3.849 \(\int \frac {(d+e x)^6}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=112 \[ \frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

[Out]

2/5*(e*x+d)^5/e/(-e^2*x^2+d^2)^(5/2)-2/3*(e*x+d)^3/e/(-e^2*x^2+d^2)^(3/2)-arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e+2
*(e*x+d)/e/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {669, 653, 217, 203} \[ \frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^6/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(2*(d + e*x)^5)/(5*e*(d^2 - e^2*x^2)^(5/2)) - (2*(d + e*x)^3)/(3*e*(d^2 - e^2*x^2)^(3/2)) + (2*(d + e*x))/(e*S
qrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 653

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

Rule 669

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx\\ &=\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\int \frac {(d+e x)^2}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 113, normalized size = 1.01 \[ \frac {(d+e x) \left (2 d \left (13 d^2-24 d e x+23 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}-15 (d-e x)^3 \sin ^{-1}\left (\frac {e x}{d}\right )\right )}{15 d e (d-e x)^2 \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^6/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(2*d*(13*d^2 - 24*d*e*x + 23*e^2*x^2)*Sqrt[1 - (e^2*x^2)/d^2] - 15*(d - e*x)^3*ArcSin[(e*x)/d]))/(1
5*d*e*(d - e*x)^2*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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fricas [A]  time = 1.01, size = 159, normalized size = 1.42 \[ \frac {2 \, {\left (13 \, e^{3} x^{3} - 39 \, d e^{2} x^{2} + 39 \, d^{2} e x - 13 \, d^{3} + 15 \, {\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (23 \, e^{2} x^{2} - 24 \, d e x + 13 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}\right )}}{15 \, {\left (e^{4} x^{3} - 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x - d^{3} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

2/15*(13*e^3*x^3 - 39*d*e^2*x^2 + 39*d^2*e*x - 13*d^3 + 15*(e^3*x^3 - 3*d*e^2*x^2 + 3*d^2*e*x - d^3)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (23*e^2*x^2 - 24*d*e*x + 13*d^2)*sqrt(-e^2*x^2 + d^2))/(e^4*x^3 - 3*d*e^3*x
^2 + 3*d^2*e^2*x - d^3*e)

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giac [A]  time = 0.36, size = 95, normalized size = 0.85 \[ -\arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) - \frac {2 \, {\left (13 \, d^{5} e^{\left (-1\right )} + {\left (15 \, d^{4} - {\left (10 \, d^{3} e - {\left (10 \, d^{2} e^{2} + {\left (23 \, x e^{4} + 45 \, d e^{3}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^(-1)*sgn(d) - 2/15*(13*d^5*e^(-1) + (15*d^4 - (10*d^3*e - (10*d^2*e^2 + (23*x*e^4 + 45*d*e^3)
*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3

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maple [B]  time = 0.09, size = 225, normalized size = 2.01 \[ \frac {e^{4} x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 d \,e^{3} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {15 d^{2} e^{2} x^{3}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{3} e \,x^{2}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {13 d^{4} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {e^{2} x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {26 d^{5}}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {23 d^{2} x}{30 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {38 x}{15 \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/5*e^4*x^5/(-e^2*x^2+d^2)^(5/2)-1/3/(-e^2*x^2+d^2)^(3/2)*e^2*x^3+38/15/(-e^2*x^2+d^2)^(1/2)*x-1/(e^2)^(1/2)*a
rctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+6*d*e^3*x^4/(-e^2*x^2+d^2)^(5/2)-4/3*d^3*e*x^2/(-e^2*x^2+d^2)^(5/2)+
26/15*d^5/e/(-e^2*x^2+d^2)^(5/2)+15/2*e^2*d^2*x^3/(-e^2*x^2+d^2)^(5/2)-13/10*d^4*x/(-e^2*x^2+d^2)^(5/2)+23/30/
(-e^2*x^2+d^2)^(3/2)*d^2*x

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maxima [B]  time = 3.14, size = 290, normalized size = 2.59 \[ \frac {1}{15} \, e^{6} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {1}{3} \, e^{4} x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {6 \, d e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {15 \, d^{2} e^{2} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {4 \, d^{3} e x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {13 \, d^{4} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {26 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {31 \, d^{2} x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {16 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e^6*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 +
d^2)^(5/2)*e^6)) - 1/3*e^4*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4)) + 6*d*
e^3*x^4/(-e^2*x^2 + d^2)^(5/2) + 15/2*d^2*e^2*x^3/(-e^2*x^2 + d^2)^(5/2) - 4/3*d^3*e*x^2/(-e^2*x^2 + d^2)^(5/2
) - 13/10*d^4*x/(-e^2*x^2 + d^2)^(5/2) + 26/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e) + 31/30*d^2*x/(-e^2*x^2 + d^2)^(
3/2) + 16/15*x/sqrt(-e^2*x^2 + d^2) - arcsin(e*x/d)/e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^6}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^6/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((d + e*x)^6/(d^2 - e^2*x^2)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{6}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**6/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**6/(-(-d + e*x)*(d + e*x))**(7/2), x)

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