3.3.45 \(\int \frac {x^4}{(d+e x) (a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=146 \[ -\frac {d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2} e^2}+\frac {a (a e+c d x)}{c^2 \sqrt {a+c x^2} \left (a e^2+c d^2\right )}+\frac {\sqrt {a+c x^2}}{c^2 e}-\frac {d^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^2 \left (a e^2+c d^2\right )^{3/2}} \]

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Rubi [A]  time = 0.31, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1647, 1654, 844, 217, 206, 725} \begin {gather*} \frac {a (a e+c d x)}{c^2 \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2} e^2}+\frac {\sqrt {a+c x^2}}{c^2 e}-\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^2 \left (a e^2+c d^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a*e + c*d*x))/(c^2*(c*d^2 + a*e^2)*Sqrt[a + c*x^2]) + Sqrt[a + c*x^2]/(c^2*e) - (d*ArcTanh[(Sqrt[c]*x)/Sqr
t[a + c*x^2]])/(c^(3/2)*e^2) - (d^4*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(e^2*(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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[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) \left (a+c x^2\right )^{3/2}} \, dx &=\frac {a (a e+c d x)}{c^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {\int \frac {\frac {a^2 d^2}{c d^2+a e^2}-a x^2}{(d+e x) \sqrt {a+c x^2}} \, dx}{a c}\\ &=\frac {a (a e+c d x)}{c^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {\sqrt {a+c x^2}}{c^2 e}-\frac {\int \frac {\frac {a^2 c d^2 e^2}{c d^2+a e^2}+a c d e x}{(d+e x) \sqrt {a+c x^2}} \, dx}{a c^2 e^2}\\ &=\frac {a (a e+c d x)}{c^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {\sqrt {a+c x^2}}{c^2 e}-\frac {d \int \frac {1}{\sqrt {a+c x^2}} \, dx}{c e^2}+\frac {d^4 \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{e^2 \left (c d^2+a e^2\right )}\\ &=\frac {a (a e+c d x)}{c^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {\sqrt {a+c x^2}}{c^2 e}-\frac {d \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{c e^2}-\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^2 \left (c d^2+a e^2\right )}\\ &=\frac {a (a e+c d x)}{c^2 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {\sqrt {a+c x^2}}{c^2 e}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2} e^2}-\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^2 \left (c d^2+a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 179, normalized size = 1.23 \begin {gather*} \frac {\frac {e \left (2 a^2 e^2+a c \left (d^2+d e x+e^2 x^2\right )+c^2 d^2 x^2\right )}{c^2 \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {\sqrt {a} d \sqrt {\frac {c x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{3/2} \sqrt {a+c x^2}}-\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 )}{\left (a e^2+c d^2\right )^{3/2}}}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*(2*a^2*e^2 + c^2*d^2*x^2 + a*c*(d^2 + d*e*x + e^2*x^2)))/(c^2*(c*d^2 + a*e^2)*Sqrt[a + c*x^2]) - (Sqrt[a]*
d*Sqrt[1 + (c*x^2)/a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(c^(3/2)*Sqrt[a + c*x^2]) - (d^4*ArcTanh[(a*e - c*d*x)/(Sq
rt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(c*d^2 + a*e^2)^(3/2))/e^2

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IntegrateAlgebraic [A]  time = 0.85, size = 223, normalized size = 1.53 \begin {gather*} \frac {2 a^2 e^2+a c d^2+a c d e x+a c e^2 x^2+c^2 d^2 x^2}{c^2 e \sqrt {a+c x^2} \left (a e^2+c d^2\right )}+\frac {d \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{c^{3/2} e^2}+\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^2 \left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4/((d + e*x)*(a + c*x^2)^(3/2)),x]

[Out]

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

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fricas [B]  time = 4.65, size = 1525, normalized size = 10.45

