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

Optimal. Leaf size=195 \[ -\frac {d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2} e^4}+\frac {\sqrt {a+c x^2} \left (11 c d^2-4 a e^2\right )}{6 c^2 e^3}-\frac {d^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4 \sqrt {a e^2+c d^2}}-\frac {7 d \sqrt {a+c x^2} (d+e x)}{6 c e^3}+\frac {\sqrt {a+c x^2} (d+e x)^2}{3 c e^3} \]

[Out]

-1/2*d*(-a*e^2+2*c*d^2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)/e^4-d^4*arctanh((-c*d*x+a*e)/(a*e^2+c*d^2)^
(1/2)/(c*x^2+a)^(1/2))/e^4/(a*e^2+c*d^2)^(1/2)+1/6*(-4*a*e^2+11*c*d^2)*(c*x^2+a)^(1/2)/c^2/e^3-7/6*d*(e*x+d)*(
c*x^2+a)^(1/2)/c/e^3+1/3*(e*x+d)^2*(c*x^2+a)^(1/2)/c/e^3

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Rubi [A]  time = 0.48, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1654, 844, 217, 206, 725} \[ \frac {\sqrt {a+c x^2} \left (11 c d^2-4 a e^2\right )}{6 c^2 e^3}-\frac {d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2} e^4}-\frac {d^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4 \sqrt {a e^2+c d^2}}-\frac {7 d \sqrt {a+c x^2} (d+e x)}{6 c e^3}+\frac {\sqrt {a+c x^2} (d+e x)^2}{3 c e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*c*d^2 - 4*a*e^2)*Sqrt[a + c*x^2])/(6*c^2*e^3) - (7*d*(d + e*x)*Sqrt[a + c*x^2])/(6*c*e^3) + ((d + e*x)^2*
Sqrt[a + c*x^2])/(3*c*e^3) - (d*(2*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(3/2)*e^4) - (d^4
*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(e^4*Sqrt[c*d^2 + a*e^2])

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

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Mathematica [A]  time = 0.23, size = 149, normalized size = 0.76 \[ \frac {-\frac {3 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}+\frac {e \sqrt {a+c x^2} \left (-4 a e^2+6 c d^2-3 c d e x+2 c e^2 x^2\right )}{c^2}-\frac {6 d^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\sqrt {a e^2+c d^2}}}{6 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Sqrt[a + c*x^2]*(6*c*d^2 - 4*a*e^2 - 3*c*d*e*x + 2*c*e^2*x^2))/c^2 - (3*d*(2*c*d^2 - a*e^2)*ArcTanh[(Sqrt[
c]*x)/Sqrt[a + c*x^2]])/c^(3/2) - (6*d^4*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/Sqrt[c*
d^2 + a*e^2])/(6*e^4)

