\(\int (d+e x)^3 \sqrt {a+c x^2} \, dx\) [526]

Optimal result

Integrand size = 19, antiderivative size = 144 \[ \int (d+e x)^3 \sqrt {a+c x^2} \, dx=\frac {d \left (4 c d^2-3 a e^2\right ) x \sqrt {a+c x^2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac {e \left (8 \left (6 c d^2-a e^2\right )+21 c d e x\right ) \left (a+c x^2\right )^{3/2}}{60 c^2}+\frac {a d \left (4 c d^2-3 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2}} \]

[Out]

1/5*e*(e*x+d)^2*(c*x^2+a)^(3/2)/c+1/60*e*(21*c*d*e*x-8*a*e^2+48*c*d^2)*(c*x^2+a)^(3/2)/c^2+1/8*a*d*(-3*a*e^2+4
*c*d^2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)+1/8*d*(-3*a*e^2+4*c*d^2)*x*(c*x^2+a)^(1/2)/c

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {757, 794, 201, 223, 212} \[ \int (d+e x)^3 \sqrt {a+c x^2} \, dx=\frac {a d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (4 c d^2-3 a e^2\right )}{8 c^{3/2}}+\frac {e \left (a+c x^2\right )^{3/2} \left (8 \left (6 c d^2-a e^2\right )+21 c d e x\right )}{60 c^2}+\frac {d x \sqrt {a+c x^2} \left (4 c d^2-3 a e^2\right )}{8 c}+\frac {e \left (a+c x^2\right )^{3/2} (d+e x)^2}{5 c} \]

[In]

Int[(d + e*x)^3*Sqrt[a + c*x^2],x]

[Out]

(d*(4*c*d^2 - 3*a*e^2)*x*Sqrt[a + c*x^2])/(8*c) + (e*(d + e*x)^2*(a + c*x^2)^(3/2))/(5*c) + (e*(8*(6*c*d^2 - a
*e^2) + 21*c*d*e*x)*(a + c*x^2)^(3/2))/(60*c^2) + (a*d*(4*c*d^2 - 3*a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]
])/(8*c^(3/2))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 757

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

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac {\int (d+e x) \left (5 c d^2-2 a e^2+7 c d e x\right ) \sqrt {a+c x^2} \, dx}{5 c} \\ & = \frac {e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac {e \left (8 \left (6 c d^2-a e^2\right )+21 c d e x\right ) \left (a+c x^2\right )^{3/2}}{60 c^2}+\frac {\left (d \left (4 c d^2-3 a e^2\right )\right ) \int \sqrt {a+c x^2} \, dx}{4 c} \\ & = \frac {d \left (4 c d^2-3 a e^2\right ) x \sqrt {a+c x^2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac {e \left (8 \left (6 c d^2-a e^2\right )+21 c d e x\right ) \left (a+c x^2\right )^{3/2}}{60 c^2}+\frac {\left (a d \left (4 c d^2-3 a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 c} \\ & = \frac {d \left (4 c d^2-3 a e^2\right ) x \sqrt {a+c x^2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac {e \left (8 \left (6 c d^2-a e^2\right )+21 c d e x\right ) \left (a+c x^2\right )^{3/2}}{60 c^2}+\frac {\left (a d \left (4 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 )}{8 c} \\ & = \frac {d \left (4 c d^2-3 a e^2\right ) x \sqrt {a+c x^2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac {e \left (8 \left (6 c d^2-a e^2\right )+21 c d e x\right ) \left (a+c x^2\right )^{3/2}}{60 c^2}+\frac {a d \left (4 c d^2-3 a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91 \[ \int (d+e x)^3 \sqrt {a+c x^2} \, dx=\frac {\sqrt {a+c x^2} \left (-16 a^2 e^3+a c e \left (120 d^2+45 d e x+8 e^2 x^2\right )+6 c^2 x \left (10 d^3+20 d^2 e x+15 d e^2 x^2+4 e^3 x^3\right )\right )+15 a \sqrt {c} d \left (-4 c d^2+3 a e^2\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{120 c^2} \]

