3.3.27 \(\int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx\)

Optimal. Leaf size=255 \[ -\frac {d \left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2} e^6}+\frac {\left (a+c x^2\right )^{3/2} \left (47 c d^2-8 a e^2\right )}{60 c^2 e^3}-\frac {d^4 \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^6}+\frac {d \sqrt {a+c x^2} \left (8 c d^3-e x \left (4 c d^2-a e^2\right )\right )}{8 c e^5}-\frac {13 d \left (a+c x^2\right )^{3/2} (d+e x)}{20 c e^3}+\frac {\left (a+c x^2\right )^{3/2} (d+e x)^2}{5 c e^3} \]

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Rubi [A]  time = 0.63, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1654, 815, 844, 217, 206, 725} \begin {gather*} -\frac {d \left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2} e^6}+\frac {\left (a+c x^2\right )^{3/2} \left (47 c d^2-8 a e^2\right )}{60 c^2 e^3}+\frac {d \sqrt {a+c x^2} \left (8 c d^3-e x \left (4 c d^2-a e^2\right )\right )}{8 c e^5}-\frac {d^4 \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^6}-\frac {13 d \left (a+c x^2\right )^{3/2} (d+e x)}{20 c e^3}+\frac {\left (a+c x^2\right )^{3/2} (d+e x)^2}{5 c e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*Sqrt[a + c*x^2])/(d + e*x),x]

[Out]

(d*(8*c*d^3 - e*(4*c*d^2 - a*e^2)*x)*Sqrt[a + c*x^2])/(8*c*e^5) + ((47*c*d^2 - 8*a*e^2)*(a + c*x^2)^(3/2))/(60
*c^2*e^3) - (13*d*(d + e*x)*(a + c*x^2)^(3/2))/(20*c*e^3) + ((d + e*x)^2*(a + c*x^2)^(3/2))/(5*c*e^3) - (d*(8*
c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(3/2)*e^6) - (d^4*Sqrt[c*d^2 + a
*e^2]*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/e^6

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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 1654

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

Rubi steps

\begin {align*} \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx &=\frac {(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac {\int \frac {\sqrt {a+c x^2} \left (-2 a d^2 e^2-d e \left (3 c d^2+4 a e^2\right ) x-e^2 \left (11 c d^2+2 a e^2\right ) x^2-13 c d e^3 x^3\right )}{d+e x} \, dx}{5 c e^4}\\ &=-\frac {13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac {(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac {\int \frac {\sqrt {a+c x^2} \left (5 a c d^2 e^5+3 c d e^4 \left (9 c d^2-a e^2\right ) x+c e^5 \left (47 c d^2-8 a e^2\right ) x^2\right )}{d+e x} \, dx}{20 c^2 e^7}\\ &=\frac {\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac {13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac {(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac {\int \frac {\left (15 a c^2 d^2 e^7-15 c^2 d e^6 \left (4 c d^2-a e^2\right ) x\right ) \sqrt {a+c x^2}}{d+e x} \, dx}{60 c^3 e^9}\\ &=\frac {d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 c e^5}+\frac {\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac {13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac {(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac {\int \frac {15 a c^3 d^2 e^7 \left (4 c d^2+a e^2\right )-15 c^3 d e^6 \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) x}{(d+e x) \sqrt {a+c x^2}} \, dx}{120 c^4 e^{11}}\\ &=\frac {d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 c e^5}+\frac {\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac {13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac {(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac {\left (d^4 \left (c d^2+a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{e^6}-\frac {\left (d \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 c e^6}\\ &=\frac {d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 c e^5}+\frac {\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac {13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac {(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}-\frac {\left (d^4 \left (c d^2+a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{e^6}-\frac {\left (d \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{8 c e^6}\\ &=\frac {d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 c e^5}+\frac {\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac {13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac {(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}-\frac {d \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2} e^6}-\frac {d^4 \sqrt {c d^2+a e^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^6}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 259, normalized size = 1.02 \begin {gather*} \frac {e \sqrt {a+c x^2} \left (-16 a^2 e^4+a c e^2 \left (40 d^2-15 d e x+8 e^2 x^2\right )+2 c^2 \left (60 d^4-30 d^3 e x+20 d^2 e^2 x^2-15 d e^3 x^3+12 e^4 x^4\right )\right )-120 c^{5/2} d^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-120 c^2 d^4 \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )+\frac {15 \sqrt {a} \sqrt {c} d e^2 \sqrt {a+c x^2} \left (a e^2-4 c d^2\right ) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {\frac {c x^2}{a}+1}}}{120 c^2 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*Sqrt[a + c*x^2])/(d + e*x),x]

