3.74 \(\int \frac {(d+e x)^3 (d^2-e^2 x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=210 \[ -\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {1}{12} d e^3 (26 d+25 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {25}{8} d^5 e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {13}{2} d^5 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-\frac {1}{8} d^3 e^3 (52 d+25 e x) \sqrt {d^2-e^2 x^2} \]

[Out]

-1/12*d*e^3*(25*e*x+26*d)*(-e^2*x^2+d^2)^(3/2)-1/30*e^2*(39*e*x+50*d)*(-e^2*x^2+d^2)^(5/2)/x-1/3*d*(-e^2*x^2+d
^2)^(7/2)/x^3-3/2*e*(-e^2*x^2+d^2)^(7/2)/x^2-25/8*d^5*e^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))+13/2*d^5*e^3*arctan
h((-e^2*x^2+d^2)^(1/2)/d)-1/8*d^3*e^3*(25*e*x+52*d)*(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 210, 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, 813, 815, 844, 217, 203, 266, 63, 208} \[ -\frac {1}{8} d^3 e^3 (52 d+25 e x) \sqrt {d^2-e^2 x^2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {1}{12} d e^3 (26 d+25 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {25}{8} d^5 e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {13}{2} d^5 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^4,x]

[Out]

-(d^3*e^3*(52*d + 25*e*x)*Sqrt[d^2 - e^2*x^2])/8 - (d*e^3*(26*d + 25*e*x)*(d^2 - e^2*x^2)^(3/2))/12 - (e^2*(50
*d + 39*e*x)*(d^2 - e^2*x^2)^(5/2))/(30*x) - (d*(d^2 - e^2*x^2)^(7/2))/(3*x^3) - (3*e*(d^2 - e^2*x^2)^(7/2))/(
2*x^2) - (25*d^5*e^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/8 + (13*d^5*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/2

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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])

Rubi steps

\begin {align*} \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}{x^4} \, dx &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{5/2} \left (-9 d^4 e-5 d^3 e^2 x-3 d^2 e^3 x^2\right )}{x^3} \, dx}{3 d^2}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}+\frac {\int \frac {\left (10 d^5 e^2-39 d^4 e^3 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x^2} \, dx}{6 d^4}\\ &=-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {\int \frac {\left (78 d^6 e^3+100 d^5 e^4 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x} \, dx}{12 d^4}\\ &=-\frac {1}{12} d e^3 (26 d+25 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}+\frac {\int \frac {\left (-312 d^8 e^5-300 d^7 e^6 x\right ) \sqrt {d^2-e^2 x^2}}{x} \, dx}{48 d^4 e^2}\\ &=-\frac {1}{8} d^3 e^3 (52 d+25 e x) \sqrt {d^2-e^2 x^2}-\frac {1}{12} d e^3 (26 d+25 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {\int \frac {624 d^{10} e^7+300 d^9 e^8 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{96 d^4 e^4}\\ &=-\frac {1}{8} d^3 e^3 (52 d+25 e x) \sqrt {d^2-e^2 x^2}-\frac {1}{12} d e^3 (26 d+25 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {1}{2} \left (13 d^6 e^3\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-\frac {1}{8} \left (25 d^5 e^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {1}{8} d^3 e^3 (52 d+25 e x) \sqrt {d^2-e^2 x^2}-\frac {1}{12} d e^3 (26 d+25 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {1}{4} \left (13 d^6 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )-\frac {1}{8} \left (25 d^5 e^4\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 {1}{8} d^3 e^3 (52 d+25 e x) \sqrt {d^2-e^2 x^2}-\frac {1}{12} d e^3 (26 d+25 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {25}{8} d^5 e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {1}{2} \left (13 d^6 e\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 )\\ &=-\frac {1}{8} d^3 e^3 (52 d+25 e x) \sqrt {d^2-e^2 x^2}-\frac {1}{12} d e^3 (26 d+25 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {e^2 (50 d+39 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 x}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{3 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {25}{8} d^5 e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {13}{2} d^5 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [C]  time = 0.27, size = 251, normalized size = 1.20 \[ -\frac {3 e^3 \left (d^2-e^2 x^2\right )^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};1-\frac {e^2 x^2}{d^2}\right )}{7 d^2}-\frac {d^7 \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{3 x^3 \sqrt {1-\frac {e^2 x^2}{d^2}}}-\frac {3 d^5 e^2 \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x \sqrt {1-\frac {e^2 x^2}{d^2}}}+\frac {1}{15} e^3 \left (\sqrt {d^2-e^2 x^2} \left (23 d^4-11 d^2 e^2 x^2+3 e^4 x^4\right )-15 d^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^4,x]

