3.91 \(\int \frac {(f+g x) (a+b \sin ^{-1}(c x))}{d+e x} \, dx\)

Optimal. Leaf size=344 \[ \frac {(e f-d g) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}-\frac {i b (e f-d g) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {i b (e f-d g) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b \sin ^{-1}(c x) (e f-d g) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b \sin ^{-1}(c x) (e f-d g) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^2}+\frac {b g \sqrt {1-c^2 x^2}}{c e}-\frac {i b \sin ^{-1}(c x)^2 (e f-d g)}{2 e^2}-\frac {b \sin ^{-1}(c x) (e f-d g) \log (d+e x)}{e^2} \]

[Out]

-1/2*I*b*(-d*g+e*f)*arcsin(c*x)^2/e^2+g*x*(a+b*arcsin(c*x))/e-b*(-d*g+e*f)*arcsin(c*x)*ln(e*x+d)/e^2+(-d*g+e*f
)*(a+b*arcsin(c*x))*ln(e*x+d)/e^2+b*(-d*g+e*f)*arcsin(c*x)*ln(1-I*e*(I*c*x+(-c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2-e
^2)^(1/2)))/e^2+b*(-d*g+e*f)*arcsin(c*x)*ln(1-I*e*(I*c*x+(-c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2-e^2)^(1/2)))/e^2-I*
b*(-d*g+e*f)*polylog(2,I*e*(I*c*x+(-c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2-e^2)^(1/2)))/e^2-I*b*(-d*g+e*f)*polylog(2,
I*e*(I*c*x+(-c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2-e^2)^(1/2)))/e^2+b*g*(-c^2*x^2+1)^(1/2)/c/e

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Rubi [A]  time = 0.64, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {43, 4753, 12, 6742, 261, 216, 2404, 4741, 4519, 2190, 2279, 2391} \[ -\frac {i b (e f-d g) \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {i b (e f-d g) \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^2}+\frac {(e f-d g) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}+\frac {b \sin ^{-1}(c x) (e f-d g) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b \sin ^{-1}(c x) (e f-d g) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^2}+\frac {b g \sqrt {1-c^2 x^2}}{c e}-\frac {i b \sin ^{-1}(c x)^2 (e f-d g)}{2 e^2}-\frac {b \sin ^{-1}(c x) (e f-d g) \log (d+e x)}{e^2} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*(a + b*ArcSin[c*x]))/(d + e*x),x]

[Out]

(b*g*Sqrt[1 - c^2*x^2])/(c*e) - ((I/2)*b*(e*f - d*g)*ArcSin[c*x]^2)/e^2 + (g*x*(a + b*ArcSin[c*x]))/e + (b*(e*
f - d*g)*ArcSin[c*x]*Log[1 - (I*e*E^(I*ArcSin[c*x]))/(c*d - Sqrt[c^2*d^2 - e^2])])/e^2 + (b*(e*f - d*g)*ArcSin
[c*x]*Log[1 - (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])])/e^2 - (b*(e*f - d*g)*ArcSin[c*x]*Log[d + e
*x])/e^2 + ((e*f - d*g)*(a + b*ArcSin[c*x])*Log[d + e*x])/e^2 - (I*b*(e*f - d*g)*PolyLog[2, (I*e*E^(I*ArcSin[c
*x]))/(c*d - Sqrt[c^2*d^2 - e^2])])/e^2 - (I*b*(e*f - d*g)*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*
d^2 - e^2])])/e^2

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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 261

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2404

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/Sqrt[(f_) + (g_.)*(x_)^2], x_Symbol] :> With[{u = Int
Hide[1/Sqrt[f + g*x^2], x]}, Simp[u*(a + b*Log[c*(d + e*x)^n]), x] - Dist[b*e*n, Int[SimplifyIntegrand[u/(d +
e*x), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && GtQ[f, 0]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4741

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Subst[Int[((a + b*x)^n*Cos[x])/
(c*d + e*Sin[x]), x], x, ArcSin[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

