3.117 \(\int (c e+d e x)^2 (a+b \sinh ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=76 \[ \frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac {b e^2 \left ((c+d x)^2+1\right )^{3/2}}{9 d}+\frac {b e^2 \sqrt {(c+d x)^2+1}}{3 d} \]

[Out]

-1/9*b*e^2*(1+(d*x+c)^2)^(3/2)/d+1/3*e^2*(d*x+c)^3*(a+b*arcsinh(d*x+c))/d+1/3*b*e^2*(1+(d*x+c)^2)^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5865, 12, 5661, 266, 43} \[ \frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac {b e^2 \left ((c+d x)^2+1\right )^{3/2}}{9 d}+\frac {b e^2 \sqrt {(c+d x)^2+1}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^2*(a + b*ArcSinh[c + d*x]),x]

[Out]

(b*e^2*Sqrt[1 + (c + d*x)^2])/(3*d) - (b*e^2*(1 + (c + d*x)^2)^(3/2))/(9*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[
c + d*x]))/(3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

\begin {align*} \int (c e+d e x)^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int e^2 x^2 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{3 d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+x}} \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{\sqrt {1+x}}+\sqrt {1+x}\right ) \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac {b e^2 \sqrt {1+(c+d x)^2}}{3 d}-\frac {b e^2 \left (1+(c+d x)^2\right )^{3/2}}{9 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 64, normalized size = 0.84 \[ \frac {e^2 \left (\frac {1}{3} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )-\frac {1}{9} b \left (c^2+2 c d x+d^2 x^2-2\right ) \sqrt {(c+d x)^2+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*(-1/9*(b*(-2 + c^2 + 2*c*d*x + d^2*x^2)*Sqrt[1 + (c + d*x)^2]) + ((c + d*x)^3*(a + b*ArcSinh[c + d*x]))/3
))/d

________________________________________________________________________________________

fricas [B]  time = 0.62, size = 168, normalized size = 2.21 \[ \frac {3 \, a d^{3} e^{2} x^{3} + 9 \, a c d^{2} e^{2} x^{2} + 9 \, a c^{2} d e^{2} x + 3 \, {\left (b d^{3} e^{2} x^{3} + 3 \, b c d^{2} e^{2} x^{2} + 3 \, b c^{2} d e^{2} x + b c^{3} e^{2}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (b d^{2} e^{2} x^{2} + 2 \, b c d e^{2} x + {\left (b c^{2} - 2 \, b\right )} e^{2}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{9 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.31, size = 403, normalized size = 5.30 \[ \frac {1}{18} \, {\left (6 \, a d^{2} x^{3} + 18 \, a c d x^{2} - 18 \, {\left (d {\left (\frac {c \log \left (-c d - {\left (x {\left | d \right |} - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} {\left | d \right |}\right )}{d {\left | d \right |}} + \frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{d^{2}}\right )} - x \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )\right )} b c^{2} + 9 \, {\left (2 \, x^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (\frac {x}{d^{2}} - \frac {3 \, c}{d^{3}}\right )} - \frac {{\left (2 \, c^{2} - 1\right )} \log \left (-c d - {\left (x {\left | d \right |} - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} {\left | d \right |}\right )}{d^{2} {\left | d \right |}}\right )} d\right )} b c d + {\left (6 \, x^{3} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (x {\left (\frac {2 \, x}{d^{2}} - \frac {5 \, c}{d^{3}}\right )} + \frac {11 \, c^{2} d - 4 \, d}{d^{5}}\right )} + \frac {3 \, {\left (2 \, c^{3} - 3 \, c\right )} \log \left (-c d - {\left (x {\left | d \right |} - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} {\left | d \right |}\right )}{d^{3} {\left | d \right |}}\right )} d\right )} b d^{2} + 18 \, a c^{2} x\right )} e^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(6*a*d^2*x^3 + 18*a*c*d*x^2 - 18*(d*(c*log(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d))/
(d*abs(d)) + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^2) - x*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)))*b*c^
2 + 9*(2*x^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x/d^2 - 3*
c/d^3) - (2*c^2 - 1)*log(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d))/(d^2*abs(d)))*d)*b*c*d
+ (6*x^3*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x*(2*x/d^2 - 5
*c/d^3) + (11*c^2*d - 4*d)/d^5) + 3*(2*c^3 - 3*c)*log(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*ab
s(d))/(d^3*abs(d)))*d)*b*d^2 + 18*a*c^2*x)*e^2

