### 3.848 $$\int \frac{(d+e x)^7}{(d^2-e^2 x^2)^{7/2}} \, dx$$

Optimal. Leaf size=138 $\frac{2 (d+e x)^6}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 (d+e x)^4}{15 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{14 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}+\frac{7 \sqrt{d^2-e^2 x^2}}{e}-\frac{7 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e}$

[Out]

(2*(d + e*x)^6)/(5*e*(d^2 - e^2*x^2)^(5/2)) - (14*(d + e*x)^4)/(15*e*(d^2 - e^2*x^2)^(3/2)) + (14*(d + e*x)^2)
/(3*e*Sqrt[d^2 - e^2*x^2]) + (7*Sqrt[d^2 - e^2*x^2])/e - (7*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e

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Rubi [A]  time = 0.0572427, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {669, 641, 217, 203} $\frac{2 (d+e x)^6}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 (d+e x)^4}{15 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{14 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}+\frac{7 \sqrt{d^2-e^2 x^2}}{e}-\frac{7 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(d + e*x)^6)/(5*e*(d^2 - e^2*x^2)^(5/2)) - (14*(d + e*x)^4)/(15*e*(d^2 - e^2*x^2)^(3/2)) + (14*(d + e*x)^2)
/(3*e*Sqrt[d^2 - e^2*x^2]) + (7*Sqrt[d^2 - e^2*x^2])/e - (7*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{(d+e x)^7}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{2 (d+e x)^6}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{7}{5} \int \frac{(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx\\ &=\frac{2 (d+e x)^6}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 (d+e x)^4}{15 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{7}{3} \int \frac{(d+e x)^3}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac{2 (d+e x)^6}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 (d+e x)^4}{15 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{14 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}-7 \int \frac{d+e x}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{2 (d+e x)^6}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 (d+e x)^4}{15 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{14 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}+\frac{7 \sqrt{d^2-e^2 x^2}}{e}-(7 d) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{2 (d+e x)^6}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 (d+e x)^4}{15 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{14 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}+\frac{7 \sqrt{d^2-e^2 x^2}}{e}-(7 d) \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)^6}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 (d+e x)^4}{15 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{14 (d+e x)^2}{3 e \sqrt{d^2-e^2 x^2}}+\frac{7 \sqrt{d^2-e^2 x^2}}{e}-\frac{7 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.240212, size = 119, normalized size = 0.86 $\frac{(d+e x) \left (\sqrt{1-\frac{e^2 x^2}{d^2}} \left (-381 d^2 e x+167 d^3+277 d e^2 x^2-15 e^3 x^3\right )-105 (d-e x)^3 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{15 e (d-e x)^2 \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((d + e*x)*(Sqrt[1 - (e^2*x^2)/d^2]*(167*d^3 - 381*d^2*e*x + 277*d*e^2*x^2 - 15*e^3*x^3) - 105*(d - e*x)^3*Arc
Sin[(e*x)/d]))/(15*e*(d - e*x)^2*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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Maple [B]  time = 0.106, size = 253, normalized size = 1.8 \begin{align*} -{{e}^{5}{x}^{6} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+27\,{\frac{{e}^{3}{d}^{2}{x}^{4}}{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{5/2}}}-{\frac{73\,e{d}^{4}{x}^{2}}{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{167\,{d}^{6}}{15\,e} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{7\,d{e}^{4}{x}^{5}}{5} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{7\,d{e}^{2}{x}^{3}}{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{176\,dx}{15}{\frac{1}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}}}-7\,{\frac{d}{\sqrt{{e}^{2}}}\arctan \left ({\frac{\sqrt{{e}^{2}}x}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}} \right ) }+{\frac{35\,{d}^{3}{e}^{2}{x}^{3}}{2} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{61\,{d}^{5}x}{10} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{71\,{d}^{3}x}{30} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^5*x^6/(-e^2*x^2+d^2)^(5/2)+27*e^3*d^2*x^4/(-e^2*x^2+d^2)^(5/2)-73/3*e*d^4*x^2/(-e^2*x^2+d^2)^(5/2)+167/15*d
^6/e/(-e^2*x^2+d^2)^(5/2)+7/5*d*e^4*x^5/(-e^2*x^2+d^2)^(5/2)-7/3*d*e^2*x^3/(-e^2*x^2+d^2)^(3/2)+176/15*d*x/(-e
^2*x^2+d^2)^(1/2)-7*d/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+35/2*d^3*e^2*x^3/(-e^2*x^2+d^2)^(
5/2)-61/10*d^5*x/(-e^2*x^2+d^2)^(5/2)+71/30*d^3*x/(-e^2*x^2+d^2)^(3/2)

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Maxima [B]  time = 2.28444, size = 443, normalized size = 3.21 \begin{align*} \frac{7}{15} \, d 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{e^{5} x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} - \frac{7}{3} \, d 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{27 \, d^{2} e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{35 \, d^{3} e^{2} x^{3}}{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} - \frac{73 \, d^{4} e x^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} - \frac{61 \, d^{5} x}{10 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{167 \, d^{6}}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} + \frac{127 \, d^{3} x}{30 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}} + \frac{22 \, d x}{15 \, \sqrt{-e^{2} x^{2} + d^{2}}} - \frac{7 \, d \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/15*d*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)) - e^5*x^6/(-e^2*x^2 + d^2)^(5/2) - 7/3*d*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)) + 27*d^2*e^3*x^4/(-e^2*x^2 + d^2)^(5/2) + 35/2*d^3*e^2*x^3/(-e^2*x^2 + d^2)^(5/
2) - 73/3*d^4*e*x^2/(-e^2*x^2 + d^2)^(5/2) - 61/10*d^5*x/(-e^2*x^2 + d^2)^(5/2) + 167/15*d^6/((-e^2*x^2 + d^2)
^(5/2)*e) + 127/30*d^3*x/(-e^2*x^2 + d^2)^(3/2) + 22/15*d*x/sqrt(-e^2*x^2 + d^2) - 7*d*arcsin(e^2*x/sqrt(d^2*e
^2))/sqrt(e^2)

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Fricas [A]  time = 2.28796, size = 378, normalized size = 2.74 \begin{align*} \frac{167 \, d e^{3} x^{3} - 501 \, d^{2} e^{2} x^{2} + 501 \, d^{3} e x - 167 \, d^{4} + 210 \,{\left (d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (15 \, e^{3} x^{3} - 277 \, d e^{2} x^{2} + 381 \, d^{2} e x - 167 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (e^{4} x^{3} - 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x - d^{3} e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(167*d*e^3*x^3 - 501*d^2*e^2*x^2 + 501*d^3*e*x - 167*d^4 + 210*(d*e^3*x^3 - 3*d^2*e^2*x^2 + 3*d^3*e*x - d
^4)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (15*e^3*x^3 - 277*d*e^2*x^2 + 381*d^2*e*x - 167*d^3)*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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{7}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{7}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34006, size = 144, normalized size = 1.04 \begin{align*} -7 \, d \arcsin \left (\frac{x e}{d}\right ) e^{\left (-1\right )} \mathrm{sgn}\left (d\right ) - \frac{{\left (167 \, d^{6} e^{\left (-1\right )} +{\left (120 \, d^{5} -{\left (365 \, d^{4} e +{\left (160 \, d^{3} e^{2} -{\left (405 \, d^{2} e^{3} -{\left (15 \, x e^{5} - 232 \, d e^{4}\right )} x\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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*d*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/15*(167*d^6*e^(-1) + (120*d^5 - (365*d^4*e + (160*d^3*e^2 - (405*d^2*e^3
- (15*x*e^5 - 232*d*e^4)*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3