3.4.11 \(\int (c e+d e x)^m (a+b \text {ArcSin}(c+d x)) \, dx\) [311]

Optimal. Leaf size=89 \[ \frac {(e (c+d x))^{1+m} (a+b \text {ArcSin}(c+d x))}{d e (1+m)}-\frac {b (e (c+d x))^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};(c+d x)^2\right )}{d e^2 (1+m) (2+m)} \]

[Out]

(e*(d*x+c))^(1+m)*(a+b*arcsin(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)

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Rubi [A]
time = 0.04, antiderivative size = 89, 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 = {4889, 4723, 371} \begin {gather*} \frac {(e (c+d x))^{m+1} (a+b \text {ArcSin}(c+d x))}{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*ArcSin[c + d*x]),x]

[Out]

((e*(c + d*x))^(1 + m)*(a + b*ArcSin[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 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSi
n[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSin[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 4889

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcSin[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 \sin ^{-1}(c+d x)\right ) \, dx &=\frac {\text {Subst}\left (\int (e x)^m \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {(e (c+d x))^{1+m} \left (a+b \sin ^{-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 \sin ^{-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*}

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Mathematica [A]
time = 0.03, size = 77, normalized size = 0.87 \begin {gather*} -\frac {(c+d x) (e (c+d x))^m \left (-((2+m) (a+b \text {ArcSin}(c+d x)))+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*ArcSin[c + d*x]),x]

[Out]

-(((c + d*x)*(e*(c + d*x))^m*(-((2 + m)*(a + b*ArcSin[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)))

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Maple [F]
time = 1.68, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{m} \left (a +b \arcsin \left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*arcsin(d*x+c)),x, algorithm="maxima")

[Out]

((d*x*e^m + c*e^m)*(d*x + c)^m*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + (d*m + d)*integrate((d
*x*e^m + c*e^m)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)*(d*x + c)^m/((d^2*m + d^2)*x^2 + c^2 + (c^2 - 1)*m + 2*(c
*d*m + c*d)*x - 1), x))*b/(d*m + d) + (d*x*e + c*e)^(m + 1)*a*e^(-1)/(d*(m + 1))

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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*arcsin(d*x+c)),x, algorithm="fricas")

[Out]

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

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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 {asin}{\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*asin(d*x+c)),x)

[Out]

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

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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*arcsin(d*x+c)),x, algorithm="giac")

[Out]

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

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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 {asin}\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*asin(c + d*x)),x)

[Out]

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

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