[Out]

(e^3*(Sqrt[d^2 - e^2*x^2]*(23*d^4 - 11*d^2*e^2*x^2 + 3*e^4*x^4) - 15*d^5*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]))/15 -
(d^7*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-5/2, -3/2, -1/2, (e^2*x^2)/d^2])/(3*x^3*Sqrt[1 - (e^2*x^2)/d^2])
- (3*d^5*e^2*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]
) - (3*e^3*(d^2 - e^2*x^2)^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, 1 - (e^2*x^2)/d^2])/(7*d^2)

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fricas [A]  time = 0.89, size = 179, normalized size = 0.85 \[ \frac {750 \, d^{5} e^{3} x^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 780 \, d^{5} e^{3} x^{3} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - 656 \, d^{5} e^{3} x^{3} + {\left (24 \, e^{7} x^{7} + 90 \, d e^{6} x^{6} + 32 \, d^{2} e^{5} x^{5} - 345 \, d^{3} e^{4} x^{4} - 656 \, d^{4} e^{3} x^{3} - 80 \, d^{5} e^{2} x^{2} - 180 \, d^{6} e x - 40 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, x^{3}} \]

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^4,x, algorithm="fricas")

[Out]

1/120*(750*d^5*e^3*x^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 780*d^5*e^3*x^3*log(-(d - sqrt(-e^2*x^2 + d
^2))/x) - 656*d^5*e^3*x^3 + (24*e^7*x^7 + 90*d*e^6*x^6 + 32*d^2*e^5*x^5 - 345*d^3*e^4*x^4 - 656*d^4*e^3*x^3 -
80*d^5*e^2*x^2 - 180*d^6*e*x - 40*d^7)*sqrt(-e^2*x^2 + d^2))/x^3

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giac [A]  time = 0.29, size = 318, normalized size = 1.51 \[ -\frac {25}{8} \, d^{5} \arcsin \left (\frac {x e}{d}\right ) e^{3} \mathrm {sgn}\relax (d) + \frac {13}{2} \, d^{5} e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right ) + \frac {{\left (d^{5} e^{8} + \frac {9 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{5} e^{6}}{x} + \frac {9 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{5} e^{4}}{x^{2}}\right )} x^{3} e}{24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3}} - \frac {1}{24} \, {\left (\frac {9 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{5} e^{16}}{x} + \frac {9 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{5} e^{14}}{x^{2}} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{5} e^{12}}{x^{3}}\right )} e^{\left (-15\right )} - \frac {1}{120} \, {\left (656 \, d^{4} e^{3} + {\left (345 \, d^{3} e^{4} - 2 \, {\left (16 \, d^{2} e^{5} + 3 \, {\left (4 \, x e^{7} + 15 \, d e^{6}\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]

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^4,x, algorithm="giac")

[Out]

-25/8*d^5*arcsin(x*e/d)*e^3*sgn(d) + 13/2*d^5*e^3*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)
) + 1/24*(d^5*e^8 + 9*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^5*e^6/x + 9*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^5*e^4/x^
2)*x^3*e/(d*e + sqrt(-x^2*e^2 + d^2)*e)^3 - 1/24*(9*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^5*e^16/x + 9*(d*e + sqrt(
-x^2*e^2 + d^2)*e)^2*d^5*e^14/x^2 + (d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^5*e^12/x^3)*e^(-15) - 1/120*(656*d^4*e^
3 + (345*d^3*e^4 - 2*(16*d^2*e^5 + 3*(4*x*e^7 + 15*d*e^6)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)

