3.1.72 \(\int \frac {(d+e x)^3 (d^2-e^2 x^2)^{5/2}}{x^2} \, dx\)

Optimal. Leaf size=193 \[ -\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {15}{16} d^7 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-3 d^7 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {3}{16} d^5 e (16 d-5 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2} \]

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Rubi [A]  time = 0.31, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1807, 1809, 815, 844, 217, 203, 266, 63, 208} \begin {gather*} \frac {3}{16} d^5 e (16 d-5 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {15}{16} d^7 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-3 d^7 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^2,x]

[Out]

(3*d^5*e*(16*d - 5*e*x)*Sqrt[d^2 - e^2*x^2])/16 + (d^3*e*(8*d - 5*e*x)*(d^2 - e^2*x^2)^(3/2))/8 + (d*e*(6*d -
5*e*x)*(d^2 - e^2*x^2)^(5/2))/10 - (e*(d^2 - e^2*x^2)^(7/2))/7 - (d*(d^2 - e^2*x^2)^(7/2))/x - (15*d^7*e*ArcTa
n[(e*x)/Sqrt[d^2 - e^2*x^2]])/16 - 3*d^7*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 208

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 815

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

Rule 844

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

\begin {align*} \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}{x^2} \, dx &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{5/2} \left (-3 d^4 e+3 d^3 e^2 x-d^2 e^3 x^2\right )}{x} \, dx}{d^2}\\ &=-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {\int \frac {\left (21 d^4 e^3-21 d^3 e^4 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x} \, dx}{7 d^2 e^2}\\ &=\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {\int \frac {\left (-126 d^6 e^5+105 d^5 e^6 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x} \, dx}{42 d^2 e^4}\\ &=\frac {1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {\int \frac {\left (504 d^8 e^7-315 d^7 e^8 x\right ) \sqrt {d^2-e^2 x^2}}{x} \, dx}{168 d^2 e^6}\\ &=\frac {3}{16} d^5 e (16 d-5 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {\int \frac {-1008 d^{10} e^9+315 d^9 e^{10} x}{x \sqrt {d^2-e^2 x^2}} \, dx}{336 d^2 e^8}\\ &=\frac {3}{16} d^5 e (16 d-5 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\left (3 d^8 e\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-\frac {1}{16} \left (15 d^7 e^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {3}{16} d^5 e (16 d-5 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {1}{2} \left (3 d^8 e\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )-\frac {1}{16} \left (15 d^7 e^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 {3}{16} d^5 e (16 d-5 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {15}{16} d^7 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {\left (3 d^8\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{e}\\ &=\frac {3}{16} d^5 e (16 d-5 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {15}{16} d^7 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-3 d^7 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [C]  time = 0.53, size = 221, normalized size = 1.15 \begin {gather*} -\frac {d^7 \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x \sqrt {1-\frac {e^2 x^2}{d^2}}}-3 d^7 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {15 d^6 e \sqrt {d^2-e^2 x^2} \sin ^{-1}\left (\frac {e x}{d}\right )}{16 \sqrt {1-\frac {e^2 x^2}{d^2}}}+\frac {1}{560} e \sqrt {d^2-e^2 x^2} \left (2496 d^6+1155 d^5 e x-992 d^4 e^2 x^2-910 d^3 e^3 x^3+96 d^2 e^4 x^4+280 d e^5 x^5+80 e^6 x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^2,x]

[Out]

(e*Sqrt[d^2 - e^2*x^2]*(2496*d^6 + 1155*d^5*e*x - 992*d^4*e^2*x^2 - 910*d^3*e^3*x^3 + 96*d^2*e^4*x^4 + 280*d*e
^5*x^5 + 80*e^6*x^6))/560 + (15*d^6*e*Sqrt[d^2 - e^2*x^2]*ArcSin[(e*x)/d])/(16*Sqrt[1 - (e^2*x^2)/d^2]) - 3*d^
7*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d] - (d^7*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-5/2, -1/2, 1/2, (e^2*x^2)/d^2
])/(x*Sqrt[1 - (e^2*x^2)/d^2])

