3.985 \(\int \frac{(d x)^m (a+b x)^n}{(c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=68 \[ -\frac{d^2 x (d x)^{m-2} (a+b x)^n \left (\frac{b x}{a}+1\right )^{-n} \, _2F_1\left (m-2,-n;m-1;-\frac{b x}{a}\right )}{c (2-m) \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0315716, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {15, 16, 66, 64} \[ -\frac{d^2 x (d x)^{m-2} (a+b x)^n \left (\frac{b x}{a}+1\right )^{-n} \, _2F_1\left (m-2,-n;m-1;-\frac{b x}{a}\right )}{c (2-m) \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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

Mathematica [A]  time = 0.015123, size = 57, normalized size = 0.84 \[ \frac{x (d x)^m (a+b x)^n \left (\frac{b x}{a}+1\right )^{-n} \, _2F_1\left (m-2,-n;m-1;-\frac{b x}{a}\right )}{(m-2) \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.034, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx \right ) ^{m} \left ( bx+a \right ) ^{n} \left ( c{x}^{2} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2}}{\left (b x + a\right )}^{n} \left (d x\right )^{m}}{c^{2} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2)*(b*x + a)^n*(d*x)^m/(c^2*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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