[Out]

(e*Sqrt[a + c*x^2]*(-16*a^2*e^4 + a*c*e^2*(40*d^2 - 15*d*e*x + 8*e^2*x^2) + 2*c^2*(60*d^4 - 30*d^3*e*x + 20*d^
2*e^2*x^2 - 15*d*e^3*x^3 + 12*e^4*x^4)) + (15*Sqrt[a]*Sqrt[c]*d*e^2*(-4*c*d^2 + a*e^2)*Sqrt[a + c*x^2]*ArcSinh
[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^2)/a] - 120*c^(5/2)*d^5*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] - 120*c^2*d^
4*Sqrt[c*d^2 + a*e^2]*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(120*c^2*e^6)

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IntegrateAlgebraic [A]  time = 0.91, size = 283, normalized size = 1.11 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-16 a^2 e^4+40 a c d^2 e^2-15 a c d e^3 x+8 a c e^4 x^2+120 c^2 d^4-60 c^2 d^3 e x+40 c^2 d^2 e^2 x^2-30 c^2 d e^3 x^3+24 c^2 e^4 x^4\right )}{120 c^2 e^5}+\frac {\left (-a^2 d e^4+4 a c d^3 e^2+8 c^2 d^5\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{8 c^{3/2} e^6}+\frac {2 d^4 \sqrt {-a e^2-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+c x^2}}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2-c d^2}}\right )}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*Sqrt[a + c*x^2])/(d + e*x),x]

[Out]

(Sqrt[a + c*x^2]*(120*c^2*d^4 + 40*a*c*d^2*e^2 - 16*a^2*e^4 - 60*c^2*d^3*e*x - 15*a*c*d*e^3*x + 40*c^2*d^2*e^2
*x^2 + 8*a*c*e^4*x^2 - 30*c^2*d*e^3*x^3 + 24*c^2*e^4*x^4))/(120*c^2*e^5) + (2*d^4*Sqrt[-(c*d^2) - a*e^2]*ArcTa
n[(Sqrt[c]*d)/Sqrt[-(c*d^2) - a*e^2] + (Sqrt[c]*e*x)/Sqrt[-(c*d^2) - a*e^2] - (e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2
) - a*e^2]])/e^6 + ((8*c^2*d^5 + 4*a*c*d^3*e^2 - a^2*d*e^4)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(8*c^(3/2)*e^
6)

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fricas [A]  time = 6.90, size = 1104, normalized size = 4.33 \begin {gather*} \left [\frac {120 \, \sqrt {c d^{2} + a e^{2}} c^{2} d^{4} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 15 \, {\left (8 \, c^{2} d^{5} + 4 \, a c d^{3} e^{2} - a^{2} d e^{4}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (24 \, c^{2} e^{5} x^{4} - 30 \, c^{2} d e^{4} x^{3} + 120 \, c^{2} d^{4} e + 40 \, a c d^{2} e^{3} - 16 \, a^{2} e^{5} + 8 \, {\left (5 \, c^{2} d^{2} e^{3} + a c e^{5}\right )} x^{2} - 15 \, {\left (4 \, c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{240 \, c^{2} e^{6}}, -\frac {240 \, \sqrt {-c d^{2} - a e^{2}} c^{2} d^{4} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) + 15 \, {\left (8 \, c^{2} d^{5} + 4 \, a c d^{3} e^{2} - a^{2} d e^{4}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (24 \, c^{2} e^{5} x^{4} - 30 \, c^{2} d e^{4} x^{3} + 120 \, c^{2} d^{4} e + 40 \, a c d^{2} e^{3} - 16 \, a^{2} e^{5} + 8 \, {\left (5 \, c^{2} d^{2} e^{3} + a c e^{5}\right )} x^{2} - 15 \, {\left (4 \, c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{240 \, c^{2} e^{6}}, \frac {60 \, \sqrt {c d^{2} + a e^{2}} c^{2} d^{4} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 15 \, {\left (8 \, c^{2} d^{5} + 4 \, a c d^{3} e^{2} - a^{2} d e^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (24 \, c^{2} e^{5} x^{4} - 30 \, c^{2} d e^{4} x^{3} + 120 \, c^{2} d^{4} e + 40 \, a c d^{2} e^{3} - 16 \, a^{2} e^{5} + 8 \, {\left (5 \, c^{2} d^{2} e^{3} + a c e^{5}\right )} x^{2} - 15 \, {\left (4 \, c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{120 \, c^{2} e^{6}}, -\frac {120 \, \sqrt {-c d^{2} - a e^{2}} c^{2} d^{4} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - 15 \, {\left (8 \, c^{2} d^{5} + 4 \, a c d^{3} e^{2} - a^{2} d e^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (24 \, c^{2} e^{5} x^{4} - 30 \, c^{2} d e^{4} x^{3} + 120 \, c^{2} d^{4} e + 40 \, a c d^{2} e^{3} - 16 \, a^{2} e^{5} + 8 \, {\left (5 \, c^{2} d^{2} e^{3} + a c e^{5}\right )} x^{2} - 15 \, {\left (4 \, c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{120 \, c^{2} e^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="fricas")

