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

Optimal. Leaf size=204 \[ \frac {\left (6 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2} e^4}-\frac {d^4 \sqrt {a+c x^2}}{e^3 (d+e x) \left (a e^2+c d^2\right )}+\frac {d^3 \left (4 a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4 \left (a e^2+c d^2\right )^{3/2}}-\frac {5 d \sqrt {a+c x^2}}{2 c e^3}+\frac {\sqrt {a+c x^2} (d+e x)}{2 c e^3} \]

[Out]

1/2*(-a*e^2+6*c*d^2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)/e^4+d^3*(4*a*e^2+3*c*d^2)*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)^(3/2)-5/2*d*(c*x^2+a)^(1/2)/c/e^3-d^4*(c*x^2+a)^(1/2)/
e^3/(a*e^2+c*d^2)/(e*x+d)+1/2*(e*x+d)*(c*x^2+a)^(1/2)/c/e^3

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Rubi [A]  time = 0.52, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1651, 1654, 844, 217, 206, 725} \[ \frac {\left (6 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2} e^4}-\frac {d^4 \sqrt {a+c x^2}}{e^3 (d+e x) \left (a e^2+c d^2\right )}+\frac {d^3 \left (4 a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4 \left (a e^2+c d^2\right )^{3/2}}-\frac {5 d \sqrt {a+c x^2}}{2 c e^3}+\frac {\sqrt {a+c x^2} (d+e x)}{2 c e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1651

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1
)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