Rule 4753

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[Px*(d
+ e*x)^m, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]
] /; FreeQ[{a, b, c, d, e, m}, x] && PolynomialQ[Px, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {(f+g x) \left (a+b \sin ^{-1}(c x)\right )}{d+e x} \, dx &=\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-(b c) \int \frac {e g x+(e f-d g) \log (d+e x)}{e^2 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {(b c) \int \frac {e g x+(e f-d g) \log (d+e x)}{\sqrt {1-c^2 x^2}} \, dx}{e^2}\\ &=\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {(b c) \int \left (\frac {e g x}{\sqrt {1-c^2 x^2}}+\frac {(e f-d g) \log (d+e x)}{\sqrt {1-c^2 x^2}}\right ) \, dx}{e^2}\\ &=\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {(b c g) \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{e}-\frac {(b c (e f-d g)) \int \frac {\log (d+e x)}{\sqrt {1-c^2 x^2}} \, dx}{e^2}\\ &=\frac {b g \sqrt {1-c^2 x^2}}{c e}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}-\frac {b (e f-d g) \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}+\frac {(b c (e f-d g)) \int \frac {\sin ^{-1}(c x)}{c d+c e x} \, dx}{e}\\ &=\frac {b g \sqrt {1-c^2 x^2}}{c e}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}-\frac {b (e f-d g) \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}+\frac {(b c (e f-d g)) \operatorname {Subst}\left (\int \frac {x \cos (x)}{c^2 d+c e \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{e}\\ &=\frac {b g \sqrt {1-c^2 x^2}}{c e}-\frac {i b (e f-d g) \sin ^{-1}(c x)^2}{2 e^2}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}-\frac {b (e f-d g) \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}+\frac {(b c (e f-d g)) \operatorname {Subst}\left (\int \frac {e^{i x} x}{c^2 d-c \sqrt {c^2 d^2-e^2}-i c e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )}{e}+\frac {(b c (e f-d g)) \operatorname {Subst}\left (\int \frac {e^{i x} x}{c^2 d+c \sqrt {c^2 d^2-e^2}-i c e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )}{e}\\ &=\frac {b g \sqrt {1-c^2 x^2}}{c e}-\frac {i b (e f-d g) \sin ^{-1}(c x)^2}{2 e^2}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}+\frac {b (e f-d g) \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b (e f-d g) \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {b (e f-d g) \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {(b (e f-d g)) \operatorname {Subst}\left (\int \log \left (1-\frac {i c e e^{i x}}{c^2 d-c \sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e^2}-\frac {(b (e f-d g)) \operatorname {Subst}\left (\int \log \left (1-\frac {i c e e^{i x}}{c^2 d+c \sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e^2}\\ &=\frac {b g \sqrt {1-c^2 x^2}}{c e}-\frac {i b (e f-d g) \sin ^{-1}(c x)^2}{2 e^2}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}+\frac {b (e f-d g) \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b (e f-d g) \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {b (e f-d g) \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}+\frac {(i b (e f-d g)) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i c e x}{c^2 d-c \sqrt {c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e^2}+\frac {(i b (e f-d g)) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i c e x}{c^2 d+c \sqrt {c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e^2}\\ &=\frac {b g \sqrt {1-c^2 x^2}}{c e}-\frac {i b (e f-d g) \sin ^{-1}(c x)^2}{2 e^2}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{e}+\frac {b (e f-d g) \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b (e f-d g) \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {b (e f-d g) \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {i b (e f-d g) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {i b (e f-d g) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 282, normalized size = 0.82 \[ \frac {(e f-d g) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )+e g x \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{2} i b (e f-d g) \left (2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )+2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )+\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+2 i \left (\log \left (1+\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )+\log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )\right )\right )\right )+\frac {b e g \sqrt {1-c^2 x^2}}{c}-b \sin ^{-1}(c x) (e f-d g) \log (d+e x)}{e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*(a + b*ArcSin[c*x]))/(d + e*x),x]

[Out]

