3.311 \(\int (c e+d e x)^m (a+b \sin ^{-1}(c+d x)) \, dx\)

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.06, 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 = {4805, 4627, 364} \[ \frac {(e (c+d x))^{m+1} \left (a+b \sin ^{-1}(c+d x)\right )}{d e (m+1)}-\frac {b (e (c+d x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};(c+d x)^2\right )}{d e^2 (m+1) (m+2)} \]

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 4627

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 4805

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 {\operatorname {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 \operatorname {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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 77, normalized size = 0.87 \[ -\frac {(c+d x) (e (c+d x))^m \left (b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};(c+d x)^2\right )-(m+2) \left (a+b \sin ^{-1}(c+d x)\right )\right )}{d (m+1) (m+2)} \]

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)))

________________________________________________________________________________________

fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \arcsin \left (d x + c\right ) + a\right )} {\left (d e x + c e\right )}^{m}, x\right ) \]

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*e*x + c*e)^m, x)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [F]  time = 1.80, 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)

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________