3.953 \(\int (d+e x)^m (d^2-e^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=59 \[ \frac{\left (d^2-e^2 x^2\right )^{5/2} (d+e x)^m \, _2F_1\left (1,m+5;m+\frac{7}{2};\frac{d+e x}{2 d}\right )}{d e (2 m+5)} \]

[Out]

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

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Rubi [A]  time = 0.0473627, antiderivative size = 83, normalized size of antiderivative = 1.41, 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} \[ -\frac{2^{m+\frac{5}{2}} \left (d^2-e^2 x^2\right )^{5/2} (d+e x)^m \left (\frac{e x}{d}+1\right )^{-m-\frac{5}{2}} \, _2F_1\left (\frac{5}{2},-m-\frac{3}{2};\frac{7}{2};\frac{d-e x}{2 d}\right )}{5 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2^(5/2 + m)*(d + e*x)^m*(1 + (e*x)/d)^(-5/2 - m)*(d^2 - e^2*x^2)^(5/2)*Hypergeometric2F1[5/2, -3/2 - m, 7/2,
(d - e*x)/(2*d)])/(5*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 \left (d^2-e^2 x^2\right )^{3/2} \, dx &=\left ((d+e x)^m \left (1+\frac{e x}{d}\right )^{-m}\right ) \int \left (1+\frac{e x}{d}\right )^m \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac{\left ((d+e x)^m \left (1+\frac{e x}{d}\right )^{-\frac{5}{2}-m} \left (d^2-e^2 x^2\right )^{5/2}\right ) \int \left (1+\frac{e x}{d}\right )^{\frac{3}{2}+m} \left (d^2-d e x\right )^{3/2} \, dx}{\left (d^2-d e x\right )^{5/2}}\\ &=-\frac{2^{\frac{5}{2}+m} (d+e x)^m \left (1+\frac{e x}{d}\right )^{-\frac{5}{2}-m} \left (d^2-e^2 x^2\right )^{5/2} \, _2F_1\left (\frac{5}{2},-\frac{3}{2}-m;\frac{7}{2};\frac{d-e x}{2 d}\right )}{5 d e}\\ \end{align*}

Mathematica [C]  time = 0.237686, size = 191, normalized size = 3.24 \[ -\frac{2^m (d+e x)^m \left (\frac{e x}{d}+1\right )^{-2 m-\frac{1}{2}} \left (e^3 x^3 \sqrt{d-e x} \sqrt{d+e x} \left (\frac{e x}{2 d}+\frac{1}{2}\right )^m F_1\left (3;-\frac{1}{2},-m-\frac{1}{2};4;\frac{e x}{d},-\frac{e x}{d}\right )+2 d^2 (d-e x) \sqrt{2-\frac{2 e x}{d}} \sqrt{d^2-e^2 x^2} \left (\frac{e x}{d}+1\right )^m \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{d-e x}{2 d}\right )\right )}{3 e \sqrt{1-\frac{e x}{d}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(2^m*(d + e*x)^m*(1 + (e*x)/d)^(-1/2 - 2*m)*(e^3*x^3*Sqrt[d - e*x]*Sqrt[d + e*x]*(1/2 + (e*x)/(2*d))^m*Appell
F1[3, -1/2, -1/2 - m, 4, (e*x)/d, -((e*x)/d)] + 2*d^2*(d - e*x)*Sqrt[2 - (2*e*x)/d]*(1 + (e*x)/d)^m*Sqrt[d^2 -
e^2*x^2]*Hypergeometric2F1[3/2, -1/2 - m, 5/2, (d - e*x)/(2*d)]))/(3*e*Sqrt[1 - (e*x)/d])

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Maple [F]  time = 0.479, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{{\frac{3}{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)^(3/2),x)

[Out]

int((e*x+d)^m*(-e^2*x^2+d^2)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{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)^(3/2),x, algorithm="maxima")

[Out]

integrate((-e^2*x^2 + d^2)^(3/2)*(e*x + d)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (e^{2} x^{2} - d^{2}\right )} \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)^(3/2),x, algorithm="fricas")

[Out]

integral(-(e^2*x^2 - d^2)*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 \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{3}{2}} \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)**(3/2),x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{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)^(3/2),x, algorithm="giac")

[Out]

integrate((-e^2*x^2 + d^2)^(3/2)*(e*x + d)^m, x)