3.4.18 \(\int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx\) [318]

Optimal. Leaf size=153 \[ \frac {d (2 d-e x) \sqrt {a+c x^2}}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 c e}-\frac {d \left (2 c d^2+a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c} e^4}-\frac {d^2 \sqrt {c d^2+a e^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^4} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1668, 12, 829, 858, 223, 212, 739} \begin {gather*} -\frac {d^2 \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^4}-\frac {d \left (a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c} e^4}+\frac {d \sqrt {a+c x^2} (2 d-e x)}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 739

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 829

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 858

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 1668

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

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Mathematica [A]
time = 0.40, size = 161, normalized size = 1.05 \begin {gather*} \frac {e \sqrt {a+c x^2} \left (6 c d^2+2 a e^2-3 c d e x+2 c e^2 x^2\right )+12 c d^2 \sqrt {-c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+3 \sqrt {c} d \left (2 c d^2+a e^2\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{6 c e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(6*c*d^2 + 2*a*e^2 - 3*c*d*e*x + 2*c*e^2*x^2) + 12*c*d^2*Sqrt[-(c*d^2) - a*e^2]*ArcTan[(Sqr
t[c]*(d + e*x) - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]] + 3*Sqrt[c]*d*(2*c*d^2 + a*e^2)*Log[-(Sqrt[c]*x) +
Sqrt[a + c*x^2]])/(6*c*e^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(322\) vs. \(2(131)=262\).
time = 0.08, size = 323, normalized size = 2.11

method	result	size
default	\(\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3 c e}-\frac {d \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{e^{2}}+\frac {d^{2} \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (a \,e^{2}+c \,d^{2}\right ) \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}\right )}{e^{3}}\)	\(323\)
risch	\(\frac {\left (2 c \,e^{2} x^{2}-3 c d e x +2 a \,e^{2}+6 c \,d^{2}\right ) \sqrt {c \,x^{2}+a}}{6 c \,e^{3}}-\frac {d \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right ) a}{2 e^{2} \sqrt {c}}-\frac {d^{3} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right ) \sqrt {c}}{e^{4}}-\frac {d^{2} \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) a}{e^{3} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}-\frac {d^{4} \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) c}{e^{5} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}\)	\(360\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^2+a)^(1/2)/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 140, normalized size = 0.92 \begin {gather*} -\sqrt {c} d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{\left (-4\right )} + \sqrt {c d^{2} e^{\left (-2\right )} + a} d^{2} \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | x e + d \right |}} - \frac {a e}{\sqrt {a c} {\left | x e + d \right |}}\right ) e^{\left (-3\right )} - \frac {1}{2} \, \sqrt {c x^{2} + a} d x e^{\left (-2\right )} - \frac {a d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{\left (-2\right )}}{2 \, \sqrt {c}} + \sqrt {c x^{2} + a} d^{2} e^{\left (-3\right )} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} e^{\left (-1\right )}}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.97, size = 745, normalized size = 4.87 \begin {gather*} \left [\frac {{\left (6 \, \sqrt {c d^{2} + a e^{2}} c d^{2} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 3 \, {\left (2 \, c d^{3} + a d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (3 \, c d x e^{2} - 6 \, c d^{2} e - 2 \, {\left (c x^{2} + a\right )} e^{3}\right )} \sqrt {c x^{2} + a}\right )} e^{\left (-4\right )}}{12 \, c}, \frac {{\left (12 \, \sqrt {-c d^{2} - a e^{2}} c d^{2} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) + 3 \, {\left (2 \, c d^{3} + a d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (3 \, c d x e^{2} - 6 \, c d^{2} e - 2 \, {\left (c x^{2} + a\right )} e^{3}\right )} \sqrt {c x^{2} + a}\right )} e^{\left (-4\right )}}{12 \, c}, \frac {{\left (3 \, \sqrt {c d^{2} + a e^{2}} c d^{2} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 3 \, {\left (2 \, c d^{3} + a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (3 \, c d x e^{2} - 6 \, c d^{2} e - 2 \, {\left (c x^{2} + a\right )} e^{3}\right )} \sqrt {c x^{2} + a}\right )} e^{\left (-4\right )}}{6 \, c}, \frac {{\left (6 \, \sqrt {-c d^{2} - a e^{2}} c d^{2} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) + 3 \, {\left (2 \, c d^{3} + a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (3 \, c d x e^{2} - 6 \, c d^{2} e - 2 \, {\left (c x^{2} + a\right )} e^{3}\right )} \sqrt {c x^{2} + a}\right )} e^{\left (-4\right )}}{6 \, c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.86, size = 157, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (c d^{4} + a d^{2} 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 (-4\right )}}{\sqrt {-c d^{2} - a e^{2}}} + \frac {{\left (2 \, c d^{3} + a d e^{2}\right )} e^{\left (-4\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, \sqrt {c}} + \frac {1}{6} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, x e^{\left (-1\right )} - 3 \, d e^{\left (-2\right )}\right )} x + \frac {2 \, {\left (3 \, c d^{2} e^{7} + a e^{9}\right )} e^{\left (-10\right )}}{c}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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