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

Optimal. Leaf size=229 \[ \frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b e}-\frac {i b \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {i b \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e} \]

[Out]

-1/2*I*(a+b*arcsin(c*x))^2/b/e+(a+b*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+(a+b*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-I*b*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-I*b*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

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Rubi [A]  time = 0.30, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4741, 4519, 2190, 2279, 2391} \[ -\frac {i b \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {i b \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(d + e*x),x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 214, normalized size = 0.93 \[ -\frac {i \left (\left (a+b \sin ^{-1}(c x)\right ) \left (a+2 i b \log \left (1+\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )+2 i b \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )+b \sin ^{-1}(c x)\right )+2 b^2 \text {Li}_2\left (-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )+2 b^2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )\right )}{2 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (c x\right ) + a}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.28, size = 759, normalized size = 3.31 \[ \frac {a \ln \left (c e x +d c \right )}{e}-\frac {i b \arcsin \left (c x \right )^{2}}{2 e}+\frac {i b e \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e -\sqrt {-c^{2} d^{2}+e^{2}}}{i d c -\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{c^{2} d^{2}-e^{2}}+\frac {i b e \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{c^{2} d^{2}-e^{2}}-\frac {b e \arcsin \left (c x \right ) \ln \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e -\sqrt {-c^{2} d^{2}+e^{2}}}{i d c -\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{c^{2} d^{2}-e^{2}}-\frac {b e \arcsin \left (c x \right ) \ln \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{c^{2} d^{2}-e^{2}}+\frac {c^{2} b \,d^{2} \arcsin \left (c x \right ) \ln \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e -\sqrt {-c^{2} d^{2}+e^{2}}}{i d c -\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e \left (c^{2} d^{2}-e^{2}\right )}+\frac {c^{2} b \,d^{2} \arcsin \left (c x \right ) \ln \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e \left (c^{2} d^{2}-e^{2}\right )}-\frac {i c^{2} b \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e -\sqrt {-c^{2} d^{2}+e^{2}}}{i d c -\sqrt {-c^{2} d^{2}+e^{2}}}\right ) d^{2}}{e \left (c^{2} d^{2}-e^{2}\right )}-\frac {i c^{2} b \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right ) d^{2}}{e \left (c^{2} d^{2}-e^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/(d + e*x),x)

[Out]

int((a + b*asin(c*x))/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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