________________________________________________________________________________________

maple [A]  time = 0.01, size = 73, normalized size = 0.96 \[ \frac {\frac {\left (d x +c \right )^{3} e^{2} a}{3}+e^{2} b \left (\frac {\left (d x +c \right )^{3} \arcsinh \left (d x +c \right )}{3}-\frac {\left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{9}+\frac {2 \sqrt {1+\left (d x +c \right )^{2}}}{9}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^2*(a+b*arcsinh(d*x+c)),x)

[Out]

1/d*(1/3*(d*x+c)^3*e^2*a+e^2*b*(1/3*(d*x+c)^3*arcsinh(d*x+c)-1/9*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+2/9*(1+(d*x+c)^
2)^(1/2)))

________________________________________________________________________________________

maxima [B]  time = 0.36, size = 445, normalized size = 5.86 \[ \frac {1}{3} \, a d^{2} e^{2} x^{3} + a c d e^{2} x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {arsinh}\left (d x + c\right ) - d {\left (\frac {3 \, c^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{3}} + \frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} x}{d^{2}} - \frac {{\left (c^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} c}{d^{3}}\right )}\right )} b c d e^{2} + \frac {1}{18} \, {\left (6 \, x^{3} \operatorname {arsinh}\left (d x + c\right ) - d {\left (\frac {2 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} x^{2}}{d^{2}} - \frac {15 \, c^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{4}} - \frac {5 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} c x}{d^{3}} + \frac {9 \, {\left (c^{2} + 1\right )} c \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{4}} + \frac {15 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} c^{2}}{d^{4}} - \frac {4 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (c^{2} + 1\right )}}{d^{4}}\right )}\right )} b d^{2} e^{2} + a c^{2} e^{2} x + \frac {{\left ({\left (d x + c\right )} \operatorname {arsinh}\left (d x + c\right ) - \sqrt {{\left (d x + c\right )}^{2} + 1}\right )} b c^{2} e^{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*d^2*e^2*x^3 + a*c*d*e^2*x^2 + 1/2*(2*x^2*arcsinh(d*x + c) - d*(3*c^2*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2
*d^2 + 4*(c^2 + 1)*d^2))/d^3 + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x/d^2 - (c^2 + 1)*arcsinh(2*(d^2*x + c*d)/sqr
t(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^3 - 3*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c/d^3))*b*c*d*e^2 + 1/18*(6*x^3*arc
sinh(d*x + c) - d*(2*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x^2/d^2 - 15*c^3*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^
2 + 4*(c^2 + 1)*d^2))/d^4 - 5*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c*x/d^3 + 9*(c^2 + 1)*c*arcsinh(2*(d^2*x + c*d
)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^4 + 15*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c^2/d^4 - 4*sqrt(d^2*x^2 + 2*
c*d*x + c^2 + 1)*(c^2 + 1)/d^4))*b*d^2*e^2 + a*c^2*e^2*x + ((d*x + c)*arcsinh(d*x + c) - sqrt((d*x + c)^2 + 1)
)*b*c^2*e^2/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^2*(a + b*asinh(c + d*x)),x)

[Out]

int((c*e + d*e*x)^2*(a + b*asinh(c + d*x)), x)

________________________________________________________________________________________

sympy [A]  time = 0.77, size = 258, normalized size = 3.39 \[ \begin {cases} a c^{2} e^{2} x + a c d e^{2} x^{2} + \frac {a d^{2} e^{2} x^{3}}{3} + \frac {b c^{3} e^{2} \operatorname {asinh}{\left (c + d x \right )}}{3 d} + b c^{2} e^{2} x \operatorname {asinh}{\left (c + d x \right )} - \frac {b c^{2} e^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9 d} + b c d e^{2} x^{2} \operatorname {asinh}{\left (c + d x \right )} - \frac {2 b c e^{2} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9} + \frac {b d^{2} e^{2} x^{3} \operatorname {asinh}{\left (c + d x \right )}}{3} - \frac {b d e^{2} x^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9} + \frac {2 b e^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9 d} & \text {for}\: d \neq 0 \\c^{2} e^{2} x \left (a + b \operatorname {asinh}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**2*(a+b*asinh(d*x+c)),x)

[Out]

Piecewise((a*c**2*e**2*x + a*c*d*e**2*x**2 + a*d**2*e**2*x**3/3 + b*c**3*e**2*asinh(c + d*x)/(3*d) + b*c**2*e*
*2*x*asinh(c + d*x) - b*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(9*d) + b*c*d*e**2*x**2*asinh(c + d*x)
- 2*b*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/9 + b*d**2*e**2*x**3*asinh(c + d*x)/3 - b*d*e**2*x**2*sqrt
(c**2 + 2*c*d*x + d**2*x**2 + 1)/9 + 2*b*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(9*d), Ne(d, 0)), (c**2*e**
2*x*(a + b*asinh(c)), True))

________________________________________________________________________________________