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

Optimal. Leaf size=211 \[ \frac {\left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2} e^5}+\frac {d^3 \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^5}-\frac {\sqrt {a+c x^2} \left (8 c d^3-e x \left (4 c d^2-a e^2\right )\right )}{8 c e^4}-\frac {7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac {\left (a+c x^2\right )^{3/2} (d+e x)}{4 c e^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 211, 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 = {1654, 815, 844, 217, 206, 725} \begin {gather*} \frac {\left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2} e^5}-\frac {\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^4}+\frac {d^3 \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^5}-\frac {7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac {\left (a+c x^2\right )^{3/2} (d+e x)}{4 c e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*c*d^3 - e*(4*c*d^2 - a*e^2)*x)*Sqrt[a + c*x^2])/(8*c*e^4) - (7*d*(a + c*x^2)^(3/2))/(12*c*e^2) + ((d + e*
x)*(a + c*x^2)^(3/2))/(4*c*e^2) + ((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^5) + (d^3*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^5

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 815

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 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^3 \sqrt {a+c x^2}}{d+e x} \, dx &=\frac {(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac {\int \frac {\sqrt {a+c x^2} \left (-a d e^2-e \left (3 c d^2+a e^2\right ) x-7 c d e^2 x^2\right )}{d+e x} \, dx}{4 c e^3}\\ &=-\frac {7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac {(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac {\int \frac {\left (-3 a c d e^4+3 c e^3 \left (4 c d^2-a e^2\right ) x\right ) \sqrt {a+c x^2}}{d+e x} \, dx}{12 c^2 e^5}\\ &=-\frac {\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^4}-\frac {7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac {(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac {\int \frac {-3 a c^2 d e^4 \left (4 c d^2+a e^2\right )+3 c^2 e^3 \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}{24 c^3 e^7}\\ &=-\frac {\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^4}-\frac {7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac {(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}-\frac {\left (d^3 \left (c d^2+a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{e^5}+\frac {\left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 c e^5}\\ &=-\frac {\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^4}-\frac {7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac {(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac {\left (d^3 \left (c d^2+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^5}+\frac {\left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{8 c e^5}\\ &=-\frac {\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^4}-\frac {7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac {(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac {\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^5}+\frac {d^3 \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^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.40, size = 225, normalized size = 1.07 \begin {gather*} \frac {24 c^{3/2} d^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+24 c d^3 \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )+e \sqrt {a+c x^2} \left (a e^2 (3 e x-8 d)+c \left (-24 d^3+12 d^2 e x-8 d e^2 x^2+6 e^3 x^3\right )\right )}{24 c e^5}-\frac {\sqrt {a} \sqrt {a+c x^2} \left (a e^2-4 c d^2\right ) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 c^{3/2} e^3 \sqrt {\frac {c x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(Sqrt[a]*(-4*c*d^2 + a*e^2)*Sqrt[a + c*x^2]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(c^(3/2)*e^3*Sqrt[1 + (c*x^2)/a
]) + (e*Sqrt[a + c*x^2]*(a*e^2*(-8*d + 3*e*x) + c*(-24*d^3 + 12*d^2*e*x - 8*d*e^2*x^2 + 6*e^3*x^3)) + 24*c^(3/
2)*d^4*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] + 24*c*d^3*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])])/(24*c*e^5)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.67, size = 236, normalized size = 1.12 \begin {gather*} \frac {\left (a^2 e^4-4 a c d^2 e^2-8 c^2 d^4\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{8 c^{3/2} e^5}-\frac {2 d^3 \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^5}+\frac {\sqrt {a+c x^2} \left (-8 a d e^2+3 a e^3 x-24 c d^3+12 c d^2 e x-8 c d e^2 x^2+6 c e^3 x^3\right )}{24 c e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-24*c*d^3 - 8*a*d*e^2 + 12*c*d^2*e*x + 3*a*e^3*x - 8*c*d*e^2*x^2 + 6*c*e^3*x^3))/(24*c*e^4)
- (2*d^3*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^5 + ((-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]])/(8*c^(3/2)*e^5)

