3.2.36 \(\int \frac {x^7}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=162 \[ \frac {x^6 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {(32 d-35 e x) \sqrt {d^2-e^2 x^2}}{10 e^8}+\frac {7 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^8}+\frac {x^2 (24 d-35 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {x^4 (6 d-7 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {850, 819, 780, 217, 203} \begin {gather*} \frac {x^6 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^4 (6 d-7 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x^2 (24 d-35 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {(32 d-35 e x) \sqrt {d^2-e^2 x^2}}{10 e^8}+\frac {7 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(x^6*(d - e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (x^4*(6*d - 7*e*x))/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (x^2*(24*
d - 35*e*x))/(15*e^6*Sqrt[d^2 - e^2*x^2]) + ((32*d - 35*e*x)*Sqrt[d^2 - e^2*x^2])/(10*e^8) + (7*d^2*ArcTan[(e*
x)/Sqrt[d^2 - e^2*x^2]])/(2*e^8)

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 850

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

Rubi steps

\begin {align*} \int \frac {x^7}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\int \frac {x^7 (d-e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {x^6 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {x^5 \left (6 d^3-7 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2}\\ &=\frac {x^6 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^4 (6 d-7 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {x^3 \left (24 d^5-35 d^4 e x\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4 e^4}\\ &=\frac {x^6 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^4 (6 d-7 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x^2 (24 d-35 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {x \left (48 d^7-105 d^6 e x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^6 e^6}\\ &=\frac {x^6 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^4 (6 d-7 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x^2 (24 d-35 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {(32 d-35 e x) \sqrt {d^2-e^2 x^2}}{10 e^8}+\frac {\left (7 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^7}\\ &=\frac {x^6 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^4 (6 d-7 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x^2 (24 d-35 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {(32 d-35 e x) \sqrt {d^2-e^2 x^2}}{10 e^8}+\frac {\left (7 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 )}{2 e^7}\\ &=\frac {x^6 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^4 (6 d-7 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x^2 (24 d-35 e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {(32 d-35 e x) \sqrt {d^2-e^2 x^2}}{10 e^8}+\frac {7 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^8}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 128, normalized size = 0.79 \begin {gather*} \frac {105 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (96 d^6-9 d^5 e x-249 d^4 e^2 x^2-4 d^3 e^3 x^3+176 d^2 e^4 x^4+15 d e^5 x^5-15 e^6 x^6\right )}{(d-e x)^2 (d+e x)^3}}{30 e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(96*d^6 - 9*d^5*e*x - 249*d^4*e^2*x^2 - 4*d^3*e^3*x^3 + 176*d^2*e^4*x^4 + 15*d*e^5*x^5 -
15*e^6*x^6))/((d - e*x)^2*(d + e*x)^3) + 105*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(30*e^8)

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IntegrateAlgebraic [A]  time = 0.64, size = 152, normalized size = 0.94 \begin {gather*} \frac {7 d^2 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{2 e^9}+\frac {\sqrt {d^2-e^2 x^2} \left (96 d^6-9 d^5 e x-249 d^4 e^2 x^2-4 d^3 e^3 x^3+176 d^2 e^4 x^4+15 d e^5 x^5-15 e^6 x^6\right )}{30 e^8 (e x-d)^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^7/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(96*d^6 - 9*d^5*e*x - 249*d^4*e^2*x^2 - 4*d^3*e^3*x^3 + 176*d^2*e^4*x^4 + 15*d*e^5*x^5 -
15*e^6*x^6))/(30*e^8*(-d + e*x)^2*(d + e*x)^3) + (7*d^2*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])
/(2*e^9)

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fricas [A]  time = 0.46, size = 274, normalized size = 1.69 \begin {gather*} \frac {96 \, d^{2} e^{5} x^{5} + 96 \, d^{3} e^{4} x^{4} - 192 \, d^{4} e^{3} x^{3} - 192 \, d^{5} e^{2} x^{2} + 96 \, d^{6} e x + 96 \, d^{7} - 210 \, {\left (d^{2} e^{5} x^{5} + d^{3} e^{4} x^{4} - 2 \, d^{4} e^{3} x^{3} - 2 \, d^{5} e^{2} x^{2} + d^{6} e x + d^{7}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (15 \, e^{6} x^{6} - 15 \, d e^{5} x^{5} - 176 \, d^{2} e^{4} x^{4} + 4 \, d^{3} e^{3} x^{3} + 249 \, d^{4} e^{2} x^{2} + 9 \, d^{5} e x - 96 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (e^{13} x^{5} + d e^{12} x^{4} - 2 \, d^{2} e^{11} x^{3} - 2 \, d^{3} e^{10} x^{2} + d^{4} e^{9} x + d^{5} e^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

