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

Optimal. Leaf size=86 \[ \frac {\left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2}}+\frac {3 d e \sqrt {a+c x^2}}{2 c}+\frac {e \sqrt {a+c x^2} (d+e x)}{2 c} \]

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Rubi [A]  time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {743, 641, 217, 206} \begin {gather*} \frac {\left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2}}+\frac {3 d e \sqrt {a+c x^2}}{2 c}+\frac {e \sqrt {a+c x^2} (d+e x)}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/Sqrt[a + c*x^2],x]

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 71, normalized size = 0.83 \begin {gather*} \frac {\left (2 c d^2-a e^2\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )+\sqrt {c} e \sqrt {a+c x^2} (4 d+e x)}{2 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.32, size = 71, normalized size = 0.83 \begin {gather*} \frac {\left (a e^2-2 c d^2\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 c^{3/2}}+\frac {\sqrt {a+c x^2} \left (4 d e+e^2 x\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 138, normalized size = 1.60 \begin {gather*} \left [-\frac {{\left (2 \, c d^{2} - a e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (c e^{2} x + 4 \, c d e\right )} \sqrt {c x^{2} + a}}{4 \, c^{2}}, -\frac {{\left (2 \, c d^{2} - a e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (c e^{2} x + 4 \, c d e\right )} \sqrt {c x^{2} + a}}{2 \, c^{2}}\right ] \end {gather*}

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]

[-1/4*((2*c*d^2 - a*e^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(c*e^2*x + 4*c*d*e)*sqrt(
c*x^2 + a))/c^2, -1/2*((2*c*d^2 - a*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (c*e^2*x + 4*c*d*e)*sqr
t(c*x^2 + a))/c^2]

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giac [A]  time = 0.25, size = 63, normalized size = 0.73 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + a} {\left (\frac {x e^{2}}{c} + \frac {4 \, d e}{c}\right )} - \frac {{\left (2 \, c d^{2} - a e^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, c^{\frac {3}{2}}} \end {gather*}

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]

1/2*sqrt(c*x^2 + a)*(x*e^2/c + 4*d*e/c) - 1/2*(2*c*d^2 - a*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 = 84, normalized size = 0.98 \begin {gather*} -\frac {a \,e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}+\frac {d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}}+\frac {\sqrt {c \,x^{2}+a}\, e^{2} x}{2 c}+\frac {2 \sqrt {c \,x^{2}+a}\, d e}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 69, normalized size = 0.80 \begin {gather*} \frac {\sqrt {c x^{2} + a} e^{2} x}{2 \, c} + \frac {d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {c}} - \frac {a e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {3}{2}}} + \frac {2 \, \sqrt {c x^{2} + a} d e}{c} \end {gather*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a + c*x^2)^(1/2),x)

[Out]

int((d + e*x)^2/(a + c*x^2)^(1/2), x)

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sympy [A]  time = 4.15, size = 158, normalized size = 1.84 \begin {gather*} \frac {\sqrt {a} e^{2} x \sqrt {1 + \frac {c x^{2}}{a}}}{2 c} - \frac {a e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 c^{\frac {3}{2}}} + d^{2} \left (\begin {cases} \frac {\sqrt {- \frac {a}{c}} \operatorname {asin}{\left (x \sqrt {- \frac {c}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge c < 0 \\\frac {\sqrt {\frac {a}{c}} \operatorname {asinh}{\left (x \sqrt {\frac {c}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge c > 0 \\\frac {\sqrt {- \frac {a}{c}} \operatorname {acosh}{\left (x \sqrt {- \frac {c}{a}} \right )}}{\sqrt {- a}} & \text {for}\: c > 0 \wedge a < 0 \end {cases}\right ) + 2 d e \left (\begin {cases} \frac {x^{2}}{2 \sqrt {a}} & \text {for}\: c = 0 \\\frac {\sqrt {a + c x^{2}}}{c} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*e**2*x*sqrt(1 + c*x**2/a)/(2*c) - a*e**2*asinh(sqrt(c)*x/sqrt(a))/(2*c**(3/2)) + d**2*Piecewise((sqrt(
-a/c)*asin(x*sqrt(-c/a))/sqrt(a), (a > 0) & (c < 0)), (sqrt(a/c)*asinh(x*sqrt(c/a))/sqrt(a), (a > 0) & (c > 0)
), (sqrt(-a/c)*acosh(x*sqrt(-c/a))/sqrt(-a), (c > 0) & (a < 0))) + 2*d*e*Piecewise((x**2/(2*sqrt(a)), Eq(c, 0)
), (sqrt(a + c*x**2)/c, True))

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