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

Optimal. Leaf size=80 \[ \frac {2^{m-\frac {1}{2}} (d+e x)^m \left (\frac {e x}{d}+1\right )^{\frac {1}{2}-m} \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {d-e x}{2 d}\right )}{d e \sqrt {d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, 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} \[ \frac {2^{m-\frac {1}{2}} (d+e x)^m \left (\frac {e x}{d}+1\right )^{\frac {1}{2}-m} \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {d-e x}{2 d}\right )}{d e \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 80, normalized size = 1.00 \[ \frac {2^{m-\frac {1}{2}} (d+e x)^m \left (\frac {e x}{d}+1\right )^{\frac {1}{2}-m} \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {d-e x}{2 d}\right )}{d e \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.19, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )}^{m}}{e^{4} x^{4} - 2 \, d^{2} e^{2} x^{2} + d^{4}}, x\right ) \]

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(sqrt(-e^2*x^2 + d^2)*(e*x + d)^m/(e^4*x^4 - 2*d^2*e^2*x^2 + d^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

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*x + d)^m/(-e^2*x^2 + d^2)^(3/2), x)

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maple [F]  time = 0.80, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{m}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

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*x + d)^m/(-e^2*x^2 + d^2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^m}{{\left (d^2-e^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{m}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}\, dx \]

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

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