1/30*(96*d^2*e^5*x^5 + 96*d^3*e^4*x^4 - 192*d^4*e^3*x^3 - 192*d^5*e^2*x^2 + 96*d^6*e*x + 96*d^7 - 210*(d^2*e^5
*x^5 + d^3*e^4*x^4 - 2*d^4*e^3*x^3 - 2*d^5*e^2*x^2 + d^6*e*x + d^7)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x))
- (15*e^6*x^6 - 15*d*e^5*x^5 - 176*d^2*e^4*x^4 + 4*d^3*e^3*x^3 + 249*d^4*e^2*x^2 + 9*d^5*e*x - 96*d^6)*sqrt(-e
^2*x^2 + d^2))/(e^13*x^5 + d*e^12*x^4 - 2*d^2*e^11*x^3 - 2*d^3*e^10*x^2 + d^4*e^9*x + d^5*e^8)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [B]  time = 0.05, size = 318, normalized size = 1.96 \begin {gather*} -\frac {x^{5}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}+\frac {d \,x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}+\frac {7 d^{2} x^{3}}{6 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5}}-\frac {5 d^{3} x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{6}}+\frac {2 d^{4} x}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{7}}-\frac {4 d^{4} x}{15 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{7}}+\frac {d^{6}}{5 \left (x +\frac {d}{e}\right ) \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{9}}+\frac {3 d^{5}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{8}}-\frac {19 d^{2} x}{6 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{7}}-\frac {8 d^{2} x}{15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{7}}+\frac {7 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x)

[Out]

-1/2/e^3*x^5/(-e^2*x^2+d^2)^(3/2)+7/6/(-e^2*x^2+d^2)^(3/2)*d^2/e^5*x^3-19/6/(-e^2*x^2+d^2)^(1/2)*d^2/e^7*x+7/2
/(e^2)^(1/2)*d^2/e^7*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+d/e^4*x^4/(-e^2*x^2+d^2)^(3/2)-5*d^3/e^6*x^2/(
-e^2*x^2+d^2)^(3/2)+3*d^5/e^8/(-e^2*x^2+d^2)^(3/2)+2/3*d^4/e^7*x/(-e^2*x^2+d^2)^(3/2)+1/5*d^6/e^9/(x+d/e)/(2*(
x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)-4/15*d^4/e^7/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x-8/15*d^2/e^7/(2*(x+d/e)*d*e
-(x+d/e)^2*e^2)^(1/2)*x

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maxima [B]  time = 1.09, size = 289, normalized size = 1.78 \begin {gather*} \frac {d^{6}}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{9} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{8}\right )}} - \frac {x^{5}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} + \frac {d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} + \frac {25 \, d^{2} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} - \frac {65 \, d^{3} x^{2}}{6 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{6}} - \frac {164 \, d^{4} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{7}} - \frac {7 \, d x^{2}}{6 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{6}} + \frac {53 \, d^{5}}{6 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{8}} + \frac {229 \, d^{2} x}{30 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{7}} + \frac {7 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{8}} - \frac {14 \, d^{3}}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{8}} - \frac {7 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{6 \, e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

1/5*d^6/((-e^2*x^2 + d^2)^(3/2)*e^9*x + (-e^2*x^2 + d^2)^(3/2)*d*e^8) - 1/2*x^5/((-e^2*x^2 + d^2)^(3/2)*e^3) +
d*x^4/((-e^2*x^2 + d^2)^(3/2)*e^4) + 25/2*d^2*x^3/((-e^2*x^2 + d^2)^(3/2)*e^5) - 65/6*d^3*x^2/((-e^2*x^2 + d^
2)^(3/2)*e^6) - 164/15*d^4*x/((-e^2*x^2 + d^2)^(3/2)*e^7) - 7/6*d*x^2/(sqrt(-e^2*x^2 + d^2)*e^6) + 53/6*d^5/((
-e^2*x^2 + d^2)^(3/2)*e^8) + 229/30*d^2*x/(sqrt(-e^2*x^2 + d^2)*e^7) + 7/2*d^2*arcsin(e*x/d)/e^8 - 14/3*d^3/(s
qrt(-e^2*x^2 + d^2)*e^8) - 7/6*sqrt(-e^2*x^2 + d^2)*d/e^8

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^7}{{\left (d^2-e^2\,x^2\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/((d^2 - e^2*x^2)^(5/2)*(d + e*x)),x)

[Out]

int(x^7/((d^2 - e^2*x^2)^(5/2)*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(x**7/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

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