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

Optimal. Leaf size=119 \[ \frac {a \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2}}+\frac {x \sqrt {a+c x^2} \left (4 c d^2-a e^2\right )}{8 c}+\frac {5 d e \left (a+c x^2\right )^{3/2}}{12 c}+\frac {e \left (a+c x^2\right )^{3/2} (d+e x)}{4 c} \]

[Out]

5/12*d*e*(c*x^2+a)^(3/2)/c+1/4*e*(e*x+d)*(c*x^2+a)^(3/2)/c+1/8*a*(-a*e^2+4*c*d^2)*arctanh(x*c^(1/2)/(c*x^2+a)^
(1/2))/c^(3/2)+1/8*(-a*e^2+4*c*d^2)*x*(c*x^2+a)^(1/2)/c

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Rubi [A]  time = 0.05, antiderivative size = 119, 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, 641, 195, 217, 206} \[ \frac {a \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2}}+\frac {x \sqrt {a+c x^2} \left (4 c d^2-a e^2\right )}{8 c}+\frac {5 d e \left (a+c x^2\right )^{3/2}}{12 c}+\frac {e \left (a+c x^2\right )^{3/2} (d+e x)}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*Sqrt[a + c*x^2],x]

[Out]

((4*c*d^2 - a*e^2)*x*Sqrt[a + c*x^2])/(8*c) + (5*d*e*(a + c*x^2)^(3/2))/(12*c) + (e*(d + e*x)*(a + c*x^2)^(3/2
))/(4*c) + (a*(4*c*d^2 - 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 641

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

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]

Rubi steps

\begin {align*} \int (d+e x)^2 \sqrt {a+c x^2} \, dx &=\frac {e (d+e x) \left (a+c x^2\right )^{3/2}}{4 c}+\frac {\int \left (4 c d^2-a e^2+5 c d e x\right ) \sqrt {a+c x^2} \, dx}{4 c}\\ &=\frac {5 d e \left (a+c x^2\right )^{3/2}}{12 c}+\frac {e (d+e x) \left (a+c x^2\right )^{3/2}}{4 c}+\frac {\left (4 c d^2-a e^2\right ) \int \sqrt {a+c x^2} \, dx}{4 c}\\ &=\frac {\left (4 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{8 c}+\frac {5 d e \left (a+c x^2\right )^{3/2}}{12 c}+\frac {e (d+e x) \left (a+c x^2\right )^{3/2}}{4 c}+\frac {\left (a \left (4 c d^2-a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 c}\\ &=\frac {\left (4 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{8 c}+\frac {5 d e \left (a+c x^2\right )^{3/2}}{12 c}+\frac {e (d+e x) \left (a+c x^2\right )^{3/2}}{4 c}+\frac {\left (a \left (4 c d^2-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 {\left (4 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{8 c}+\frac {5 d e \left (a+c x^2\right )^{3/2}}{12 c}+\frac {e (d+e x) \left (a+c x^2\right )^{3/2}}{4 c}+\frac {a \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 99, normalized size = 0.83 \[ \frac {\sqrt {c} \sqrt {a+c x^2} \left (a e (16 d+3 e x)+2 c x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )-3 a \left (a e^2-4 c d^2\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )}{24 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*Sqrt[a + c*x^2]*(a*e*(16*d + 3*e*x) + 2*c*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2)) - 3*a*(-4*c*d^2 + a*e^2)*L
og[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(24*c^(3/2))

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fricas [A]  time = 0.90, size = 214, normalized size = 1.80 \[ \left [-\frac {3 \, {\left (4 \, a c d^{2} - a^{2} e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (6 \, c^{2} e^{2} x^{3} + 16 \, c^{2} d e x^{2} + 16 \, a c d e + 3 \, {\left (4 \, c^{2} d^{2} + a c e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{48 \, c^{2}}, -\frac {3 \, {\left (4 \, a c d^{2} - a^{2} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (6 \, c^{2} e^{2} x^{3} + 16 \, c^{2} d e x^{2} + 16 \, a c d e + 3 \, {\left (4 \, c^{2} d^{2} + a c e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{24 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(4*a*c*d^2 - a^2*e^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(6*c^2*e^2*x^3 + 1
6*c^2*d*e*x^2 + 16*a*c*d*e + 3*(4*c^2*d^2 + a*c*e^2)*x)*sqrt(c*x^2 + a))/c^2, -1/24*(3*(4*a*c*d^2 - a^2*e^2)*s
qrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (6*c^2*e^2*x^3 + 16*c^2*d*e*x^2 + 16*a*c*d*e + 3*(4*c^2*d^2 + a*c
*e^2)*x)*sqrt(c*x^2 + a))/c^2]

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giac [A]  time = 0.20, size = 96, normalized size = 0.81 \[ \frac {1}{24} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left (3 \, x e^{2} + 8 \, d e\right )} x + \frac {3 \, {\left (4 \, c^{2} d^{2} + a c e^{2}\right )}}{c^{2}}\right )} x + \frac {16 \, a d e}{c}\right )} - \frac {{\left (4 \, a c d^{2} - a^{2} e^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{8 \, c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + a)*((2*(3*x*e^2 + 8*d*e)*x + 3*(4*c^2*d^2 + a*c*e^2)/c^2)*x + 16*a*d*e/c) - 1/8*(4*a*c*d^2 -
a^2*e^2)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)

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maple [A]  time = 0.05, size = 122, normalized size = 1.03 \[ -\frac {a^{2} e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}+\frac {a \,d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}-\frac {\sqrt {c \,x^{2}+a}\, a \,e^{2} x}{8 c}+\frac {\sqrt {c \,x^{2}+a}\, d^{2} x}{2}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} e^{2} x}{4 c}+\frac {2 \left (c \,x^{2}+a \right )^{\frac {3}{2}} d e}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.38, size = 107, normalized size = 0.90 \[ \frac {1}{2} \, \sqrt {c x^{2} + a} d^{2} x + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} e^{2} x}{4 \, c} - \frac {\sqrt {c x^{2} + a} a e^{2} x}{8 \, c} + \frac {a d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {c}} - \frac {a^{2} e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, c^{\frac {3}{2}}} + \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d e}{3 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^(1/2)*(d + e*x)^2,x)

[Out]

int((a + c*x^2)^(1/2)*(d + e*x)^2, x)

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sympy [A]  time = 6.59, size = 184, normalized size = 1.55 \[ \frac {a^{\frac {3}{2}} e^{2} x}{8 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {\sqrt {a} d^{2} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {3 \sqrt {a} e^{2} x^{3}}{8 \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {a^{2} e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{8 c^{\frac {3}{2}}} + \frac {a d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 \sqrt {c}} + 2 d e \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + \frac {c e^{2} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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