\(\int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx\) [316]

Optimal result

Integrand size = 22, antiderivative size = 255 \[ \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx=\frac {d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt {a+c x^2}}{8 c e^5}+\frac {\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac {13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac {(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}-\frac {d \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2} e^6}-\frac {d^4 \sqrt {c d^2+a e^2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^6} \]

[Out]

1/60*(-8*a*e^2+47*c*d^2)*(c*x^2+a)^(3/2)/c^2/e^3-13/20*d*(e*x+d)*(c*x^2+a)^(3/2)/c/e^3+1/5*(e*x+d)^2*(c*x^2+a)
^(3/2)/c/e^3-1/8*d*(-a^2*e^4+4*a*c*d^2*e^2+8*c^2*d^4)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)/e^6-d^4*arcta
nh((-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))*(a*e^2+c*d^2)^(1/2)/e^6+1/8*d*(8*c*d^3-e*(-a*e^2+4*c*d^2)
*x)*(c*x^2+a)^(1/2)/c/e^5

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1668, 829, 858, 223, 212, 739} \[ \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx=-\frac {d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right )}{8 c^{3/2} e^6}-\frac {d^4 \sqrt {a e^2+c d^2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^6}+\frac {\left (a+c x^2\right )^{3/2} \left (47 c d^2-8 a e^2\right )}{60 c^2 e^3}+\frac {d \sqrt {a+c x^2} \left (8 c d^3-e x \left (4 c d^2-a e^2\right )\right )}{8 c e^5}-\frac {13 d \left (a+c x^2\right )^{3/2} (d+e x)}{20 c e^3}+\frac {\left (a+c x^2\right )^{3/2} (d+e x)^2}{5 c e^3} \]

[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 212

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

Rule 223

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 739

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 829

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 858

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 1668

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*} \text {integral}& = \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 ) \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.90 \[ \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx=\frac {e \sqrt {a+c x^2} \left (-16 a^2 e^4+a c e^2 \left (40 d^2-15 d e x+8 e^2 x^2\right )+2 c^2 \left (60 d^4-30 d^3 e x+20 d^2 e^2 x^2-15 d e^3 x^3+12 e^4 x^4\right )\right )+240 c^2 d^4 \sqrt {-c d^2-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+15 \sqrt {c} d \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{120 c^2 e^6} \]

[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)) + 240*c^2*d^4*Sqrt[-(c*d^2) - a*e^2]*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqr
t[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]] + 15*Sqrt[c]*d*(8*c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*Log[-(Sqrt[c]*x) +
Sqrt[a + c*x^2]])/(120*c^2*e^6)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.22

method	result	size
risch	\(-\frac {\left (-24 e^{4} c^{2} x^{4}+30 d \,e^{3} c^{2} x^{3}-8 a c \,e^{4} x^{2}-40 c^{2} d^{2} e^{2} x^{2}+15 a c d \,e^{3} x +60 c^{2} d^{3} e x +16 a^{2} e^{4}-40 a c \,d^{2} e^{2}-120 c^{2} d^{4}\right ) \sqrt {c \,x^{2}+a}}{120 c^{2} e^{5}}+\frac {d \left (\frac {\left (a^{2} e^{4}-4 a c \,d^{2} e^{2}-8 c^{2} d^{4}\right ) \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{e \sqrt {c}}-\frac {8 d^{3} \left (e^{2} a +c \,d^{2}\right ) c \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{8 e^{5} c}\)	\(312\)
default	\(\frac {\frac {x^{2} \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{5 c}-\frac {2 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{15 c^{2}}}{e}+\frac {d^{2} \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3 e^{3} c}-\frac {d^{3} \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{e^{4}}-\frac {d \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4 c}-\frac {a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4 c}\right )}{e^{2}}+\frac {d^{4} \left (\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{e^{5}}\)	\(428\)

[In]

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

[Out]

-1/120*(-24*c^2*e^4*x^4+30*c^2*d*e^3*x^3-8*a*c*e^4*x^2-40*c^2*d^2*e^2*x^2+15*a*c*d*e^3*x+60*c^2*d^3*e*x+16*a^2
*e^4-40*a*c*d^2*e^2-120*c^2*d^4)*(c*x^2+a)^(1/2)/c^2/e^5+1/8*d/e^5/c*((a^2*e^4-4*a*c*d^2*e^2-8*c^2*d^4)/e*ln(x
*c^(1/2)+(c*x^2+a)^(1/2))/c^(1/2)-8*d^3*(a*e^2+c*d^2)*c/e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-
2/e*c*d*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c-2/e*c*d*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e)))

Fricas [A] (verification not implemented)

none

Time = 4.06 (sec) , antiderivative size = 1104, normalized size of antiderivative = 4.33 \[ \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx=\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 ] \]

[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)]

Sympy [F]

\[ \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx=\int \frac {x^{4} \sqrt {a + c x^{2}}}{d + e x}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F(-2)]

Exception generated. \[ \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \sqrt {a+c x^2}}{d+e x} \, dx=\int \frac {x^4\,\sqrt {c\,x^2+a}}{d+e\,x} \,d x \]

[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)