3.847 \(\int \frac {(d+e x)^8}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=173 \[ \frac {2 (d+e x)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {42 (d+e x)^3}{5 e \sqrt {d^2-e^2 x^2}}+\frac {21 \sqrt {d^2-e^2 x^2} (d+e x)}{2 e}+\frac {63 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {63 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

[Out]

2/5*(e*x+d)^7/e/(-e^2*x^2+d^2)^(5/2)-6/5*(e*x+d)^5/e/(-e^2*x^2+d^2)^(3/2)-63/2*d^2*arctan(e*x/(-e^2*x^2+d^2)^(
1/2))/e+42/5*(e*x+d)^3/e/(-e^2*x^2+d^2)^(1/2)+63/2*d*(-e^2*x^2+d^2)^(1/2)/e+21/2*(e*x+d)*(-e^2*x^2+d^2)^(1/2)/
e

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Rubi [A]  time = 0.08, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {2 (d+e x)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {42 (d+e x)^3}{5 e \sqrt {d^2-e^2 x^2}}+\frac {21 \sqrt {d^2-e^2 x^2} (d+e x)}{2 e}+\frac {63 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {63 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^8/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(2*(d + e*x)^7)/(5*e*(d^2 - e^2*x^2)^(5/2)) - (6*(d + e*x)^5)/(5*e*(d^2 - e^2*x^2)^(3/2)) + (42*(d + e*x)^3)/(
5*e*Sqrt[d^2 - e^2*x^2]) + (63*d*Sqrt[d^2 - e^2*x^2])/(2*e) + (21*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(2*e) - (63*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)^8}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {2 (d+e x)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {9}{5} \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx\\ &=\frac {2 (d+e x)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {21}{5} \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {2 (d+e x)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {42 (d+e x)^3}{5 e \sqrt {d^2-e^2 x^2}}-21 \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {2 (d+e x)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {42 (d+e x)^3}{5 e \sqrt {d^2-e^2 x^2}}+\frac {21 (d+e x) \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{2} (63 d) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {2 (d+e x)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {42 (d+e x)^3}{5 e \sqrt {d^2-e^2 x^2}}+\frac {63 d \sqrt {d^2-e^2 x^2}}{2 e}+\frac {21 (d+e x) \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{2} \left (63 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {2 (d+e x)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {42 (d+e x)^3}{5 e \sqrt {d^2-e^2 x^2}}+\frac {63 d \sqrt {d^2-e^2 x^2}}{2 e}+\frac {21 (d+e x) \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{2} \left (63 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)^7}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {42 (d+e x)^3}{5 e \sqrt {d^2-e^2 x^2}}+\frac {63 d \sqrt {d^2-e^2 x^2}}{2 e}+\frac {21 (d+e x) \sqrt {d^2-e^2 x^2}}{2 e}-\frac {63 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 = 0.28, size = 131, normalized size = 0.76 \[ \frac {(d+e x) \left (\sqrt {1-\frac {e^2 x^2}{d^2}} \left (496 d^4-1163 d^3 e x+801 d^2 e^2 x^2-65 d e^3 x^3-5 e^4 x^4\right )-315 d (d-e x)^3 \sin ^{-1}\left (\frac {e x}{d}\right )\right )}{10 e (d-e x)^2 \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^8/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(Sqrt[1 - (e^2*x^2)/d^2]*(496*d^4 - 1163*d^3*e*x + 801*d^2*e^2*x^2 - 65*d*e^3*x^3 - 5*e^4*x^4) - 31
5*d*(d - e*x)^3*ArcSin[(e*x)/d]))/(10*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 = 0.91, size = 190, normalized size = 1.10 \[ \frac {496 \, d^{2} e^{3} x^{3} - 1488 \, d^{3} e^{2} x^{2} + 1488 \, d^{4} e x - 496 \, d^{5} + 630 \, {\left (d^{2} e^{3} x^{3} - 3 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x - d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (5 \, e^{4} x^{4} + 65 \, d e^{3} x^{3} - 801 \, d^{2} e^{2} x^{2} + 1163 \, d^{3} e x - 496 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{10 \, {\left (e^{4} x^{3} - 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x - d^{3} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^8/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/10*(496*d^2*e^3*x^3 - 1488*d^3*e^2*x^2 + 1488*d^4*e*x - 496*d^5 + 630*(d^2*e^3*x^3 - 3*d^3*e^2*x^2 + 3*d^4*e
*x - d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (5*e^4*x^4 + 65*d*e^3*x^3 - 801*d^2*e^2*x^2 + 1163*d^3*e
*x - 496*d^4)*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 = 0.33, size = 118, normalized size = 0.68 \[ -\frac {63}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) - \frac {{\left (496 \, d^{7} e^{\left (-1\right )} + {\left (325 \, d^{6} - {\left (1200 \, d^{5} e + {\left (655 \, d^{4} e^{2} - {\left (1040 \, d^{3} e^{3} + {\left (591 \, d^{2} e^{4} - 5 \, {\left (x e^{6} + 16 \, d e^{5}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{10 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^8/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-63/2*d^2*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/10*(496*d^7*e^(-1) + (325*d^6 - (1200*d^5*e + (655*d^4*e^2 - (1040*d
^3*e^3 + (591*d^2*e^4 - 5*(x*e^6 + 16*d*e^5)*x)*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3