((b*e*g*Sqrt[1 - c^2*x^2])/c + e*g*x*(a + b*ArcSin[c*x]) - b*(e*f - d*g)*ArcSin[c*x]*Log[d + e*x] + (e*f - d*g
)*(a + b*ArcSin[c*x])*Log[d + e*x] - (I/2)*b*(e*f - d*g)*(ArcSin[c*x]*(ArcSin[c*x] + (2*I)*(Log[1 + (I*e*E^(I*
ArcSin[c*x]))/(-(c*d) + Sqrt[c^2*d^2 - e^2])] + Log[1 - (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])]))
+ 2*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d - Sqrt[c^2*d^2 - e^2])] + 2*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c
*d + Sqrt[c^2*d^2 - e^2])]))/e^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))/(e*x+d),x, algorithm="fricas")

[Out]

integral((a*g*x + a*f + (b*g*x + b*f)*arcsin(c*x))/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))/(e*x+d),x, algorithm="giac")

[Out]

integrate((g*x + f)*(b*arcsin(c*x) + a)/(e*x + d), x)

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maple [B]  time = 0.71, size = 1566, normalized size = 4.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(a+b*arcsin(c*x))/(e*x+d),x)

[Out]

a*g/e*x-a/e^2*ln(c*e*x+c*d)*d*g+a/e*ln(c*e*x+c*d)*f-c^2*b/e^2*d^3*g*arcsin(c*x)/(c^2*d^2-e^2)*ln((-I*d*c-(I*c*
x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(-I*d*c+(-c^2*d^2+e^2)^(1/2)))+c^2*b/e*f*arcsin(c*x)/(c^2*d^2-e^
2)*ln((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))*d^2+b*d*g*arcsin
(c*x)/(c^2*d^2-e^2)*ln((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))
+b*d*g*arcsin(c*x)/(c^2*d^2-e^2)*ln((-I*d*c-(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(-I*d*c+(-c^2*d
^2+e^2)^(1/2)))-b*e*f*arcsin(c*x)/(c^2*d^2-e^2)*ln((-I*d*c-(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/
(-I*d*c+(-c^2*d^2+e^2)^(1/2)))+I*b*e*f/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e-(-c^2*d^2+e^2)^
(1/2))/(I*d*c-(-c^2*d^2+e^2)^(1/2)))+1/2*I*b*arcsin(c*x)^2/e^2*d*g+I*c^2*b/e^2*d^3*g/(c^2*d^2-e^2)*dilog((I*d*
c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))-I*b*d*g/(c^2*d^2-e^2)*dilog
((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e-(-c^2*d^2+e^2)^(1/2))/(I*d*c-(-c^2*d^2+e^2)^(1/2)))+c^2*b/e*f*arcsin(c*x)
/(c^2*d^2-e^2)*ln((-I*d*c-(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(-I*d*c+(-c^2*d^2+e^2)^(1/2)))*d^
2-c^2*b/e^2*d^3*g*arcsin(c*x)/(c^2*d^2-e^2)*ln((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*
c+(-c^2*d^2+e^2)^(1/2)))-I*c^2*b/e*f/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e-(-c^2*d^2+e^2)^(1
/2))/(I*d*c-(-c^2*d^2+e^2)^(1/2)))*d^2+I*b*e*f/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d
^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))-I*c^2*b/e*f/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))
*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))*d^2+I*c^2*b/e^2*d^3*g/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x
+(-c^2*x^2+1)^(1/2))*e-(-c^2*d^2+e^2)^(1/2))/(I*d*c-(-c^2*d^2+e^2)^(1/2)))-b*e*f*arcsin(c*x)/(c^2*d^2-e^2)*ln(
(I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))-I*b*d*g/(c^2*d^2-e^2)*
dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))+b*arcsin(c*x)*g/
e*x-1/2*I*b*arcsin(c*x)^2/e*f+b*g*(-c^2*x^2+1)^(1/2)/c/e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a g {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} + \frac {a f \log \left (e x + d\right )}{e} + \int \frac {{\left (b g x + b f\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))/(e*x+d),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(a + b*asin(c*x)))/(d + e*x),x)

[Out]

int(((f + g*x)*(a + b*asin(c*x)))/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*asin(c*x))/(e*x+d),x)

[Out]

Integral((a + b*asin(c*x))*(f + g*x)/(d + e*x), x)

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