3.2.14 \(\int (c e+d e x)^m (a+b \sinh ^{-1}(c+d x)) \, dx\) [114]

Optimal. Leaf size=91 \[ \frac {(e (c+d x))^{1+m} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e (1+m)}-\frac {b (e (c+d x))^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-(c+d x)^2\right )}{d e^2 (1+m) (2+m)} \]

[Out]

(e*(d*x+c))^(1+m)*(a+b*arcsinh(d*x+c))/d/e/(1+m)-b*(e*(d*x+c))^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],-(d*x+
c)^2)/d/e^2/(1+m)/(2+m)

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5859, 5776, 371} \begin {gather*} \frac {(e (c+d x))^{m+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e (m+1)}-\frac {b (e (c+d x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-(c+d x)^2\right )}{d e^2 (m+1) (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*(c + d*x))^(1 + m)*(a + b*ArcSinh[c + d*x]))/(d*e*(1 + m)) - (b*(e*(c + d*x))^(2 + m)*Hypergeometric2F1[1/
2, (2 + m)/2, (4 + m)/2, -(c + d*x)^2])/(d*e^2*(1 + m)*(2 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 5776

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 5859

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)^m \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac {\text {Subst}\left (\int (e x)^m \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {(e (c+d x))^{1+m} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e (1+m)}-\frac {b \text {Subst}\left (\int \frac {(e x)^{1+m}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e (1+m)}\\ &=\frac {(e (c+d x))^{1+m} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e (1+m)}-\frac {b (e (c+d x))^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-(c+d x)^2\right )}{d e^2 (1+m) (2+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 79, normalized size = 0.87 \begin {gather*} -\frac {(c+d x) (e (c+d x))^m \left (-\left ((2+m) \left (a+b \sinh ^{-1}(c+d x)\right )\right )+b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-(c+d x)^2\right )\right )}{d (1+m) (2+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((c + d*x)*(e*(c + d*x))^m*(-((2 + m)*(a + b*ArcSinh[c + d*x])) + b*(c + d*x)*Hypergeometric2F1[1/2, (2 + m)
/2, (4 + m)/2, -(c + d*x)^2]))/(d*(1 + m)*(2 + m)))

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{m} \left (a +b \arcsinh \left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsinh(d*x + c) + a)*((d*x + c)*e)^m, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \left (c + d x\right )\right )^{m} \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*(c + d*x))**m*(a + b*asinh(c + d*x)), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(d*x + c) + a)*(d*e*x + c*e)^m, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________