________________________________________________________________________________________

fricas [A]  time = 7.04, size = 963, normalized size = 4.56 \begin {gather*} \left [\frac {24 \, \sqrt {c d^{2} + a e^{2}} c^{2} d^{3} \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 (8 \, c^{2} d^{4} + 4 \, a c d^{2} e^{2} - a^{2} 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} e^{4} x^{3} - 8 \, c^{2} d e^{3} x^{2} - 24 \, c^{2} d^{3} e - 8 \, a c d e^{3} + 3 \, {\left (4 \, c^{2} d^{2} e^{2} + a c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{48 \, c^{2} e^{5}}, \frac {48 \, \sqrt {-c d^{2} - a e^{2}} c^{2} d^{3} \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 (8 \, c^{2} d^{4} + 4 \, a c d^{2} e^{2} - a^{2} 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} e^{4} x^{3} - 8 \, c^{2} d e^{3} x^{2} - 24 \, c^{2} d^{3} e - 8 \, a c d e^{3} + 3 \, {\left (4 \, c^{2} d^{2} e^{2} + a c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{48 \, c^{2} e^{5}}, \frac {12 \, \sqrt {c d^{2} + a e^{2}} c^{2} d^{3} \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 (8 \, c^{2} d^{4} + 4 \, a c d^{2} e^{2} - a^{2} e^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (6 \, c^{2} e^{4} x^{3} - 8 \, c^{2} d e^{3} x^{2} - 24 \, c^{2} d^{3} e - 8 \, a c d e^{3} + 3 \, {\left (4 \, c^{2} d^{2} e^{2} + a c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{24 \, c^{2} e^{5}}, \frac {24 \, \sqrt {-c d^{2} - a e^{2}} c^{2} d^{3} \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 (8 \, c^{2} d^{4} + 4 \, a c d^{2} e^{2} - a^{2} e^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (6 \, c^{2} e^{4} x^{3} - 8 \, c^{2} d e^{3} x^{2} - 24 \, c^{2} d^{3} e - 8 \, a c d e^{3} + 3 \, {\left (4 \, c^{2} d^{2} e^{2} + a c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{24 \, c^{2} e^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(24*sqrt(c*d^2 + a*e^2)*c^2*d^3*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*s
qrt(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*(8*c^2*d^4 + 4*a*c*d^2*e^2 -
a^2*e^4)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(6*c^2*e^4*x^3 - 8*c^2*d*e^3*x^2 - 24*c^2
*d^3*e - 8*a*c*d*e^3 + 3*(4*c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^2*e^5), 1/48*(48*sqrt(-c*d^2 - a*e^2
)*c^2*d^3*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*(8*c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(6*
c^2*e^4*x^3 - 8*c^2*d*e^3*x^2 - 24*c^2*d^3*e - 8*a*c*d*e^3 + 3*(4*c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(
c^2*e^5), 1/24*(12*sqrt(c*d^2 + a*e^2)*c^2*d^3*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*(8*c^2*d^4 + 4*a*c*d
^2*e^2 - a^2*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (6*c^2*e^4*x^3 - 8*c^2*d*e^3*x^2 - 24*c^2*d^3*
e - 8*a*c*d*e^3 + 3*(4*c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^2*e^5), 1/24*(24*sqrt(-c*d^2 - a*e^2)*c^2
*d^3*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*(8*c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (6*c^2*e^4*x^3 - 8*c^2
*d*e^3*x^2 - 24*c^2*d^3*e - 8*a*c*d*e^3 + 3*(4*c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^2*e^5)]