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-16*a^2*e^3 + a*c*e*(120*d^2 + 45*d*e*x + 8*e^2*x^2) + 6*c^2*x*(10*d^3 + 20*d^2*e*x + 15*d*e
^2*x^2 + 4*e^3*x^3)) + 15*a*Sqrt[c]*d*(-4*c*d^2 + 3*a*e^2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(120*c^2)

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.92

method	result	size
risch	\(-\frac {\left (-24 c^{2} x^{4} e^{3}-90 d \,e^{2} c^{2} x^{3}-8 a c \,e^{3} x^{2}-120 c^{2} d^{2} e \,x^{2}-45 a d \,e^{2} x c -60 c^{2} d^{3} x +16 a^{2} e^{3}-120 d^{2} e a c \right ) \sqrt {c \,x^{2}+a}}{120 c^{2}}-\frac {a d \left (3 e^{2} a -4 c \,d^{2}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}\)	\(132\)
default	\(d^{3} \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )+e^{3} \left (\frac {x^{2} \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{5 c}-\frac {2 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{15 c^{2}}\right )+3 d \,e^{2} \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4 c}-\frac {a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4 c}\right )+\frac {d^{2} e \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{c}\)	\(158\)

[In]

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

[Out]

-1/120*(-24*c^2*e^3*x^4-90*c^2*d*e^2*x^3-8*a*c*e^3*x^2-120*c^2*d^2*e*x^2-45*a*c*d*e^2*x-60*c^2*d^3*x+16*a^2*e^
3-120*a*c*d^2*e)*(c*x^2+a)^(1/2)/c^2-1/8*a/c^(3/2)*d*(3*a*e^2-4*c*d^2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.99 \[ \int (d+e x)^3 \sqrt {a+c x^2} \, dx=\left [-\frac {15 \, {\left (4 \, a c d^{3} - 3 \, a^{2} d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (24 \, c^{2} e^{3} x^{4} + 90 \, c^{2} d e^{2} x^{3} + 120 \, a c d^{2} e - 16 \, a^{2} e^{3} + 8 \, {\left (15 \, c^{2} d^{2} e + a c e^{3}\right )} x^{2} + 15 \, {\left (4 \, c^{2} d^{3} + 3 \, a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{240 \, c^{2}}, -\frac {15 \, {\left (4 \, a c d^{3} - 3 \, a^{2} d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (24 \, c^{2} e^{3} x^{4} + 90 \, c^{2} d e^{2} x^{3} + 120 \, a c d^{2} e - 16 \, a^{2} e^{3} + 8 \, {\left (15 \, c^{2} d^{2} e + a c e^{3}\right )} x^{2} + 15 \, {\left (4 \, c^{2} d^{3} + 3 \, a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{120 \, c^{2}}\right ] \]

[In]

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

[Out]

[-1/240*(15*(4*a*c*d^3 - 3*a^2*d*e^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(24*c^2*e^3*
x^4 + 90*c^2*d*e^2*x^3 + 120*a*c*d^2*e - 16*a^2*e^3 + 8*(15*c^2*d^2*e + a*c*e^3)*x^2 + 15*(4*c^2*d^3 + 3*a*c*d
*e^2)*x)*sqrt(c*x^2 + a))/c^2, -1/120*(15*(4*a*c*d^3 - 3*a^2*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)
) - (24*c^2*e^3*x^4 + 90*c^2*d*e^2*x^3 + 120*a*c*d^2*e - 16*a^2*e^3 + 8*(15*c^2*d^2*e + a*c*e^3)*x^2 + 15*(4*c
^2*d^3 + 3*a*c*d*e^2)*x)*sqrt(c*x^2 + a))/c^2]