[Out]

[1/240*(120*sqrt(c*d^2 + a*e^2)*c^2*d^4*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2
*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) - 15*(8*c^2*d^5 + 4*a*c*d^3*e^2
- a^2*d*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(24*c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 +
120*c^2*d^4*e + 40*a*c*d^2*e^3 - 16*a^2*e^5 + 8*(5*c^2*d^2*e^3 + a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + a*c*d*e^4
)*x)*sqrt(c*x^2 + a))/(c^2*e^6), -1/240*(240*sqrt(-c*d^2 - a*e^2)*c^2*d^4*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x -
a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) + 15*(8*c^2*d^5 + 4*a*c*d^3*e^2 - a^2*d*e
^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(24*c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 + 120*c^2*d
^4*e + 40*a*c*d^2*e^3 - 16*a^2*e^5 + 8*(5*c^2*d^2*e^3 + a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(
c*x^2 + a))/(c^2*e^6), 1/120*(60*sqrt(c*d^2 + a*e^2)*c^2*d^4*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d
^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + 15*(8*c^
2*d^5 + 4*a*c*d^3*e^2 - a^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (24*c^2*e^5*x^4 - 30*c^2*d*e^
4*x^3 + 120*c^2*d^4*e + 40*a*c*d^2*e^3 - 16*a^2*e^5 + 8*(5*c^2*d^2*e^3 + a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + a*
c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^2*e^6), -1/120*(120*sqrt(-c*d^2 - a*e^2)*c^2*d^4*arctan(sqrt(-c*d^2 - a*e^2)*(
c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - 15*(8*c^2*d^5 + 4*a*c*d^3*e^2 -
a^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (24*c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 + 120*c^2*d^4*e +
40*a*c*d^2*e^3 - 16*a^2*e^5 + 8*(5*c^2*d^2*e^3 + a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 +
a))/(c^2*e^6)]

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giac [A]  time = 0.21, size = 252, normalized size = 0.99 \begin {gather*} \frac {2 \, {\left (c d^{6} + a d^{4} e^{2}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{\left (-6\right )}}{\sqrt {-c d^{2} - a e^{2}}} + \frac {1}{120} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (4 \, x e^{\left (-1\right )} - 5 \, d e^{\left (-2\right )}\right )} x + \frac {4 \, {\left (5 \, c^{3} d^{2} e^{18} + a c^{2} e^{20}\right )} e^{\left (-21\right )}}{c^{3}}\right )} x - \frac {15 \, {\left (4 \, c^{3} d^{3} e^{17} + a c^{2} d e^{19}\right )} e^{\left (-21\right )}}{c^{3}}\right )} x + \frac {8 \, {\left (15 \, c^{3} d^{4} e^{16} + 5 \, a c^{2} d^{2} e^{18} - 2 \, a^{2} c e^{20}\right )} e^{\left (-21\right )}}{c^{3}}\right )} + \frac {{\left (8 \, c^{\frac {5}{2}} d^{5} + 4 \, a c^{\frac {3}{2}} d^{3} e^{2} - a^{2} \sqrt {c} d e^{4}\right )} e^{\left (-6\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{8 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="giac")

