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

Optimal result

Integrand size = 22, antiderivative size = 153 \[ \int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx=\frac {d (2 d-e x) \sqrt {a+c x^2}}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 c e}-\frac {d \left (2 c d^2+a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c} e^4}-\frac {d^2 \sqrt {c d^2+a e^2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^4} \]

[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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1668, 12, 829, 858, 223, 212, 739} \[ \int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx=-\frac {d^2 \sqrt {a e^2+c d^2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4}-\frac {d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (a e^2+2 c d^2\right )}{2 \sqrt {c} e^4}+\frac {d \sqrt {a+c x^2} (2 d-e x)}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 c e} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx=\frac {e \sqrt {a+c x^2} \left (6 c d^2+2 a e^2-3 c d e x+2 c e^2 x^2\right )+12 c d^2 \sqrt {-c d^2-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+3 \sqrt {c} d \left (2 c d^2+a e^2\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{6 c e^4} \]

[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)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.48

method	result	size
risch	\(\frac {\left (2 c \,e^{2} x^{2}-3 c d e x +2 e^{2} a +6 c \,d^{2}\right ) \sqrt {c \,x^{2}+a}}{6 c \,e^{3}}-\frac {d \left (\frac {\left (e^{2} a +2 c \,d^{2}\right ) \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{e \sqrt {c}}+\frac {2 d \left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{2 e^{3}}\)	\(227\)
default	\(\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3 c e}-\frac {d \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{e^{2}}+\frac {d^{2} \left (\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{e^{3}}\)	\(323\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 776, normalized size of antiderivative = 5.07 \[ \int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx=\left [\frac {6 \, \sqrt {c d^{2} + a e^{2}} c d^{2} \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 (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 (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 2 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{12 \, c e^{4}}, -\frac {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}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{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 (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 2 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{12 \, c e^{4}}, \frac {3 \, \sqrt {c d^{2} + a e^{2}} c d^{2} \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 (2 \, c d^{3} + a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 2 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{6 \, c e^{4}}, -\frac {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}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{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 (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 2 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{6 \, c e^{4}}\right ] \]

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

Sympy [F]

\[ \int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx=\int \frac {x^{2} \sqrt {a + c x^{2}}}{d + e x}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F(-2)]

Exception generated. \[ \int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \sqrt {a+c x^2}}{d+e x} \, dx=\int \frac {x^2\,\sqrt {c\,x^2+a}}{d+e\,x} \,d x \]

[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)