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maple [A]  time = 0.19, size = 284, normalized size = 1.64 \[ -\frac {e^{6} x^{7}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {8 d \,e^{5} x^{6}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {63 d^{2} e^{4} x^{5}}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {104 d^{3} e^{3} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {35 d^{4} e^{2} x^{3}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {120 d^{5} e \,x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {76 d^{6} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {21 d^{2} e^{2} x^{3}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {248 d^{7}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {27 d^{4} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {423 d^{2} x}{10 \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {63 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^8/(-e^2*x^2+d^2)^(7/2),x)

[Out]

-21/2/(-e^2*x^2+d^2)^(3/2)*d^2*e^2*x^3+423/10/(-e^2*x^2+d^2)^(1/2)*d^2*x+35*d^4*e^2*x^3/(-e^2*x^2+d^2)^(5/2)-8
*e^5*d*x^6/(-e^2*x^2+d^2)^(5/2)+104*e^3*d^3*x^4/(-e^2*x^2+d^2)^(5/2)-120*e*d^5*x^2/(-e^2*x^2+d^2)^(5/2)+63/10*
e^4*d^2*x^5/(-e^2*x^2+d^2)^(5/2)-63/2/(e^2)^(1/2)*d^2*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+27/5/(-e^2*x^
2+d^2)^(3/2)*d^4*x-76/5*d^6*x/(-e^2*x^2+d^2)^(5/2)-1/2*e^6*x^7/(-e^2*x^2+d^2)^(5/2)+248/5*d^7/e/(-e^2*x^2+d^2)
^(5/2)

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maxima [B]  time = 3.11, size = 349, normalized size = 2.02 \[ -\frac {e^{6} x^{7}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {21}{10} \, d^{2} e^{6} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {8 \, d e^{5} x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {21}{2} \, d^{2} e^{4} x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {104 \, d^{3} e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {35 \, d^{4} e^{2} x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {120 \, d^{5} e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {76 \, d^{6} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {248 \, d^{7}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {69 \, d^{4} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {39 \, d^{2} x}{10 \, \sqrt {-e^{2} x^{2} + d^{2}}} - \frac {63 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^8/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

-1/2*e^6*x^7/(-e^2*x^2 + d^2)^(5/2) + 21/10*d^2*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)) - 8*d*e^5*x^6/(-e^2*x^2 + d^2)^(5/2) - 21/2*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)) + 104*d^3*e^3*x^4/(-e^2*x^2 + d
^2)^(5/2) + 35*d^4*e^2*x^3/(-e^2*x^2 + d^2)^(5/2) - 120*d^5*e*x^2/(-e^2*x^2 + d^2)^(5/2) - 76/5*d^6*x/(-e^2*x^
2 + d^2)^(5/2) + 248/5*d^7/((-e^2*x^2 + d^2)^(5/2)*e) + 69/5*d^4*x/(-e^2*x^2 + d^2)^(3/2) - 39/10*d^2*x/sqrt(-
e^2*x^2 + d^2) - 63/2*d^2*arcsin(e*x/d)/e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^8}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^8/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((d + e*x)^8/(d^2 - e^2*x^2)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{8}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**8/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**8/(-(-d + e*x)*(d + e*x))**(7/2), x)

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