[Out]

2*(c*d^6 + a*d^4*e^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-6)/sqrt(
-c*d^2 - a*e^2) + 1/120*sqrt(c*x^2 + a)*((2*(3*(4*x*e^(-1) - 5*d*e^(-2))*x + 4*(5*c^3*d^2*e^18 + a*c^2*e^20)*e
^(-21)/c^3)*x - 15*(4*c^3*d^3*e^17 + a*c^2*d*e^19)*e^(-21)/c^3)*x + 8*(15*c^3*d^4*e^16 + 5*a*c^2*d^2*e^18 - 2*
a^2*c*e^20)*e^(-21)/c^3) + 1/8*(8*c^(5/2)*d^5 + 4*a*c^(3/2)*d^3*e^2 - a^2*sqrt(c)*d*e^4)*e^(-6)*log(abs(-sqrt(
c)*x + sqrt(c*x^2 + a)))/c^2

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maple [B]  time = 0.02, size = 560, normalized size = 2.20 \begin {gather*} -\frac {a \,d^{4} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{5}}-\frac {c \,d^{6} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{7}}+\frac {a^{2} d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {3}{2}} e^{2}}-\frac {a \,d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}\, e^{4}}-\frac {\sqrt {c}\, d^{5} \ln \left (\frac {-\frac {c d}{e}+\left (x +\frac {d}{e}\right ) c}{\sqrt {c}}+\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\right )}{e^{6}}+\frac {\sqrt {c \,x^{2}+a}\, a d x}{8 c \,e^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} x^{2}}{5 c e}-\frac {\sqrt {c \,x^{2}+a}\, d^{3} x}{2 e^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} d x}{4 c \,e^{2}}+\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{4}}{e^{5}}-\frac {2 \left (c \,x^{2}+a \right )^{\frac {3}{2}} a}{15 c^{2} e}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} d^{2}}{3 c \,e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(c*x^2+a)^(1/2)/(e*x+d),x)

[Out]

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

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maxima [A]  time = 0.68, size = 249, normalized size = 0.98 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} x^{2}}{5 \, c e} - \frac {\sqrt {c x^{2} + a} d^{3} x}{2 \, e^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} d x}{4 \, c e^{2}} + \frac {\sqrt {c x^{2} + a} a d x}{8 \, c e^{2}} - \frac {\sqrt {c} d^{5} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{e^{6}} - \frac {a d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {c} e^{4}} + \frac {a^{2} d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, c^{\frac {3}{2}} e^{2}} + \frac {\sqrt {a + \frac {c d^{2}}{e^{2}}} d^{4} \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{e^{5}} + \frac {\sqrt {c x^{2} + a} d^{4}}{e^{5}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{2}}{3 \, c e^{3}} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a}{15 \, c^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="maxima")

[Out]

1/5*(c*x^2 + a)^(3/2)*x^2/(c*e) - 1/2*sqrt(c*x^2 + a)*d^3*x/e^4 - 1/4*(c*x^2 + a)^(3/2)*d*x/(c*e^2) + 1/8*sqrt
(c*x^2 + a)*a*d*x/(c*e^2) - sqrt(c)*d^5*arcsinh(c*x/sqrt(a*c))/e^6 - 1/2*a*d^3*arcsinh(c*x/sqrt(a*c))/(sqrt(c)
*e^4) + 1/8*a^2*d*arcsinh(c*x/sqrt(a*c))/(c^(3/2)*e^2) + sqrt(a + c*d^2/e^2)*d^4*arcsinh(c*d*x/(sqrt(a*c)*abs(
e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/e^5 + sqrt(c*x^2 + a)*d^4/e^5 + 1/3*(c*x^2 + a)^(3/2)*d^2/(c*e^3) -
2/15*(c*x^2 + a)^(3/2)*a/(c^2*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,\sqrt {c\,x^2+a}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + c*x^2)^(1/2))/(d + e*x),x)

[Out]

int((x^4*(a + c*x^2)^(1/2))/(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \sqrt {a + c x^{2}}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(c*x**2+a)**(1/2)/(e*x+d),x)

[Out]

Integral(x**4*sqrt(a + c*x**2)/(d + e*x), x)

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