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)/(c*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.27, size = 299, normalized size = 2.05 \begin {gather*} \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 )}{{\left (c d^{2} e^{2} + a e^{4}\right )} \sqrt {-c d^{2} - a e^{2}}} + \frac {d e^{\left (-2\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{c^{\frac {3}{2}}} + \frac {{\left (\frac {{\left (c^{4} d^{4} e^{5} + 2 \, a c^{3} d^{2} e^{7} + a^{2} c^{2} e^{9}\right )} x}{c^{5} d^{4} e^{6} + 2 \, a c^{4} d^{2} e^{8} + a^{2} c^{3} e^{10}} + \frac {a c^{3} d^{3} e^{6} + a^{2} c^{2} d e^{8}}{c^{5} d^{4} e^{6} + 2 \, a c^{4} d^{2} e^{8} + a^{2} c^{3} e^{10}}\right )} x + \frac {a c^{3} d^{4} e^{5} + 3 \, a^{2} c^{2} d^{2} e^{7} + 2 \, a^{3} c e^{9}}{c^{5} d^{4} e^{6} + 2 \, a c^{4} d^{2} e^{8} + a^{2} c^{3} e^{10}}}{\sqrt {c x^{2} + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)/(c*x^2+a)^(3/2),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))/((c*d^2*e^2 + a*e^4)*sqrt(-c
*d^2 - a*e^2)) + d*e^(-2)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2) + (((c^4*d^4*e^5 + 2*a*c^3*d^2*e^7 +
a^2*c^2*e^9)*x/(c^5*d^4*e^6 + 2*a*c^4*d^2*e^8 + a^2*c^3*e^10) + (a*c^3*d^3*e^6 + a^2*c^2*d*e^8)/(c^5*d^4*e^6 +
2*a*c^4*d^2*e^8 + a^2*c^3*e^10))*x + (a*c^3*d^4*e^5 + 3*a^2*c^2*d^2*e^7 + 2*a^3*c*e^9)/(c^5*d^4*e^6 + 2*a*c^4
*d^2*e^8 + a^2*c^3*e^10))/sqrt(c*x^2 + a)

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maple [B]  time = 0.02, size = 396, normalized size = 2.71 \begin {gather*} \frac {c \,d^{5} x}{\left (a \,e^{2}+c \,d^{2}\right ) \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}}}\, a \,e^{4}}-\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 )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{3}}+\frac {d^{4}}{\left (a \,e^{2}+c \,d^{2}\right ) \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}}}\, e^{3}}+\frac {x^{2}}{\sqrt {c \,x^{2}+a}\, c e}-\frac {d^{3} x}{\sqrt {c \,x^{2}+a}\, a \,e^{4}}+\frac {d x}{\sqrt {c \,x^{2}+a}\, c \,e^{2}}-\frac {d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}} e^{2}}+\frac {2 a}{\sqrt {c \,x^{2}+a}\, c^{2} e}-\frac {d^{2}}{\sqrt {c \,x^{2}+a}\, c \,e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(e*x+d)/(c*x^2+a)^(3/2),x)

[Out]

1/e*x^2/c/(c*x^2+a)^(1/2)+2/e*a/c^2/(c*x^2+a)^(1/2)+d/e^2*x/c/(c*x^2+a)^(1/2)-d/e^2/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*x/a/(c*x^2+a)^(1/2)+d^4/e^3/(a*e^2+c*d^2)/(-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^4/(a*e^2+c*d^2)/a/(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2
)*c*x-d^4/e^3/(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 = 0.62, size = 251, normalized size = 1.72 \begin {gather*} \frac {c d^{5} x}{\sqrt {c x^{2} + a} a c d^{2} e^{4} + \sqrt {c x^{2} + a} a^{2} e^{6}} + \frac {d^{4}}{\sqrt {c x^{2} + a} c d^{2} e^{3} + \sqrt {c x^{2} + a} a e^{5}} + \frac {x^{2}}{\sqrt {c x^{2} + a} c e} - \frac {d^{3} x}{\sqrt {c x^{2} + a} a e^{4}} + \frac {d x}{\sqrt {c x^{2} + a} c e^{2}} - \frac {d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{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 )}{{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} e^{5}} - \frac {d^{2}}{\sqrt {c x^{2} + a} c e^{3}} + \frac {2 \, a}{\sqrt {c x^{2} + a} c^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)/(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

c*d^5*x/(sqrt(c*x^2 + a)*a*c*d^2*e^4 + sqrt(c*x^2 + a)*a^2*e^6) + d^4/(sqrt(c*x^2 + a)*c*d^2*e^3 + sqrt(c*x^2
+ a)*a*e^5) + x^2/(sqrt(c*x^2 + a)*c*e) - d^3*x/(sqrt(c*x^2 + a)*a*e^4) + d*x/(sqrt(c*x^2 + a)*c*e^2) - d*arcs
inh(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)))/
((a + c*d^2/e^2)^(3/2)*e^5) - d^2/(sqrt(c*x^2 + a)*c*e^3) + 2*a/(sqrt(c*x^2 + a)*c^2*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{{\left (c\,x^2+a\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((a + c*x^2)^(3/2)*(d + e*x)),x)

[Out]

int(x^4/((a + c*x^2)^(3/2)*(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}}{\left (a + c x^{2}\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(e*x+d)/(c*x**2+a)**(3/2),x)

[Out]

Integral(x**4/((a + c*x**2)**(3/2)*(d + e*x)), x)

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