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

Optimal. Leaf size=154 \[ \frac {(d+e x)^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{e (m+1)}-\frac {b c \sqrt {1-\frac {c (d+e x)}{c d-e}} \sqrt {1-\frac {c (d+e x)}{c d+e}} (d+e x)^{m+2} F_1\left (m+2;\frac {1}{2},\frac {1}{2};m+3;\frac {c (d+e x)}{c d-e},\frac {c (d+e x)}{c d+e}\right )}{e^2 (m+1) (m+2) \sqrt {1-c^2 x^2}} \]

[Out]

(e*x+d)^(1+m)*(a+b*arcsin(c*x))/e/(1+m)-b*c*(e*x+d)^(2+m)*AppellF1(2+m,1/2,1/2,3+m,c*(e*x+d)/(c*d-e),c*(e*x+d)
/(c*d+e))*(1-c*(e*x+d)/(c*d-e))^(1/2)*(1-c*(e*x+d)/(c*d+e))^(1/2)/e^2/(1+m)/(2+m)/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {4743, 760, 133} \[ \frac {(d+e x)^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{e (m+1)}-\frac {b c \sqrt {1-\frac {c (d+e x)}{c d-e}} \sqrt {1-\frac {c (d+e x)}{c d+e}} (d+e x)^{m+2} F_1\left (m+2;\frac {1}{2},\frac {1}{2};m+3;\frac {c (d+e x)}{c d-e},\frac {c (d+e x)}{c d+e}\right )}{e^2 (m+1) (m+2) \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*(d + e*x)^(2 + m)*Sqrt[1 - (c*(d + e*x))/(c*d - e)]*Sqrt[1 - (c*(d + e*x))/(c*d + e)]*AppellF1[2 + m, 1
/2, 1/2, 3 + m, (c*(d + e*x))/(c*d - e), (c*(d + e*x))/(c*d + e)])/(e^2*(1 + m)*(2 + m)*Sqrt[1 - c^2*x^2])) +
((d + e*x)^(1 + m)*(a + b*ArcSin[c*x]))/(e*(1 + m))

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 760

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[(a + c*x^
2)^p/(e*(1 - (d + e*x)/(d + (e*q)/c))^p*(1 - (d + e*x)/(d - (e*q)/c))^p), Subst[Int[x^m*Simp[1 - x/(d + (e*q)/
c), x]^p*Simp[1 - x/(d - (e*q)/c), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a
*e^2, 0] &&  !IntegerQ[p]

Rule 4743

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

Rubi steps

\begin {align*} \int (d+e x)^m \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {(d+e x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{e (1+m)}-\frac {(b c) \int \frac {(d+e x)^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{e (1+m)}\\ &=\frac {(d+e x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{e (1+m)}-\frac {\left (b c \sqrt {1-\frac {d+e x}{d-\frac {e}{c}}} \sqrt {1-\frac {d+e x}{d+\frac {e}{c}}}\right ) \operatorname {Subst}\left (\int \frac {x^{1+m}}{\sqrt {1-\frac {c x}{c d-e}} \sqrt {1-\frac {c x}{c d+e}}} \, dx,x,d+e x\right )}{e^2 (1+m) \sqrt {1-c^2 x^2}}\\ &=-\frac {b c (d+e x)^{2+m} \sqrt {1-\frac {c (d+e x)}{c d-e}} \sqrt {1-\frac {c (d+e x)}{c d+e}} F_1\left (2+m;\frac {1}{2},\frac {1}{2};3+m;\frac {c (d+e x)}{c d-e},\frac {c (d+e x)}{c d+e}\right )}{e^2 (1+m) (2+m) \sqrt {1-c^2 x^2}}+\frac {(d+e x)^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{e (1+m)}\\ \end {align*}

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Mathematica [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int (d+e x)^m \left (a+b \sin ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \arcsin \left (c x\right ) + a\right )} {\left (e x + d\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{m} \left (a +b \arcsin \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left ({\left (e x + d\right )} {\left (e x + d\right )}^{m} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + \frac {{\left (e m + e\right )} {\left (e \int \frac {\sqrt {-c x + 1} {\left (e x + d\right )}^{m} x}{\sqrt {c x + 1} c x - \sqrt {c x + 1}}\,{d x} + d \int \frac {\sqrt {-c x + 1} {\left (e x + d\right )}^{m}}{\sqrt {c x + 1} c x - \sqrt {c x + 1}}\,{d x}\right )} c}{e {\left (m + 1\right )}}\right )} b}{e m + e} + \frac {{\left (e x + d\right )}^{m + 1} a}{e {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(d + e*x)^m,x)

[Out]

int((a + b*asin(c*x))*(d + e*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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