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

Optimal. Leaf size=144 \[ \frac {a d \left (4 c d^2-3 a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2}}+\frac {e \left (a+c x^2\right )^{3/2} \left (8 \left (6 c d^2-a e^2\right )+21 c d e x\right )}{60 c^2}+\frac {d x \sqrt {a+c x^2} \left (4 c d^2-3 a e^2\right )}{8 c}+\frac {e \left (a+c x^2\right )^{3/2} (d+e x)^2}{5 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {743, 780, 195, 217, 206} \begin {gather*} \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 {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}}+\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} \end {gather*}

Antiderivative was successfully verified.

[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 195

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 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 743

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 780

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*} \int (d+e x)^3 \sqrt {a+c x^2} \, dx &=\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 ) \operatorname {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]  time = 0.08, size = 132, normalized size = 0.92 \begin {gather*} \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} \sqrt {a+c x^2}+c x\right )}{120 c^2} \end {gather*}

Antiderivative was successfully verified.

[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[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(120*c^2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.43, size = 146, normalized size = 1.01 \begin {gather*} \frac {\left (3 a^2 d e^2-4 a c d^3\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{8 c^{3/2}}+\frac {\sqrt {a+c x^2} \left (-16 a^2 e^3+120 a c d^2 e+45 a c d e^2 x+8 a c e^3 x^2+60 c^2 d^3 x+120 c^2 d^2 e x^2+90 c^2 d e^2 x^3+24 c^2 e^3 x^4\right )}{120 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(120*a*c*d^2*e - 16*a^2*e^3 + 60*c^2*d^3*x + 45*a*c*d*e^2*x + 120*c^2*d^2*e*x^2 + 8*a*c*e^3*x
^2 + 90*c^2*d*e^2*x^3 + 24*c^2*e^3*x^4))/(120*c^2) + ((-4*a*c*d^3 + 3*a^2*d*e^2)*Log[-(Sqrt[c]*x) + Sqrt[a + c
*x^2]])/(8*c^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 286, normalized size = 1.99 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

________________________________________________________________________________________

giac [A]  time = 0.33, size = 144, normalized size = 1.00 \begin {gather*} \frac {1}{120} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (4 \, x e^{3} + 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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*x*e^3 + 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)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 164, normalized size = 1.14 \begin {gather*} -\frac {3 a^{2} d \,e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}+\frac {a \,d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}-\frac {3 \sqrt {c \,x^{2}+a}\, a d \,e^{2} x}{8 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} e^{3} x^{2}}{5 c}+\frac {\sqrt {c \,x^{2}+a}\, d^{3} x}{2}+\frac {3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} d \,e^{2} x}{4 c}-\frac {2 \left (c \,x^{2}+a \right )^{\frac {3}{2}} a \,e^{3}}{15 c^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} d^{2} e}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.36, size = 149, normalized size = 1.03 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {c\,x^2+a}\,{\left (d+e\,x\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 7.08, size = 265, normalized size = 1.84 \begin {gather*} \frac {3 a^{\frac {3}{2}} d e^{2} x}{8 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {\sqrt {a} d^{3} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {9 \sqrt {a} d e^{2} x^{3}}{8 \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 a^{2} d e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{8 c^{\frac {3}{2}}} + \frac {a d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 \sqrt {c}} + 3 d^{2} e \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {a x^{2} \sqrt {a + c x^{2}}}{15 c} + \frac {x^{4} \sqrt {a + c x^{2}}}{5} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + \frac {3 c d e^{2} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a**(3/2)*d*e**2*x/(8*c*sqrt(1 + c*x**2/a)) + sqrt(a)*d**3*x*sqrt(1 + c*x**2/a)/2 + 9*sqrt(a)*d*e**2*x**3/(8*
sqrt(1 + c*x**2/a)) - 3*a**2*d*e**2*asinh(sqrt(c)*x/sqrt(a))/(8*c**(3/2)) + a*d**3*asinh(sqrt(c)*x/sqrt(a))/(2
*sqrt(c)) + 3*d**2*e*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True)) + e**3*Piecewise
((-2*a**2*sqrt(a + c*x**2)/(15*c**2) + a*x**2*sqrt(a + c*x**2)/(15*c) + x**4*sqrt(a + c*x**2)/5, Ne(c, 0)), (s
qrt(a)*x**4/4, True)) + 3*c*d*e**2*x**5/(4*sqrt(a)*sqrt(1 + c*x**2/a))

________________________________________________________________________________________