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

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Mathematica [A]  time = 0.37, size = 208, normalized size = 1.02 \[ \frac {\frac {\left (6 c d^2-a e^2\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )}{c^{3/2}}+e \sqrt {a+c x^2} \left (\frac {e x-4 d}{c}-\frac {2 d^4}{(d+e x) \left (a e^2+c d^2\right )}\right )+\frac {2 d^3 \left (4 a e^2+3 c d^2\right ) \log \left (\sqrt {a+c x^2} \sqrt {a e^2+c d^2}+a e-c d x\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac {2 d^3 \left (4 a e^2+3 c d^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}}}{2 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 152.35, size = 1786, normalized size = 8.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((6*c^3*d^7 + 11*a*c^2*d^5*e^2 + 4*a^2*c*d^3*e^4 - a^3*d*e^6 + (6*c^3*d^6*e + 11*a*c^2*d^4*e^3 + 4*a^2*c
*d^2*e^5 - a^3*e^7)*x)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(3*c^3*d^6 + 4*a*c^2*d^4*e^
2 + (3*c^3*d^5*e + 4*a*c^2*d^3*e^3)*x)*sqrt(c*d^2 + a*e^2)*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)) + 2*(6*c^3*d
^6*e + 10*a*c^2*d^4*e^3 + 4*a^2*c*d^2*e^5 - (c^3*d^4*e^3 + 2*a*c^2*d^2*e^5 + a^2*c*e^7)*x^2 + 3*(c^3*d^5*e^2 +
2*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x)*sqrt(c*x^2 + a))/(c^4*d^5*e^4 + 2*a*c^3*d^3*e^6 + a^2*c^2*d*e^8 + (c^4*d^4*
e^5 + 2*a*c^3*d^2*e^7 + a^2*c^2*e^9)*x), 1/4*(4*(3*c^3*d^6 + 4*a*c^2*d^4*e^2 + (3*c^3*d^5*e + 4*a*c^2*d^3*e^3)
*x)*sqrt(-c*d^2 - a*e^2)*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)) - (6*c^3*d^7 + 11*a*c^2*d^5*e^2 + 4*a^2*c*d^3*e^4 - a^3*d*e^6 + (6*c^3*d^6*e + 11*a*c^2*d^
4*e^3 + 4*a^2*c*d^2*e^5 - a^3*e^7)*x)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(6*c^3*d^6*e
+ 10*a*c^2*d^4*e^3 + 4*a^2*c*d^2*e^5 - (c^3*d^4*e^3 + 2*a*c^2*d^2*e^5 + a^2*c*e^7)*x^2 + 3*(c^3*d^5*e^2 + 2*a
*c^2*d^3*e^4 + a^2*c*d*e^6)*x)*sqrt(c*x^2 + a))/(c^4*d^5*e^4 + 2*a*c^3*d^3*e^6 + a^2*c^2*d*e^8 + (c^4*d^4*e^5
+ 2*a*c^3*d^2*e^7 + a^2*c^2*e^9)*x), -1/2*((6*c^3*d^7 + 11*a*c^2*d^5*e^2 + 4*a^2*c*d^3*e^4 - a^3*d*e^6 + (6*c^
3*d^6*e + 11*a*c^2*d^4*e^3 + 4*a^2*c*d^2*e^5 - a^3*e^7)*x)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (3*c^
3*d^6 + 4*a*c^2*d^4*e^2 + (3*c^3*d^5*e + 4*a*c^2*d^3*e^3)*x)*sqrt(c*d^2 + a*e^2)*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)) + (6*c^3*d^6*e + 10*a*c^2*d^4*e^3 + 4*a^2*c*d^2*e^5 - (c^3*d^4*e^3 + 2*a*c^2*d^2*e^5 + a^2*c*e^7)*x^
2 + 3*(c^3*d^5*e^2 + 2*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x)*sqrt(c*x^2 + a))/(c^4*d^5*e^4 + 2*a*c^3*d^3*e^6 + a^2*c
^2*d*e^8 + (c^4*d^4*e^5 + 2*a*c^3*d^2*e^7 + a^2*c^2*e^9)*x), 1/2*(2*(3*c^3*d^6 + 4*a*c^2*d^4*e^2 + (3*c^3*d^5*
e + 4*a*c^2*d^3*e^3)*x)*sqrt(-c*d^2 - a*e^2)*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)) - (6*c^3*d^7 + 11*a*c^2*d^5*e^2 + 4*a^2*c*d^3*e^4 - a^3*d*e^6 + (6*c^3
*d^6*e + 11*a*c^2*d^4*e^3 + 4*a^2*c*d^2*e^5 - a^3*e^7)*x)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (6*c^3
*d^6*e + 10*a*c^2*d^4*e^3 + 4*a^2*c*d^2*e^5 - (c^3*d^4*e^3 + 2*a*c^2*d^2*e^5 + a^2*c*e^7)*x^2 + 3*(c^3*d^5*e^2
+ 2*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x)*sqrt(c*x^2 + a))/(c^4*d^5*e^4 + 2*a*c^3*d^3*e^6 + a^2*c^2*d*e^8 + (c^4*d^
4*e^5 + 2*a*c^3*d^2*e^7 + a^2*c^2*e^9)*x)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.01, size = 435, normalized size = 2.13 \[ -\frac {c \,d^{5} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{5}}-\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{4}}{\left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) e^{4}}+\frac {4 d^{3} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{5}}-\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {3}{2}} e^{2}}+\frac {3 d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}\, e^{4}}+\frac {\sqrt {c \,x^{2}+a}\, x}{2 c \,e^{2}}-\frac {2 \sqrt {c \,x^{2}+a}\, d}{c \,e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/e^2*x/c*(c*x^2+a)^(1/2)-1/2/e^2*a/c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))-2*d*(c*x^2+a)^(1/2)/c/e^3+3*d^2/e^
4*ln(c^(1/2)*x+(c*x^2+a)^(1/2))/c^(1/2)+4/e^5*d^3/((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))-d^4/e^4/(a
*e^2+c*d^2)/(x+d/e)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)-d^5/e^5*c/(a*e^2+c*d^2)/((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 = 0.57, size = 233, normalized size = 1.14 \[ -\frac {\sqrt {c x^{2} + a} d^{4}}{c d^{2} e^{4} x + a e^{6} x + c d^{3} e^{3} + a d e^{5}} + \frac {\sqrt {c x^{2} + a} x}{2 \, c e^{2}} + \frac {3 \, d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {c} e^{4}} - \frac {a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {3}{2}} e^{2}} + \frac {c d^{5} \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} e^{7}} - \frac {4 \, d^{3} \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{\sqrt {a + \frac {c d^{2}}{e^{2}}} e^{5}} - \frac {2 \, \sqrt {c x^{2} + a} d}{c e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(c*x^2 + a)*d^4/(c*d^2*e^4*x + a*e^6*x + c*d^3*e^3 + a*d*e^5) + 1/2*sqrt(c*x^2 + a)*x/(c*e^2) + 3*d^2*arc
sinh(c*x/sqrt(a*c))/(sqrt(c)*e^4) - 1/2*a*arcsinh(c*x/sqrt(a*c))/(c^(3/2)*e^2) + c*d^5*arcsinh(c*d*x/(sqrt(a*c
)*abs(e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/((a + c*d^2/e^2)^(3/2)*e^7) - 4*d^3*arcsinh(c*d*x/(sqrt(a*c)*a
bs(e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/(sqrt(a + c*d^2/e^2)*e^5) - 2*sqrt(c*x^2 + a)*d/(c*e^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {a + c x^{2}} \left (d + e x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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