3.9.36 \(\int \frac {(d+e x)^6}{(d^2-e^2 x^2)^{5/2}} \, dx\) [836]

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

[Out]

2/3*(e*x+d)^5/e/(-e^2*x^2+d^2)^(3/2)+35/2*d^2*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-14/3*(e*x+d)^3/e/(-e^2*x^2+d^
2)^(1/2)-35/2*d*(-e^2*x^2+d^2)^(1/2)/e-35/6*(e*x+d)*(-e^2*x^2+d^2)^(1/2)/e

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Rubi [A]
time = 0.04, 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 = {683, 685, 655, 223, 209} \begin {gather*} \frac {35 d^2 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\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} \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 209

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

Rule 223

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 655

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 683

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 685

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 ) \text {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*}

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Mathematica [A]
time = 0.42, size = 109, normalized size = 0.76 \begin {gather*} \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 (-d+e x)^2}-\frac {35 d^2 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 \sqrt {-e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(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*Log[-(S
qrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*Sqrt[-e^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(482\) vs. \(2(123)=246\).
time = 0.50, size = 483, normalized size = 3.38

method	result	size
risch	\(-\frac {\left (e x +12 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e}+\frac {35 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}+\frac {16 d^{3} \sqrt {-e^{2} \left (x -\frac {d}{e}\right )^{2}-2 \left (x -\frac {d}{e}\right ) d e}}{3 e^{3} \left (x -\frac {d}{e}\right )^{2}}+\frac {80 d^{2} \sqrt {-e^{2} \left (x -\frac {d}{e}\right )^{2}-2 \left (x -\frac {d}{e}\right ) d e}}{3 e^{2} \left (x -\frac {d}{e}\right )}\)	\(156\)
default	\(e^{6} \left (-\frac {x^{5}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {5 d^{2} \left (\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}\right )}{2 e^{2}}\right )+6 d \,e^{5} \left (-\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {4 d^{2} \left (\frac {x^{2}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\right )}{e^{2}}\right )+15 d^{2} e^{4} \left (\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}\right )+20 d^{3} e^{3} \left (\frac {x^{2}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\right )+15 d^{4} e^{2} \left (\frac {x}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2}}\right )+\frac {2 d^{5}}{e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+d^{6} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )\)	\(483\)

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,method=_RETURNVERBOSE)

[Out]

e^6*(-1/2*x^5/e^2/(-e^2*x^2+d^2)^(3/2)+5/2*d^2/e^2*(1/3*x^3/e^2/(-e^2*x^2+d^2)^(3/2)-1/e^2*(x/e^2/(-e^2*x^2+d^
2)^(1/2)-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))))+6*d*e^5*(-x^4/e^2/(-e^2*x^2+d^2)^(3/2
)+4*d^2/e^2*(x^2/e^2/(-e^2*x^2+d^2)^(3/2)-2/3*d^2/e^4/(-e^2*x^2+d^2)^(3/2)))+15*d^2*e^4*(1/3*x^3/e^2/(-e^2*x^2
+d^2)^(3/2)-1/e^2*(x/e^2/(-e^2*x^2+d^2)^(1/2)-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))))+2
0*d^3*e^3*(x^2/e^2/(-e^2*x^2+d^2)^(3/2)-2/3*d^2/e^4/(-e^2*x^2+d^2)^(3/2))+15*d^4*e^2*(1/2*x/e^2/(-e^2*x^2+d^2)
^(3/2)-1/2*d^2/e^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2)))+2*d^5/e/(-e^2*x^2+d^2)^(3/
2)+d^6*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2))

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Maxima [A]
time = 0.52, size = 187, normalized size = 1.31 \begin {gather*} \frac {35}{6} \, {\left (\frac {3 \, x^{2} e^{\left (-2\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, d^{2} e^{\left (-4\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}}\right )} d^{2} x e^{4} - \frac {x^{5} e^{4}}{2 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {6 \, d x^{4} e^{3}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {44 \, d^{3} x^{2} e}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {82 \, d^{5} e^{\left (-1\right )}}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {35}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} + \frac {16 \, d^{4} x}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {61 \, d^{2} x}{6 \, \sqrt {-x^{2} e^{2} + d^{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="maxima")

[Out]

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

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Fricas [A]
time = 2.71, size = 138, normalized size = 0.97 \begin {gather*} -\frac {164 \, d^{2} x^{2} e^{2} - 328 \, d^{3} x e + 164 \, d^{4} + 210 \, {\left (d^{2} x^{2} e^{2} - 2 \, d^{3} x e + d^{4}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (3 \, x^{3} e^{3} + 30 \, d x^{2} e^{2} - 229 \, d^{2} x e + 164 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{6 \, {\left (x^{2} e^{3} - 2 \, d x e^{2} + 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*x^2*e^2 - 328*d^3*x*e + 164*d^4 + 210*(d^2*x^2*e^2 - 2*d^3*x*e + d^4)*arctan(-(d - sqrt(-x^2*e^2
+ d^2))*e^(-1)/x) + (3*x^3*e^3 + 30*d*x^2*e^2 - 229*d^2*x*e + 164*d^3)*sqrt(-x^2*e^2 + d^2))/(x^2*e^3 - 2*d*x
*e^2 + d^2*e)

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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)

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Giac [A]
time = 0.98, size = 145, normalized size = 1.01 \begin {gather*} \frac {35}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\left (d\right ) - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (12 \, d e^{\left (-1\right )} + x\right )} + \frac {32 \, {\left (\frac {9 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{2} e^{\left (-2\right )}}{x} - \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{2} e^{\left (-4\right )}}{x^{2}} - 4 \, d^{2}\right )} e^{\left (-1\right )}}{3 \, {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - 1\right )}^{3}} \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) - 1/2*sqrt(-x^2*e^2 + d^2)*(12*d*e^(-1) + x) + 32/3*(9*(d*e + sqrt(-x^2*e
^2 + d^2)*e)*d^2*e^(-2)/x - 3*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^2*e^(-4)/x^2 - 4*d^2)*e^(-1)/((d*e + sqrt(-x^
2*e^2 + d^2)*e)*e^(-2)/x - 1)^3

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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)

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