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IntegrateAlgebraic [A]  time = 0.57, size = 187, normalized size = 0.97 \begin {gather*} -\frac {15}{16} d^7 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )+6 d^7 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-560 d^7+2496 d^6 e x+525 d^5 e^2 x^2-992 d^4 e^3 x^3-770 d^3 e^4 x^4+96 d^2 e^5 x^5+280 d e^6 x^6+80 e^7 x^7\right )}{560 x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^2,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-560*d^7 + 2496*d^6*e*x + 525*d^5*e^2*x^2 - 992*d^4*e^3*x^3 - 770*d^3*e^4*x^4 + 96*d^2*e
^5*x^5 + 280*d*e^6*x^6 + 80*e^7*x^7))/(560*x) + 6*d^7*e*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d] - (1
5*d^7*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/16

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fricas [A]  time = 0.43, size = 167, normalized size = 0.87 \begin {gather*} \frac {1050 \, d^{7} e x \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 1680 \, d^{7} e x \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + 2496 \, d^{7} e x + {\left (80 \, e^{7} x^{7} + 280 \, d e^{6} x^{6} + 96 \, d^{2} e^{5} x^{5} - 770 \, d^{3} e^{4} x^{4} - 992 \, d^{4} e^{3} x^{3} + 525 \, d^{5} e^{2} x^{2} + 2496 \, d^{6} e x - 560 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{560 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(-e^2*x^2+d^2)^(5/2)/x^2,x, algorithm="fricas")

[Out]

1/560*(1050*d^7*e*x*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 1680*d^7*e*x*log(-(d - sqrt(-e^2*x^2 + d^2))/x
) + 2496*d^7*e*x + (80*e^7*x^7 + 280*d*e^6*x^6 + 96*d^2*e^5*x^5 - 770*d^3*e^4*x^4 - 992*d^4*e^3*x^3 + 525*d^5*
e^2*x^2 + 2496*d^6*e*x - 560*d^7)*sqrt(-e^2*x^2 + d^2))/x

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giac [A]  time = 0.24, size = 199, normalized size = 1.03 \begin {gather*} -\frac {15}{16} \, d^{7} \arcsin \left (\frac {x e}{d}\right ) e \mathrm {sgn}\relax (d) - 3 \, d^{7} e \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right ) + \frac {d^{7} x e^{3}}{2 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{7} e^{\left (-1\right )}}{2 \, x} + \frac {1}{560} \, {\left (2496 \, d^{6} e + {\left (525 \, d^{5} e^{2} - 2 \, {\left (496 \, d^{4} e^{3} + {\left (385 \, d^{3} e^{4} - 4 \, {\left (12 \, d^{2} e^{5} + 5 \, {\left (2 \, x e^{7} + 7 \, d e^{6}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(-e^2*x^2+d^2)^(5/2)/x^2,x, algorithm="giac")

[Out]

-15/16*d^7*arcsin(x*e/d)*e*sgn(d) - 3*d^7*e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/
2*d^7*x*e^3/(d*e + sqrt(-x^2*e^2 + d^2)*e) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^7*e^(-1)/x + 1/560*(2496*d^6
*e + (525*d^5*e^2 - 2*(496*d^4*e^3 + (385*d^3*e^4 - 4*(12*d^2*e^5 + 5*(2*x*e^7 + 7*d*e^6)*x)*x)*x)*x)*x)*sqrt(
-x^2*e^2 + d^2)

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maple [A]  time = 0.01, size = 243, normalized size = 1.26 \begin {gather*} -\frac {3 d^{8} e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}-\frac {15 d^{7} e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}}-\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5} e^{2} x}{16}+3 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{6} e -\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} e^{2} x}{8}+\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4} e -\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d \,e^{2} x}{2}+\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2} e}{5}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{7}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(-e^2*x^2+d^2)^(5/2)/x^2,x)

[Out]

-1/7*e*(-e^2*x^2+d^2)^(7/2)-1/2*d*e^2*x*(-e^2*x^2+d^2)^(5/2)-5/8*d^3*e^2*x*(-e^2*x^2+d^2)^(3/2)-15/16*d^5*e^2*
x*(-e^2*x^2+d^2)^(1/2)-15/16*d^7*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-d*(-e^2*x^2+d^2)^(
7/2)/x+3/5*d^2*e*(-e^2*x^2+d^2)^(5/2)+d^4*e*(-e^2*x^2+d^2)^(3/2)+3*d^6*e*(-e^2*x^2+d^2)^(1/2)-3*d^8*e/(d^2)^(1
/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

