3.7.77 \(\int \frac {(d+e x)^6}{(d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=143 \[ \frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 \sqrt {d^2-e^2 x^2} (d+e x)}{6 e}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}+\frac {35 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {669, 671, 641, 217, 203} \begin {gather*} \frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 \sqrt {d^2-e^2 x^2} (d+e x)}{6 e}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}+\frac {35 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x]

[Out]

(2*(d + e*x)^5)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (14*(d + e*x)^3)/(3*e*Sqrt[d^2 - e^2*x^2]) - (35*d*Sqrt[d^2 - e^
2*x^2])/(2*e) - (35*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(6*e) + (35*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 641

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

Rule 671

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*(m + 2*p + 1)), x] + Dist[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rubi steps

\begin {align*} \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {7}{3} \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}+\frac {35}{3} \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {1}{2} (35 d) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {1}{2} \left (35 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {1}{2} \left (35 d^2\right ) \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}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {35 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 121, normalized size = 0.85 \begin {gather*} \frac {(d+e x) \left (\sqrt {1-\frac {e^2 x^2}{d^2}} \left (164 d^3-229 d^2 e x+30 d e^2 x^2+3 e^3 x^3\right )-105 d (d-e x)^2 \sin ^{-1}\left (\frac {e x}{d}\right )\right )}{6 e (e x-d) \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x]

[Out]

((d + e*x)*(Sqrt[1 - (e^2*x^2)/d^2]*(164*d^3 - 229*d^2*e*x + 30*d*e^2*x^2 + 3*e^3*x^3) - 105*d*(d - e*x)^2*Arc
Sin[(e*x)/d]))/(6*e*(-d + e*x)*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.48, size = 112, normalized size = 0.78 \begin {gather*} \frac {35 d^2 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{2 e^2}+\frac {\sqrt {d^2-e^2 x^2} \left (-164 d^3+229 d^2 e x-30 d e^2 x^2-3 e^3 x^3\right )}{6 e (e x-d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-164*d^3 + 229*d^2*e*x - 30*d*e^2*x^2 - 3*e^3*x^3))/(6*e*(-d + e*x)^2) + (35*d^2*Sqrt[-e
^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*e^2)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 143, normalized size = 1.00 \begin {gather*} -\frac {164 \, d^{2} e^{2} x^{2} - 328 \, d^{3} e x + 164 \, d^{4} + 210 \, {\left (d^{2} e^{2} x^{2} - 2 \, d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (3 \, e^{3} x^{3} + 30 \, d e^{2} x^{2} - 229 \, d^{2} e x + 164 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{3} x^{2} - 2 \, d e^{2} x + d^{2} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(164*d^2*e^2*x^2 - 328*d^3*e*x + 164*d^4 + 210*(d^2*e^2*x^2 - 2*d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2
+ d^2))/(e*x)) + (3*e^3*x^3 + 30*d*e^2*x^2 - 229*d^2*e*x + 164*d^3)*sqrt(-e^2*x^2 + d^2))/(e^3*x^2 - 2*d*e^2*
x + d^2*e)

________________________________________________________________________________________

giac [A]  time = 0.31, size = 97, normalized size = 0.68 \begin {gather*} \frac {35}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) - \frac {{\left (164 \, d^{5} e^{\left (-1\right )} + {\left (99 \, d^{4} - {\left (264 \, d^{3} e + {\left (166 \, d^{2} e^{2} - 3 \, {\left (x e^{4} + 12 \, d e^{3}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{6 \, {\left (x^{2} e^{2} - d^{2}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

35/2*d^2*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/6*(164*d^5*e^(-1) + (99*d^4 - (264*d^3*e + (166*d^2*e^2 - 3*(x*e^4 +
12*d*e^3)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^2

________________________________________________________________________________________

maple [A]  time = 0.10, size = 189, normalized size = 1.32 \begin {gather*} -\frac {e^{4} x^{5}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {6 d \,e^{3} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {35 d^{2} e^{2} x^{3}}{6 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {44 d^{3} e \,x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {16 d^{4} x}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {82 d^{5}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}-\frac {131 d^{2} x}{6 \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {35 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x)

[Out]

-1/2*e^4*x^5/(-e^2*x^2+d^2)^(3/2)+35/6*e^2*d^2*x^3/(-e^2*x^2+d^2)^(3/2)-131/6*d^2*x/(-e^2*x^2+d^2)^(1/2)+35/2/
(e^2)^(1/2)*d^2*arctan((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)^(3/2)+44*d^3*e*x^2/(-e^2
*x^2+d^2)^(3/2)-82/3*d^5/e/(-e^2*x^2+d^2)^(3/2)+16/3*d^4*x/(-e^2*x^2+d^2)^(3/2)

________________________________________________________________________________________

maxima [A]  time = 3.12, size = 200, normalized size = 1.40 \begin {gather*} \frac {35}{6} \, d^{2} 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 {e^{4} x^{5}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {6 \, d e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {44 \, d^{3} e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {16 \, d^{4} x}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {82 \, d^{5}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} - \frac {61 \, d^{2} x}{6 \, \sqrt {-e^{2} x^{2} + d^{2}}} + \frac {35 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

35/6*d^2*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)) - 1/2*e^4*x^5/(-e^2*x
^2 + d^2)^(3/2) - 6*d*e^3*x^4/(-e^2*x^2 + d^2)^(3/2) + 44*d^3*e*x^2/(-e^2*x^2 + d^2)^(3/2) + 16/3*d^4*x/(-e^2*
x^2 + d^2)^(3/2) - 82/3*d^5/((-e^2*x^2 + d^2)^(3/2)*e) - 61/6*d^2*x/sqrt(-e^2*x^2 + d^2) + 35/2*d^2*arcsin(e*x
/d)/e

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^6}{{\left (d^2-e^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x)

[Out]

int((d + e*x)^6/(d^2 - e^2*x^2)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{6}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**6/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral((d + e*x)**6/(-(-d + e*x)*(d + e*x))**(5/2), x)

________________________________________________________________________________________