________________________________________________________________________________________

giac [A]  time = 0.22, size = 201, normalized size = 0.95 \begin {gather*} -\frac {2 \, {\left (c d^{5} + a d^{3} e^{2}\right )} \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 (-5\right )}}{\sqrt {-c d^{2} - a e^{2}}} + \frac {1}{24} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left (3 \, x e^{\left (-1\right )} - 4 \, d e^{\left (-2\right )}\right )} x + \frac {3 \, {\left (4 \, c^{2} d^{2} e^{12} + a c e^{14}\right )} e^{\left (-15\right )}}{c^{2}}\right )} x - \frac {8 \, {\left (3 \, c^{2} d^{3} e^{11} + a c d e^{13}\right )} e^{\left (-15\right )}}{c^{2}}\right )} - \frac {{\left (8 \, c^{2} d^{4} + 4 \, a c d^{2} e^{2} - a^{2} e^{4}\right )} e^{\left (-5\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{8 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(c*d^5 + a*d^3*e^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-5)/sqrt
(-c*d^2 - a*e^2) + 1/24*sqrt(c*x^2 + a)*((2*(3*x*e^(-1) - 4*d*e^(-2))*x + 3*(4*c^2*d^2*e^12 + a*c*e^14)*e^(-15
)/c^2)*x - 8*(3*c^2*d^3*e^11 + a*c*d*e^13)*e^(-15)/c^2) - 1/8*(8*c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*e^(-5)*log
(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)

________________________________________________________________________________________

maple [B]  time = 0.01, size = 515, normalized size = 2.44 \begin {gather*} \frac {a \,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^{4}}+\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 )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{6}}-\frac {a^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {3}{2}} e}+\frac {a \,d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}\, e^{3}}+\frac {\sqrt {c}\, d^{4} \ln \left (\frac {-\frac {c d}{e}+\left (x +\frac {d}{e}\right ) c}{\sqrt {c}}+\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}}}\right )}{e^{5}}-\frac {\sqrt {c \,x^{2}+a}\, a x}{8 c e}+\frac {\sqrt {c \,x^{2}+a}\, d^{2} x}{2 e^{3}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} x}{4 c e}-\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}}{e^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} d}{3 c \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/e*x*(c*x^2+a)^(3/2)/c-1/8/e*a/c*x*(c*x^2+a)^(1/2)-1/8/e*a^2/c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))-1/3*d*(c
*x^2+a)^(3/2)/c/e^2+1/2*d^2/e^3*x*(c*x^2+a)^(1/2)+1/2*d^2/e^3*a/c^(1/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))-d^3/e^4*
(-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^5*c^(1/2)*ln((-c*d/e+(x+d/e)*c)/c^(1/2)+(-2*(x+d/
e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2))+d^3/e^4/((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))*a+d^5
/e^6/((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))*c

________________________________________________________________________________________

maxima [A]  time = 0.58, size = 207, normalized size = 0.98 \begin {gather*} \frac {\sqrt {c x^{2} + a} d^{2} x}{2 \, e^{3}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} x}{4 \, c e} - \frac {\sqrt {c x^{2} + a} a x}{8 \, c e} + \frac {\sqrt {c} d^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{e^{5}} + \frac {a d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {c} e^{3}} - \frac {a^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, c^{\frac {3}{2}} e} - \frac {\sqrt {a + \frac {c d^{2}}{e^{2}}} 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 )}{e^{4}} - \frac {\sqrt {c x^{2} + a} d^{3}}{e^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} d}{3 \, c e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c*x^2 + a)*d^2*x/e^3 + 1/4*(c*x^2 + a)^(3/2)*x/(c*e) - 1/8*sqrt(c*x^2 + a)*a*x/(c*e) + sqrt(c)*d^4*ar
csinh(c*x/sqrt(a*c))/e^5 + 1/2*a*d^2*arcsinh(c*x/sqrt(a*c))/(sqrt(c)*e^3) - 1/8*a^2*arcsinh(c*x/sqrt(a*c))/(c^
(3/2)*e) - sqrt(a + c*d^2/e^2)*d^3*arcsinh(c*d*x/(sqrt(a*c)*abs(e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/e^4
- sqrt(c*x^2 + a)*d^3/e^4 - 1/3*(c*x^2 + a)^(3/2)*d/(c*e^2)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sqrt {a + c x^{2}}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________