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maxima [A]  time = 1.00, size = 217, normalized size = 1.12 \begin {gather*} -\frac {15}{16} \, d^{7} e \arcsin \left (\frac {e x}{d}\right ) - 3 \, d^{7} e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {15}{16} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5} e^{2} x + 3 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6} e - \frac {5}{8} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} e + \frac {1}{2} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d e^{2} x + \frac {3}{5} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} e - \frac {1}{7} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(-e^2*x^2+d^2)^(5/2)/x^2,x, algorithm="maxima")

[Out]

-15/16*d^7*e*arcsin(e*x/d) - 3*d^7*e*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) - 15/16*sqrt(-e^2*x^2
+ d^2)*d^5*e^2*x + 3*sqrt(-e^2*x^2 + d^2)*d^6*e - 5/8*(-e^2*x^2 + d^2)^(3/2)*d^3*e^2*x + (-e^2*x^2 + d^2)^(3/
2)*d^4*e + 1/2*(-e^2*x^2 + d^2)^(5/2)*d*e^2*x + 3/5*(-e^2*x^2 + d^2)^(5/2)*d^2*e - 1/7*(-e^2*x^2 + d^2)^(7/2)*
e - (-e^2*x^2 + d^2)^(5/2)*d^3/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}\,{\left (d+e\,x\right )}^3}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d^2 - e^2*x^2)^(5/2)*(d + e*x)^3)/x^2,x)

[Out]

int(((d^2 - e^2*x^2)^(5/2)*(d + e*x)^3)/x^2, x)

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sympy [C]  time = 19.88, size = 1057, normalized size = 5.48

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(-e**2*x**2+d**2)**(5/2)/x**2,x)

[Out]

d**7*Piecewise((I*d/(x*sqrt(-1 + e**2*x**2/d**2)) + I*e*acosh(e*x/d) - I*e**2*x/(d*sqrt(-1 + e**2*x**2/d**2)),
Abs(e**2*x**2/d**2) > 1), (-d/(x*sqrt(1 - e**2*x**2/d**2)) - e*asin(e*x/d) + e**2*x/(d*sqrt(1 - e**2*x**2/d**
2)), True)) + 3*d**6*e*Piecewise((d**2/(e*x*sqrt(d**2/(e**2*x**2) - 1)) - d*acosh(d/(e*x)) - e*x/sqrt(d**2/(e*
*2*x**2) - 1), Abs(d**2/(e**2*x**2)) > 1), (-I*d**2/(e*x*sqrt(-d**2/(e**2*x**2) + 1)) + I*d*asin(d/(e*x)) + I*
e*x/sqrt(-d**2/(e**2*x**2) + 1), True)) + d**5*e**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 +
e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**2*asin(e*x/d)/(
2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) - 5*d**4*e**3*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**
2 - e**2*x**2)**(3/2)/(3*e**2), True)) - 5*d**3*e**4*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e*
*2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**
2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**4*asin(e*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*
d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + d**2*e**5*Piecewise((
-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*sqrt(d**2 - e**2*
x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) + 3*d*e**6*Piecewise((-I*d**6*acosh(e*x/d)/(16*e**5) + I*d**5*x
/(16*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**3/(48*e**2*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*x**5/(24*sqrt(-
1 + e**2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**6*asin(e*x/d
)/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2*x**2/d**2)) + d**3*x**3/(48*e**2*sqrt(1 - e**2*x**2/d**2)) + 5*d*x
**5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x**7/(6*d*sqrt(1 - e**2*x**2/d**2)), True)) + e**7*Piecewise((-8*d**6
*sqrt(d**2 - e**2*x**2)/(105*e**6) - 4*d**4*x**2*sqrt(d**2 - e**2*x**2)/(105*e**4) - d**2*x**4*sqrt(d**2 - e**
2*x**2)/(35*e**2) + x**6*sqrt(d**2 - e**2*x**2)/7, Ne(e, 0)), (x**6*sqrt(d**2)/6, True))

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