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maple [A]  time = 0.02, size = 277, normalized size = 1.32 \[ \frac {13 d^{6} e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}-\frac {25 d^{5} e^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {25 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} e^{4} x}{8}-\frac {13 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} e^{3}}{2}-\frac {25 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d \,e^{4} x}{12}-\frac {13 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2} e^{3}}{6}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4} x}{3 d}-\frac {13 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}{10}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{3 d x}-\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{2 x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d}{3 x^{3}} \]

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^4,x)

[Out]

-1/3*d*(-e^2*x^2+d^2)^(7/2)/x^3-5/3/d*e^2/x*(-e^2*x^2+d^2)^(7/2)-5/3/d*e^4*x*(-e^2*x^2+d^2)^(5/2)-25/12*d*e^4*
x*(-e^2*x^2+d^2)^(3/2)-25/8*d^3*e^4*x*(-e^2*x^2+d^2)^(1/2)-25/8*d^5*e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x
^2+d^2)^(1/2)*x)-3/2*e*(-e^2*x^2+d^2)^(7/2)/x^2-13/10*e^3*(-e^2*x^2+d^2)^(5/2)-13/6*d^2*e^3*(-e^2*x^2+d^2)^(3/
2)-13/2*d^4*e^3*(-e^2*x^2+d^2)^(1/2)+13/2*d^6*e^3/(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 = 1.01, size = 226, normalized size = 1.08 \[ -\frac {25}{8} \, d^{5} e^{3} \arcsin \left (\frac {e x}{d}\right ) + \frac {13}{2} \, d^{5} e^{3} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {25}{8} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} e^{4} x - \frac {13}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} e^{3} - \frac {25}{12} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{4} x - \frac {13}{6} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{3} - \frac {13}{10} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d e^{2}}{3 \, x} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e}{2 \, x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} d}{3 \, x^{3}} \]

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^4,x, algorithm="maxima")

[Out]

-25/8*d^5*e^3*arcsin(e*x/d) + 13/2*d^5*e^3*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) - 25/8*sqrt(-e^
2*x^2 + d^2)*d^3*e^4*x - 13/2*sqrt(-e^2*x^2 + d^2)*d^4*e^3 - 25/12*(-e^2*x^2 + d^2)^(3/2)*d*e^4*x - 13/6*(-e^2
*x^2 + d^2)^(3/2)*d^2*e^3 - 13/10*(-e^2*x^2 + d^2)^(5/2)*e^3 - 5/3*(-e^2*x^2 + d^2)^(5/2)*d*e^2/x - 3/2*(-e^2*
x^2 + d^2)^(7/2)*e/x^2 - 1/3*(-e^2*x^2 + d^2)^(7/2)*d/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}\,{\left (d+e\,x\right )}^3}{x^4} \,d x \]

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^4,x)

[Out]

int(((d^2 - e^2*x^2)^(5/2)*(d + e*x)^3)/x^4, x)

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sympy [C]  time = 15.74, size = 911, normalized size = 4.34 \[ \text {result too large to display} \]

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**4,x)

[Out]

d**7*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x**2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**2), Abs(d**2/(e
**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*x**2) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**2), Tru
e)) + 3*d**6*e*Piecewise((-d**2/(2*e*x**3*sqrt(d**2/(e**2*x**2) - 1)) + e/(2*x*sqrt(d**2/(e**2*x**2) - 1)) + e
**2*acosh(d/(e*x))/(2*d), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(2*x) - I*e**2*asin(d/
(e*x))/(2*d), True)) + d**5*e**2*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)) - 5*d**4*e**3*Piecewise((d**2/(e*x*sqrt(d**2/(e**2*x**2) - 1)) - d*aco
sh(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)) - 5*d**3*e**4*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)) + d**2*e**5*Piecewise((x**2*sqr
t(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + 3*d*e**6*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*sq
rt(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)), Tr
ue)) + e**7*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))

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