3.708 \(\int x^m (a+b x)^{3/2} \, dx\)

Optimal. Leaf size=48 \[ \frac{2 x^m (a+b x)^{5/2} \left (-\frac{b x}{a}\right )^{-m} \, _2F_1\left (\frac{5}{2},-m;\frac{7}{2};\frac{b x}{a}+1\right )}{5 b} \]

[Out]

(2*x^m*(a + b*x)^(5/2)*Hypergeometric2F1[5/2, -m, 7/2, 1 + (b*x)/a])/(5*b*(-((b*x)/a))^m)

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Rubi [A]  time = 0.0115617, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {67, 65} \[ \frac{2 x^m (a+b x)^{5/2} \left (-\frac{b x}{a}\right )^{-m} \, _2F_1\left (\frac{5}{2},-m;\frac{7}{2};\frac{b x}{a}+1\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x)^(3/2),x]

[Out]

(2*x^m*(a + b*x)^(5/2)*Hypergeometric2F1[5/2, -m, 7/2, 1 + (b*x)/a])/(5*b*(-((b*x)/a))^m)

Rule 67

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

Rule 65

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

Rubi steps

\begin{align*} \int x^m (a+b x)^{3/2} \, dx &=\left (x^m \left (-\frac{b x}{a}\right )^{-m}\right ) \int \left (-\frac{b x}{a}\right )^m (a+b x)^{3/2} \, dx\\ &=\frac{2 x^m \left (-\frac{b x}{a}\right )^{-m} (a+b x)^{5/2} \, _2F_1\left (\frac{5}{2},-m;\frac{7}{2};1+\frac{b x}{a}\right )}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0399808, size = 48, normalized size = 1. \[ \frac{2 x^m (a+b x)^{5/2} \left (-\frac{b x}{a}\right )^{-m} \, _2F_1\left (\frac{5}{2},-m;\frac{7}{2};\frac{b x}{a}+1\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x)^(3/2),x]

[Out]

(2*x^m*(a + b*x)^(5/2)*Hypergeometric2F1[5/2, -m, 7/2, 1 + (b*x)/a])/(5*b*(-((b*x)/a))^m)

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( bx+a \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{\frac{3}{2}} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^(3/2)*x^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x + a\right )}^{\frac{3}{2}} x^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral((b*x + a)^(3/2)*x^m, x)

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Sympy [C]  time = 5.22225, size = 37, normalized size = 0.77 \begin{align*} \frac{a^{\frac{3}{2}} x x^{m} \Gamma \left (m + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{b x e^{i \pi }}{a}} \right )}}{\Gamma \left (m + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*x*x**m*gamma(m + 1)*hyper((-3/2, m + 1), (m + 2,), b*x*exp_polar(I*pi)/a)/gamma(m + 2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{\frac{3}{2}} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^(3/2)*x^m, x)