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fricas [A]  time = 5.99, size = 1060, normalized size = 5.44 \[ \left [\frac {6 \, \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 ) - 3 \, {\left (2 \, c^{2} d^{5} + 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 (6 \, c^{2} d^{4} e + 2 \, a c d^{2} e^{3} - 4 \, a^{2} e^{5} + 2 \, {\left (c^{2} d^{2} e^{3} + a c e^{5}\right )} x^{2} - 3 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{12 \, {\left (c^{3} d^{2} e^{4} + a c^{2} e^{6}\right )}}, -\frac {12 \, \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 ) + 3 \, {\left (2 \, c^{2} d^{5} + 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 (6 \, c^{2} d^{4} e + 2 \, a c d^{2} e^{3} - 4 \, a^{2} e^{5} + 2 \, {\left (c^{2} d^{2} e^{3} + a c e^{5}\right )} x^{2} - 3 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{12 \, {\left (c^{3} d^{2} e^{4} + a c^{2} e^{6}\right )}}, \frac {3 \, \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 ) + 3 \, {\left (2 \, c^{2} d^{5} + 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 (6 \, c^{2} d^{4} e + 2 \, a c d^{2} e^{3} - 4 \, a^{2} e^{5} + 2 \, {\left (c^{2} d^{2} e^{3} + a c e^{5}\right )} x^{2} - 3 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{6 \, {\left (c^{3} d^{2} e^{4} + a c^{2} e^{6}\right )}}, -\frac {6 \, \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 ) - 3 \, {\left (2 \, c^{2} d^{5} + 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 (6 \, c^{2} d^{4} e + 2 \, a c d^{2} e^{3} - 4 \, a^{2} e^{5} + 2 \, {\left (c^{2} d^{2} e^{3} + a c e^{5}\right )} x^{2} - 3 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{6 \, {\left (c^{3} d^{2} e^{4} + a c^{2} e^{6}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(6*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*sq
rt(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)) - 3*(2*c^2*d^5 + 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*(6*c^2*d^4*e + 2*a*c*d^2*e^3 - 4*a^2*e^5 +
2*(c^2*d^2*e^3 + a*c*e^5)*x^2 - 3*(c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^4 + a*c^2*e^6), -1
/12*(12*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)) + 3*(2*c^2*d^5 + 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*(6*c^2*d^4*e + 2*a*c*d^2*e^3 - 4*a^2*e^5 + 2*(c^2*d^2*e^3 + a*c*e^5)*x^2 - 3*(c^2*d^3*
e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^4 + a*c^2*e^6), 1/6*(3*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)) + 3*(2*c^2*d^5 + a*c*d^3*e^2 - a^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 +
a)) + (6*c^2*d^4*e + 2*a*c*d^2*e^3 - 4*a^2*e^5 + 2*(c^2*d^2*e^3 + a*c*e^5)*x^2 - 3*(c^2*d^3*e^2 + a*c*d*e^4)*x
)*sqrt(c*x^2 + a))/(c^3*d^2*e^4 + a*c^2*e^6), -1/6*(6*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)) - 3*(2*c^2*d^5 + a*c*d^3*e^2 - a
^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (6*c^2*d^4*e + 2*a*c*d^2*e^3 - 4*a^2*e^5 + 2*(c^2*d^2*
e^3 + a*c*e^5)*x^2 - 3*(c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^2*e^4 + a*c^2*e^6)]

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giac [A]  time = 0.21, size = 163, normalized size = 0.84 \[ \frac {2 \, d^{4} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{\left (-4\right )}}{\sqrt {-c d^{2} - a e^{2}}} + \frac {1}{6} \, \sqrt {c x^{2} + a} {\left (x {\left (\frac {2 \, x e^{\left (-1\right )}}{c} - \frac {3 \, d e^{\left (-2\right )}}{c}\right )} + \frac {2 \, {\left (3 \, c^{2} d^{2} e^{7} - 2 \, a c e^{9}\right )} e^{\left (-10\right )}}{c^{3}}\right )} + \frac {{\left (2 \, c^{\frac {3}{2}} d^{3} - a \sqrt {c} d e^{2}\right )} e^{\left (-4\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^4*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-4)/sqrt(-c*d^2 - a*e^2)
+ 1/6*sqrt(c*x^2 + a)*(x*(2*x*e^(-1)/c - 3*d*e^(-2)/c) + 2*(3*c^2*d^2*e^7 - 2*a*c*e^9)*e^(-10)/c^3) + 1/2*(2*c
^(3/2)*d^3 - a*sqrt(c)*d*e^2)*e^(-4)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^2

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maple [A]  time = 0.02, size = 260, normalized size = 1.33 \[ \frac {\sqrt {c \,x^{2}+a}\, x^{2}}{3 c e}-\frac {d^{4} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{5}}+\frac {a d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {3}{2}} e^{2}}-\frac {d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}\, e^{4}}-\frac {\sqrt {c \,x^{2}+a}\, d x}{2 c \,e^{2}}-\frac {2 \sqrt {c \,x^{2}+a}\, a}{3 c^{2} e}+\frac {\sqrt {c \,x^{2}+a}\, d^{2}}{c \,e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 171, normalized size = 0.88 \[ \frac {\sqrt {c x^{2} + a} x^{2}}{3 \, c e} - \frac {\sqrt {c x^{2} + a} d x}{2 \, c e^{2}} - \frac {d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {c} e^{4}} + \frac {a d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {3}{2}} e^{2}} + \frac {d^{4} \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{\sqrt {a + \frac {c d^{2}}{e^{2}}} e^{5}} + \frac {\sqrt {c x^{2} + a} d^{2}}{c e^{3}} - \frac {2 \, \sqrt {c x^{2} + a} a}{3 \, c^{2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{\sqrt {c\,x^2+a}\,\left (d+e\,x\right )} \,d x \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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