3.10.55 \(\int \frac {(d+e x)^m}{\sqrt {d^2-e^2 x^2}} \, dx\) [955]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 81, normalized size of antiderivative = 1.21, 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 = {694, 692, 71} \begin {gather*} -\frac {2^{m+\frac {1}{2}} \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 {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {d-e x}{2 d}\right )}{d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 692

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/e)*x)^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 694

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])

Rubi steps

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

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Mathematica [A]
time = 0.36, size = 84, normalized size = 1.25 \begin {gather*} -\frac {2^{\frac {1}{2}+m} (d-e x) (d+e x)^m \left (1+\frac {e x}{d}\right )^{\frac {1}{2}-m} \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {d-e x}{2 d}\right )}{e \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m}}{\sqrt {-e^{2} x^{2}+d^{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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((x*e + d)^m/sqrt(-x^2*e^2 + d^2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{m}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}\, dx \end {gather*}

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((x*e + d)^m/sqrt(-x^2*e^2 + d^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^m}{\sqrt {d^2-e^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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