3.6.62 \(\int \frac {(d+e x)^3}{\sqrt {a+c x^2}} \, dx\) [562]

Optimal. Leaf size=110 \[ \frac {e (d+e x)^2 \sqrt {a+c x^2}}{3 c}+\frac {e \left (4 \left (4 c d^2-a e^2\right )+5 c d e x\right ) \sqrt {a+c x^2}}{6 c^2}+\frac {d \left (2 c d^2-3 a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2}} \]

[Out]

1/2*d*(-3*a*e^2+2*c*d^2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)+1/3*e*(e*x+d)^2*(c*x^2+a)^(1/2)/c+1/6*e*(5
*c*d*e*x-4*a*e^2+16*c*d^2)*(c*x^2+a)^(1/2)/c^2

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Rubi [A]
time = 0.05, antiderivative size = 110, 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 = {757, 794, 223, 212} \begin {gather*} \frac {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 c^{3/2}}+\frac {e \sqrt {a+c x^2} \left (4 \left (4 c d^2-a e^2\right )+5 c d e x\right )}{6 c^2}+\frac {e \sqrt {a+c x^2} (d+e x)^2}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/Sqrt[a + c*x^2],x]

[Out]

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

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*} \int \frac {(d+e x)^3}{\sqrt {a+c x^2}} \, dx &=\frac {e (d+e x)^2 \sqrt {a+c x^2}}{3 c}+\frac {\int \frac {(d+e x) \left (3 c d^2-2 a e^2+5 c d e x\right )}{\sqrt {a+c x^2}} \, dx}{3 c}\\ &=\frac {e (d+e x)^2 \sqrt {a+c x^2}}{3 c}+\frac {e \left (4 \left (4 c d^2-a e^2\right )+5 c d e x\right ) \sqrt {a+c x^2}}{6 c^2}+\frac {\left (d \left (2 c d^2-3 a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 c}\\ &=\frac {e (d+e x)^2 \sqrt {a+c x^2}}{3 c}+\frac {e \left (4 \left (4 c d^2-a e^2\right )+5 c d e x\right ) \sqrt {a+c x^2}}{6 c^2}+\frac {\left (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 c}\\ &=\frac {e (d+e x)^2 \sqrt {a+c x^2}}{3 c}+\frac {e \left (4 \left (4 c d^2-a e^2\right )+5 c d e x\right ) \sqrt {a+c x^2}}{6 c^2}+\frac {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 c^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 91, normalized size = 0.83 \begin {gather*} \frac {e \sqrt {a+c x^2} \left (-4 a e^2+c \left (18 d^2+9 d e x+2 e^2 x^2\right )\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 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(-4*a*e^2 + c*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) + 3*Sqrt[c]*d*(-2*c*d^2 + 3*a*e^2)*Log[-(Sqrt
[c]*x) + Sqrt[a + c*x^2]])/(6*c^2)

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Maple [A]
time = 0.43, size = 124, normalized size = 1.13

method	result	size
risch	\(-\frac {e \left (-2 c \,e^{2} x^{2}-9 c d e x +4 e^{2} a -18 c \,d^{2}\right ) \sqrt {c \,x^{2}+a}}{6 c^{2}}-\frac {3 d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right ) e^{2} a}{2 c^{\frac {3}{2}}}+\frac {d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}}\)	\(94\)
default	\(e^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+a}}{3 c}-\frac {2 a \sqrt {c \,x^{2}+a}}{3 c^{2}}\right )+3 d \,e^{2} \left (\frac {x \sqrt {c \,x^{2}+a}}{2 c}-\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}\right )+\frac {3 d^{2} e \sqrt {c \,x^{2}+a}}{c}+\frac {d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{\sqrt {c}}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 108, normalized size = 0.98 \begin {gather*} \frac {d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {c}} + \frac {\sqrt {c x^{2} + a} x^{2} e^{3}}{3 \, c} + \frac {3 \, \sqrt {c x^{2} + a} d x e^{2}}{2 \, c} - \frac {3 \, a d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{2}}{2 \, c^{\frac {3}{2}}} + \frac {3 \, \sqrt {c x^{2} + a} d^{2} e}{c} - \frac {2 \, \sqrt {c x^{2} + a} a e^{3}}{3 \, 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]

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

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Fricas [A]
time = 1.80, size = 174, normalized size = 1.58 \begin {gather*} \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 ) - 2 \, {\left (9 \, c d x e^{2} + 18 \, c d^{2} e + 2 \, {\left (c x^{2} - 2 \, a\right )} e^{3}\right )} \sqrt {c x^{2} + a}}{12 \, c^{2}}, -\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 ) - {\left (9 \, c d x e^{2} + 18 \, c d^{2} e + 2 \, {\left (c x^{2} - 2 \, a\right )} e^{3}\right )} \sqrt {c x^{2} + a}}{6 \, 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]

[-1/12*(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) - 2*(9*c*d*x*e^2 + 18*
c*d^2*e + 2*(c*x^2 - 2*a)*e^3)*sqrt(c*x^2 + a))/c^2, -1/6*(3*(2*c*d^3 - 3*a*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/
sqrt(c*x^2 + a)) - (9*c*d*x*e^2 + 18*c*d^2*e + 2*(c*x^2 - 2*a)*e^3)*sqrt(c*x^2 + a))/c^2]

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Sympy [A]
time = 2.27, size = 216, normalized size = 1.96 \begin {gather*} \frac {3 \sqrt {a} d e^{2} x \sqrt {1 + \frac {c x^{2}}{a}}}{2 c} - \frac {3 a d e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 c^{\frac {3}{2}}} + d^{3} \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 ) + 3 d^{2} 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 ) + e^{3} \left (\begin {cases} - \frac {2 a \sqrt {a + c x^{2}}}{3 c^{2}} + \frac {x^{2} \sqrt {a + c x^{2}}}{3 c} & \text {for}\: c \neq 0 \\\frac {x^{4}}{4 \sqrt {a}} & \text {otherwise} \end {cases}\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)

[Out]

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

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Giac [A]
time = 1.59, size = 90, normalized size = 0.82 \begin {gather*} \frac {1}{6} \, \sqrt {c x^{2} + a} {\left (x {\left (\frac {2 \, x e^{3}}{c} + \frac {9 \, d e^{2}}{c}\right )} + \frac {2 \, {\left (9 \, c^{2} d^{2} e - 2 \, a c e^{3}\right )}}{c^{3}}\right )} - \frac {{\left (2 \, c d^{3} - 3 \, a d 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)^3/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/6*sqrt(c*x^2 + a)*(x*(2*x*e^3/c + 9*d*e^2/c) + 2*(9*c^2*d^2*e - 2*a*c*e^3)/c^3) - 1/2*(2*c*d^3 - 3*a*d*e^2)*
log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + c*x^2)^(1/2),x)

[Out]

int((d + e*x)^3/(a + c*x^2)^(1/2), x)

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