3.3.55 \(\int \frac {x^3}{(d+e x)^2 \sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=160 \[ -\frac {d^2 \left (3 a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^3 \left (a e^2+c d^2\right )^{3/2}}+\frac {d^3 \sqrt {a+c x^2}}{e^2 (d+e x) \left (a e^2+c d^2\right )}-\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} e^3}+\frac {\sqrt {a+c x^2}}{c e^2} \]

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Rubi [A]  time = 0.33, antiderivative size = 160, 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 = {1651, 1654, 844, 217, 206, 725} \begin {gather*} \frac {d^3 \sqrt {a+c x^2}}{e^2 (d+e x) \left (a e^2+c d^2\right )}-\frac {d^2 \left (3 a e^2+2 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^3 \left (a e^2+c d^2\right )^{3/2}}-\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} e^3}+\frac {\sqrt {a+c x^2}}{c e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a + c*x^2]/(c*e^2) + (d^3*Sqrt[a + c*x^2])/(e^2*(c*d^2 + a*e^2)*(d + e*x)) - (2*d*ArcTanh[(Sqrt[c]*x)/Sqr
t[a + c*x^2]])/(Sqrt[c]*e^3) - (d^2*(2*c*d^2 + 3*a*e^2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*
x^2])])/(e^3*(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^3}{(d+e x)^2 \sqrt {a+c x^2}} \, dx &=\frac {d^3 \sqrt {a+c x^2}}{e^2 \left (c d^2+a e^2\right ) (d+e x)}-\frac {\int \frac {-\frac {a d^2}{e}+d \left (a+\frac {c d^2}{e^2}\right ) x-\frac {\left (c d^2+a e^2\right ) x^2}{e}}{(d+e x) \sqrt {a+c x^2}} \, dx}{c d^2+a e^2}\\ &=\frac {\sqrt {a+c x^2}}{c e^2}+\frac {d^3 \sqrt {a+c x^2}}{e^2 \left (c d^2+a e^2\right ) (d+e x)}-\frac {\int \frac {-a c d^2 e+2 c d \left (c d^2+a e^2\right ) x}{(d+e x) \sqrt {a+c x^2}} \, dx}{c e^2 \left (c d^2+a e^2\right )}\\ &=\frac {\sqrt {a+c x^2}}{c e^2}+\frac {d^3 \sqrt {a+c x^2}}{e^2 \left (c d^2+a e^2\right ) (d+e x)}-\frac {(2 d) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{e^3}+\frac {\left (d^2 \left (2 c d^2+3 a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{e^3 \left (c d^2+a e^2\right )}\\ &=\frac {\sqrt {a+c x^2}}{c e^2}+\frac {d^3 \sqrt {a+c x^2}}{e^2 \left (c d^2+a e^2\right ) (d+e x)}-\frac {(2 d) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{e^3}-\frac {\left (d^2 \left (2 c d^2+3 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^3 \left (c d^2+a e^2\right )}\\ &=\frac {\sqrt {a+c x^2}}{c e^2}+\frac {d^3 \sqrt {a+c x^2}}{e^2 \left (c d^2+a e^2\right ) (d+e x)}-\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} e^3}-\frac {d^2 \left (2 c d^2+3 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^3 \left (c d^2+a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 184, normalized size = 1.15 \begin {gather*} \frac {-\frac {d^2 \left (3 a e^2+2 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 {d^2 \left (3 a e^2+2 c d^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}}+e \sqrt {a+c x^2} \left (\frac {d^3}{(d+e x) \left (a e^2+c d^2\right )}+\frac {1}{c}\right )-\frac {2 d \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )}{\sqrt {c}}}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(c^(-1) + d^3/((c*d^2 + a*e^2)*(d + e*x))) + (d^2*(2*c*d^2 + 3*a*e^2)*Log[d + e*x])/(c*d^2
+ a*e^2)^(3/2) - (2*d*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/Sqrt[c] - (d^2*(2*c*d^2 + 3*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))/e^3

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IntegrateAlgebraic [A]  time = 1.12, size = 197, normalized size = 1.23 \begin {gather*} \frac {2 \sqrt {-a e^2-c d^2} \left (3 a d^2 e^2+2 c d^4\right ) \tan ^{-1}\left (\frac {-e \sqrt {a+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}\right )}{e^3 \left (a e^2+c d^2\right )^2}+\frac {\sqrt {a+c x^2} \left (a d e^2+a e^3 x+2 c d^3+c d^2 e x\right )}{c e^2 (d+e x) \left (a e^2+c d^2\right )}+\frac {2 d \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{\sqrt {c} e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 6.77, size = 1449, normalized size = 9.06

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Unable to divide, perhaps due to rounding error%%%{%%%{1,[0,1,0,0]%%%},[4,1]%%%}+%%%{%%%{-2,[0,1,1
,0]%%%},[2,1]%%%}+%%%{%%%{1,[0,1,2,0]%%%},[0,1]%%%} / %%%{%%%{1,[1,2,0,0]%%%}+%%%{1,[0,0,1,2]%%%},[4,0]%%%}+%%
%{%%%{-2,[1,2,1,0]%%%}+%%%{-2,[0,0,2,2]%%%},[2,0]%%%}+%%%{%%%{1,[1,2,2,0]%%%}+%%%{1,[0,0,3,2]%%%},[0,0]%%%} Er
ror: Bad Argument Value

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maple [B]  time = 0.01, size = 386, normalized size = 2.41 \begin {gather*} \frac {c \,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^{4}}+\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^{3}}{\left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) e^{3}}-\frac {3 d^{2} \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^{4}}-\frac {2 d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}\, e^{3}}+\frac {\sqrt {c \,x^{2}+a}}{c \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(e*x+d)^2/(c*x^2+a)^(1/2),x)

[Out]

(c*x^2+a)^(1/2)/c/e^2-2/e^3*d*ln(c^(1/2)*x+(c*x^2+a)^(1/2))/c^(1/2)-3/e^4*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))+d^3/e^3/(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^4/e^4
*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 = 0.54, size = 193, normalized size = 1.21 \begin {gather*} \frac {\sqrt {c x^{2} + a} d^{3}}{c d^{2} e^{3} x + a e^{5} x + c d^{3} e^{2} + a d e^{4}} - \frac {2 \, d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {c} e^{3}} - \frac {c 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^{6}} + \frac {3 \, d^{2} \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^{4}} + \frac {\sqrt {c x^{2} + a}}{c e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((a + c*x^2)^(1/2)*(d + e*x)^2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(e*x+d)**2/(c*x**2+a)**(1/2),x)

[Out]

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

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