3.958 \(\int \frac {(d+e x)^m}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {(d-e x) (d+e x)^{m+1} \, _2F_1\left (1,m-5;m-\frac {3}{2};\frac {d+e x}{2 d}\right )}{d e (5-2 m) \left (d^2-e^2 x^2\right )^{7/2}} \]

[Out]

-(-e*x+d)*(e*x+d)^(1+m)*hypergeom([1, -5+m],[-3/2+m],1/2*(e*x+d)/d)/d/e/(5-2*m)/(-e^2*x^2+d^2)^(7/2)

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Rubi [A]  time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.22, 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} \[ \frac {2^{m-\frac {5}{2}} (d+e x)^m \left (\frac {e x}{d}+1\right )^{\frac {5}{2}-m} \, _2F_1\left (-\frac {5}{2},\frac {7}{2}-m;-\frac {3}{2};\frac {d-e x}{2 d}\right )}{5 d e \left (d^2-e^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-5/2 + m)*(d + e*x)^m*(1 + (e*x)/d)^(5/2 - m)*Hypergeometric2F1[-5/2, 7/2 - m, -3/2, (d - e*x)/(2*d)])/(5*
d*e*(d^2 - e^2*x^2)^(5/2))

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

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Mathematica [A]  time = 0.09, size = 91, normalized size = 1.34 \[ \frac {2^{m-\frac {5}{2}} (d+e x)^m \left (\frac {e x}{d}+1\right )^{\frac {1}{2}-m} \, _2F_1\left (-\frac {5}{2},\frac {7}{2}-m;-\frac {3}{2};\frac {d-e x}{2 d}\right )}{5 d^3 e (d-e x)^2 \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-5/2 + m)*(d + e*x)^m*(1 + (e*x)/d)^(1/2 - m)*Hypergeometric2F1[-5/2, 7/2 - m, -3/2, (d - e*x)/(2*d)])/(5*
d^3*e*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )}^{m}}{e^{8} x^{8} - 4 \, d^{2} e^{6} x^{6} + 6 \, d^{4} e^{4} x^{4} - 4 \, d^{6} e^{2} x^{2} + d^{8}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(-e^2*x^2 + d^2)*(e*x + d)^m/(e^8*x^8 - 4*d^2*e^6*x^6 + 6*d^4*e^4*x^4 - 4*d^6*e^2*x^2 + d^8), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^m/(-e^2*x^2 + d^2)^(7/2), x)

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maple [F]  time = 0.81, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{m}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m/(-e^2*x^2+d^2)^(7/2),x)

[Out]

int((e*x+d)^m/(-e^2*x^2+d^2)^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^m/(-e^2*x^2 + d^2)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^m}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((d + e*x)^m/(d^2 - e^2*x^2)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{m}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**m/(-(-d + e*x)*(d + e*x))**(7/2), x)

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