### 3.954 $$\int (d+e x)^m \sqrt{d^2-e^2 x^2} \, dx$$

Optimal. Leaf size=67 $\frac{(d-e x) \sqrt{d^2-e^2 x^2} (d+e x)^{m+1} \, _2F_1\left (1,m+3;m+\frac{5}{2};\frac{d+e x}{2 d}\right )}{d e (2 m+3)}$

[Out]

((d - e*x)*(d + e*x)^(1 + m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[1, 3 + m, 5/2 + m, (d + e*x)/(2*d)])/(d*e*(
3 + 2*m))

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Rubi [A]  time = 0.0469089, antiderivative size = 83, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {680, 678, 69} $-\frac{2^{m+\frac{3}{2}} \left (d^2-e^2 x^2\right )^{3/2} (d+e x)^m \left (\frac{e x}{d}+1\right )^{-m-\frac{3}{2}} \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{d-e x}{2 d}\right )}{3 d e}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m*Sqrt[d^2 - e^2*x^2],x]

[Out]

-(2^(3/2 + m)*(d + e*x)^m*(1 + (e*x)/d)^(-3/2 - m)*(d^2 - e^2*x^2)^(3/2)*Hypergeometric2F1[3/2, -1/2 - m, 5/2,
(d - e*x)/(2*d)])/(3*d*e)

Rule 680

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPart[m]*(d + e*x)^FracPart[m]
)/(1 + (e*x)/d)^FracPart[m], Int[(1 + (e*x)/d)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && EqQ[c*d
^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] || GtQ[d, 0])

Rule 678

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^(m - 1)*(a + c*x^2)^(p + 1))/((1
+ (e*x)/d)^(p + 1)*(a/d + (c*x)/e)^(p + 1)), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 69

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

Rubi steps

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

Mathematica [A]  time = 0.0639009, size = 86, normalized size = 1.28 $-\frac{2^{m+\frac{3}{2}} (d-e x) \sqrt{d^2-e^2 x^2} (d+e x)^m \left (\frac{e x}{d}+1\right )^{-m-\frac{1}{2}} \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{d-e x}{2 d}\right )}{3 e}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^m*Sqrt[d^2 - e^2*x^2],x]

[Out]

-(2^(3/2 + m)*(d - e*x)*(d + e*x)^m*(1 + (e*x)/d)^(-1/2 - m)*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[3/2, -1/2 -
m, 5/2, (d - e*x)/(2*d)])/(3*e)

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Maple [F]  time = 0.481, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x)

[Out]

int((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-e^2*x^2 + d^2)*(e*x + d)^m, x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-e^2*x^2 + d^2)*(e*x + d)^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{m}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**m, x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-e^2*x^2 + d^2)*(e*x + d)^m, x)