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.33 \[ \int (d+e x)^3 \sqrt {a+c x^2} \, dx=\begin {cases} \sqrt {a + c x^{2}} \cdot \left (\frac {3 d e^{2} x^{3}}{4} + \frac {e^{3} x^{4}}{5} + \frac {x^{2} \left (\frac {a e^{3}}{5} + 3 c d^{2} e\right )}{3 c} + \frac {x \left (\frac {3 a d e^{2}}{4} + c d^{3}\right )}{2 c} + \frac {3 a d^{2} e - \frac {2 a \left (\frac {a e^{3}}{5} + 3 c d^{2} e\right )}{3 c}}{c}\right ) + \left (a d^{3} - \frac {a \left (\frac {3 a d e^{2}}{4} + c d^{3}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\sqrt {a} \left (\begin {cases} d^{3} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{4}}{4 e} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a + c*x**2)*(3*d*e**2*x**3/4 + e**3*x**4/5 + x**2*(a*e**3/5 + 3*c*d**2*e)/(3*c) + x*(3*a*d*e**
2/4 + c*d**3)/(2*c) + (3*a*d**2*e - 2*a*(a*e**3/5 + 3*c*d**2*e)/(3*c))/c) + (a*d**3 - a*(3*a*d*e**2/4 + c*d**3
)/(2*c))*Piecewise((log(2*sqrt(c)*sqrt(a + c*x**2) + 2*c*x)/sqrt(c), Ne(a, 0)), (x*log(x)/sqrt(c*x**2), True))
, Ne(c, 0)), (sqrt(a)*Piecewise((d**3*x, Eq(e, 0)), ((d + e*x)**4/(4*e), True)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03 \[ \int (d+e x)^3 \sqrt {a+c x^2} \, dx=\frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} e^{3} x^{2}}{5 \, c} + \frac {1}{2} \, \sqrt {c x^{2} + a} d^{3} x + \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d e^{2} x}{4 \, c} - \frac {3 \, \sqrt {c x^{2} + a} a d e^{2} x}{8 \, c} + \frac {a d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {c}} - \frac {3 \, a^{2} d e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, c^{\frac {3}{2}}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{2} e}{c} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a e^{3}}{15 \, c^{2}} \]

[In]

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

[Out]

1/5*(c*x^2 + a)^(3/2)*e^3*x^2/c + 1/2*sqrt(c*x^2 + a)*d^3*x + 3/4*(c*x^2 + a)^(3/2)*d*e^2*x/c - 3/8*sqrt(c*x^2
+ a)*a*d*e^2*x/c + 1/2*a*d^3*arcsinh(c*x/sqrt(a*c))/sqrt(c) - 3/8*a^2*d*e^2*arcsinh(c*x/sqrt(a*c))/c^(3/2) +
(c*x^2 + a)^(3/2)*d^2*e/c - 2/15*(c*x^2 + a)^(3/2)*a*e^3/c^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.03 \[ \int (d+e x)^3 \sqrt {a+c x^2} \, dx=\frac {1}{120} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (4 \, e^{3} x + 15 \, d e^{2}\right )} x + \frac {4 \, {\left (15 \, c^{3} d^{2} e + a c^{2} e^{3}\right )}}{c^{3}}\right )} x + \frac {15 \, {\left (4 \, c^{3} d^{3} + 3 \, a c^{2} d e^{2}\right )}}{c^{3}}\right )} x + \frac {8 \, {\left (15 \, a c^{2} d^{2} e - 2 \, a^{2} c e^{3}\right )}}{c^{3}}\right )} - \frac {{\left (4 \, a c d^{3} - 3 \, a^{2} d e^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{8 \, c^{\frac {3}{2}}} \]

[In]

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

[Out]

1/120*sqrt(c*x^2 + a)*((2*(3*(4*e^3*x + 15*d*e^2)*x + 4*(15*c^3*d^2*e + a*c^2*e^3)/c^3)*x + 15*(4*c^3*d^3 + 3*
a*c^2*d*e^2)/c^3)*x + 8*(15*a*c^2*d^2*e - 2*a^2*c*e^3)/c^3) - 1/8*(4*a*c*d^3 - 3*a^2*d*e^2)*log(abs(-sqrt(c)*x
+ sqrt(c*x^2 + a)))/c^(3/2)

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^3 \sqrt {a+c x^2} \, dx=\int \sqrt {c\,x^2+a}\,{\left (d+e\,x\right )}^3 \,d x \]

[In]

int((a + c*x^2)^(1/2)*(d + e*x)^3,x)

[Out]

int((a + c*x^2)^(1/2)*(d + e*x)^3, x)