3.876 \(\int (\frac{a+b+c x^2}{d})^m \, dx\)

Optimal. Leaf size=49 \[ \frac{d x \left (\frac{a+b}{d}+\frac{c x^2}{d}\right )^{m+1} \, _2F_1\left (1,m+\frac{3}{2};\frac{3}{2};-\frac{c x^2}{a+b}\right )}{a+b} \]

[Out]

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

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Rubi [A]  time = 0.0167575, antiderivative size = 57, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {1972, 246, 245} \[ x \left (\frac{c x^2}{a+b}+1\right )^{-m} \left (\frac{a+b}{d}+\frac{c x^2}{d}\right )^m \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};-\frac{c x^2}{a+b}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((a + b + c*x^2)/d)^m,x]

[Out]

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

Rule 1972

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && BinomialQ[u, x] &&  !BinomialMatchQ[
u, x]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.0120852, size = 53, normalized size = 1.08 \[ x \left (\frac{c x^2}{a+b}+1\right )^{-m} \left (\frac{a+b+c x^2}{d}\right )^m \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};-\frac{c x^2}{a+b}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b + c*x^2)/d)^m,x]

[Out]

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

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int \left ({\frac{c{x}^{2}+a+b}{d}} \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*x^2 + a + b)/d)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((c*x^2 + a + b)/d)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(((a + b + c*x**2)/d)**m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*x^2 + a + b)/d)^m, x)