (b x)^{7/2} (c+d x)^{1+n}}{b d (9+2 n)}+\frac{\left (-\frac{7}{2} b c f+b d e \left (\frac{9}{2}+n\right )\right ) \int (b x)^{5/2} (c+d x)^n \, dx}{b d \left (\frac{9}{2}+n\right )}\\ &=\frac{2 f (b x)^{7/2} (c+d x)^{1+n}}{b d (9+2 n)}+\frac{\left (\left (-\frac{7}{2} b c f+b d e \left (\frac{9}{2}+n\right )\right ) (c+d x)^n \left (1+\frac{d x}{c}\right )^{-n}\right ) \int (b x)^{5/2} \left (1+\frac{d x}{c}\right )^n \, dx}{b d \left (\frac{9}{2}+n\right )}\\ &=\frac{2 f (b x)^{7/2} (c+d x)^{1+n}}{b d (9+2 n)}-\frac{2 (7 c f-d e (9+2 n)) (b x)^{7/2} (c+d x)^n \left (1+\frac{d x}{c}\right )^{-n} \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};-\frac{d x}{c}\right )}{7 b d (9+2 n)}\\ \end{align*}

Mathematica [A]  time = 0.0494336, size = 91, normalized size = 0.85 \[ \frac{2 x (b x)^{5/2} (c+d x)^n \left (\frac{d x}{c}+1\right )^{-n} \left ((d e (2 n+9)-7 c f) \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};-\frac{d x}{c}\right )+7 f (c+d x) \left (\frac{d x}{c}+1\right )^n\right )}{7 d (2 n+9)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(b*x)^(5/2)*(c + d*x)^n*(7*f*(c + d*x)*(1 + (d*x)/c)^n + (-7*c*f + d*e*(9 + 2*n))*Hypergeometric2F1[7/2,
-n, 9/2, -((d*x)/c)]))/(7*d*(9 + 2*n)*(1 + (d*x)/c)^n)

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int \left ( bx \right ) ^{{\frac{5}{2}}} \left ( dx+c \right ) ^{n} \left ( fx+e \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{\frac{5}{2}}{\left (f x + e\right )}{\left (d x + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} f x^{3} + b^{2} e x^{2}\right )} \sqrt{b x}{\left (d x + c\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*f*x^3 + b^2*e*x^2)*sqrt(b*x)*(d*x + c)^n, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{\frac{5}{2}}{\left (f x + e\right )}{\left (d x + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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