\(\int \frac {(a+c x^2)^{3/2}}{d+e x} \, dx\) [538]

Optimal result

Integrand size = 19, antiderivative size = 159 \[ \int \frac {\left (a+c x^2\right )^{3/2}}{d+e x} \, dx=\frac {\left (2 \left (c d^2+a e^2\right )-c d e x\right ) \sqrt {a+c x^2}}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 e}-\frac {\sqrt {c} d \left (2 c d^2+3 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 e^4}-\frac {\left (c d^2+a e^2\right )^{3/2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {749, 829, 858, 223, 212, 739} \[ \int \frac {\left (a+c x^2\right )^{3/2}}{d+e x} \, dx=-\frac {\left (a e^2+c d^2\right )^{3/2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^4}-\frac {\sqrt {c} d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a e^2+2 c d^2\right )}{2 e^4}+\frac {\sqrt {a+c x^2} \left (2 \left (a e^2+c d^2\right )-c d e x\right )}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 e} \]

[In]

Int[(a + c*x^2)^(3/2)/(d + e*x),x]

[Out]

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

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 749

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] + Dist[2*(p/(e*(m + 2*p + 1))), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+c x^2\right )^{3/2}}{3 e}+\frac {\int \frac {(a e-c d x) \sqrt {a+c x^2}}{d+e x} \, dx}{e} \\ & = \frac {\left (2 \left (c d^2+a e^2\right )-c d e x\right ) \sqrt {a+c x^2}}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 e}+\frac {\int \frac {a c e \left (c d^2+2 a e^2\right )-c^2 d \left (2 c d^2+3 a e^2\right ) x}{(d+e x) \sqrt {a+c x^2}} \, dx}{2 c e^3} \\ & = \frac {\left (2 \left (c d^2+a e^2\right )-c d e x\right ) \sqrt {a+c x^2}}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 e}+\frac {\left (c d^2+a e^2\right )^2 \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{e^4}-\frac {\left (c d \left (2 c d^2+3 a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 e^4} \\ & = \frac {\left (2 \left (c d^2+a e^2\right )-c d e x\right ) \sqrt {a+c x^2}}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 e}-\frac {\left (c d^2+a e^2\right )^2 \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 (c d \left (2 c d^2+3 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 {\left (2 \left (c d^2+a e^2\right )-c d e x\right ) \sqrt {a+c x^2}}{2 e^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 e}-\frac {\sqrt {c} d \left (2 c d^2+3 a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 e^4}-\frac {\left (c d^2+a e^2\right )^{3/2} \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 = 0.68 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+c x^2\right )^{3/2}}{d+e x} \, dx=\frac {e \sqrt {a+c x^2} \left (6 c d^2+8 a e^2-3 c d e x+2 c e^2 x^2\right )-12 \left (-c d^2-a e^2\right )^{3/2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+3 \sqrt {c} d \left (2 c d^2+3 a e^2\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{6 e^4} \]

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(6*c*d^2 + 8*a*e^2 - 3*c*d*e*x + 2*c*e^2*x^2) - 12*(-(c*d^2) - a*e^2)^(3/2)*ArcTan[(Sqrt[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 + 3*a*e^2)*Log[-(Sqrt[c]*x) + S
qrt[a + c*x^2]])/(6*e^4)

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.51

method	result	size
risch	\(\frac {\left (2 c \,x^{2} e^{2}-3 x c d e +8 e^{2} a +6 c \,d^{2}\right ) \sqrt {c \,x^{2}+a}}{6 e^{3}}-\frac {\frac {\sqrt {c}\, d \left (3 e^{2} a +2 c \,d^{2}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{e}-\frac {\left (-2 a^{2} e^{4}-4 a c \,d^{2} e^{2}-2 c^{2} d^{4}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{2 e^{3}}\)	\(240\)
default	\(\frac {\frac {\left (c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{3}-\frac {c d \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )-\frac {2 c d}{e}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (e^{2} a +c \,d^{2}\right )}{e^{2}}-\frac {4 c^{2} d^{2}}{e^{2}}\right ) \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{e}+\frac {\left (e^{2} a +c \,d^{2}\right ) \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{e^{2}}}{e}\)	\(496\)

[In]

int((c*x^2+a)^(3/2)/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 1.62 (sec) , antiderivative size = 774, normalized size of antiderivative = 4.87 \[ \int \frac {\left (a+c x^2\right )^{3/2}}{d+e x} \, dx=\left [\frac {3 \, {\left (2 \, c d^{3} + 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 ) + 6 \, {\left (c d^{2} + a e^{2}\right )}^{\frac {3}{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 ) + 2 \, {\left (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 8 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{12 \, e^{4}}, \frac {3 \, {\left (2 \, c d^{3} + 3 \, a d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + 3 \, {\left (c d^{2} + a e^{2}\right )}^{\frac {3}{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 ) + {\left (2 \, c e^{3} x^{2} - 3 \, c d e^{2} x + 6 \, c d^{2} e + 8 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{6 \, e^{4}}, -\frac {12 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {-c d^{2} - a e^{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} + 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 + 8 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{12 \, e^{4}}, -\frac {6 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {-c d^{2} - a e^{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} + 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 + 8 \, a e^{3}\right )} \sqrt {c x^{2} + a}}{6 \, e^{4}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (a+c x^2\right )^{3/2}}{d+e x} \, dx=\int \frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{d + e x}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+c x^2\right )^{3/2}}{d+e x} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^2+a)^(3/2)/(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. \[ \int \frac {\left (a+c x^2\right )^{3/2}}{d+e x} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c*x^2+a)^(3/2)/(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. \[ \int \frac {\left (a+c x^2\right )^{3/2}}{d+e x} \, dx=\int \frac {{\left (c\,x^2+a\right )}^{3/2}}{d+e\,x} \,d x \]

[In]

int((a + c*x^2)^(3/2)/(d + e*x),x)

[Out]

int((a + c*x^2)^(3/2)/(d + e*x), x)