3.521 \(\int \frac{a+b \log (c (d (e+f x)^p)^q)}{\sqrt{2-h x} \sqrt{2+h x}} \, dx\)

Optimal. Leaf size=287 \[ \frac{i b p q \text{PolyLog}\left (2,-\frac{2 f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{-\sqrt{4 f^2-e^2 h^2}+i e h}\right )}{h}+\frac{i b p q \text{PolyLog}\left (2,-\frac{2 f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{\sqrt{4 f^2-e^2 h^2}+i e h}\right )}{h}+\frac{\sin ^{-1}\left (\frac{h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{-\sqrt{4 f^2-e^2 h^2}+i e h}\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{\sqrt{4 f^2-e^2 h^2}+i e h}\right )}{h}+\frac{i b p q \sin ^{-1}\left (\frac{h x}{2}\right )^2}{2 h} \]

[Out]

((I/2)*b*p*q*ArcSin[(h*x)/2]^2)/h - (b*p*q*ArcSin[(h*x)/2]*Log[1 + (2*E^(I*ArcSin[(h*x)/2])*f)/(I*e*h - Sqrt[4
*f^2 - e^2*h^2])])/h - (b*p*q*ArcSin[(h*x)/2]*Log[1 + (2*E^(I*ArcSin[(h*x)/2])*f)/(I*e*h + Sqrt[4*f^2 - e^2*h^
2])])/h + (ArcSin[(h*x)/2]*(a + b*Log[c*(d*(e + f*x)^p)^q]))/h + (I*b*p*q*PolyLog[2, (-2*E^(I*ArcSin[(h*x)/2])
*f)/(I*e*h - Sqrt[4*f^2 - e^2*h^2])])/h + (I*b*p*q*PolyLog[2, (-2*E^(I*ArcSin[(h*x)/2])*f)/(I*e*h + Sqrt[4*f^2
- e^2*h^2])])/h

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Rubi [A]  time = 1.05943, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {216, 2405, 4741, 4521, 2190, 2279, 2391, 2445} \[ \frac{i b p q \text{PolyLog}\left (2,-\frac{2 f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{-\sqrt{4 f^2-e^2 h^2}+i e h}\right )}{h}+\frac{i b p q \text{PolyLog}\left (2,-\frac{2 f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{\sqrt{4 f^2-e^2 h^2}+i e h}\right )}{h}+\frac{\sin ^{-1}\left (\frac{h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{-\sqrt{4 f^2-e^2 h^2}+i e h}\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{\sqrt{4 f^2-e^2 h^2}+i e h}\right )}{h}+\frac{i b p q \sin ^{-1}\left (\frac{h x}{2}\right )^2}{2 h} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])/(Sqrt[2 - h*x]*Sqrt[2 + h*x]),x]

[Out]

((I/2)*b*p*q*ArcSin[(h*x)/2]^2)/h - (b*p*q*ArcSin[(h*x)/2]*Log[1 + (2*E^(I*ArcSin[(h*x)/2])*f)/(I*e*h - Sqrt[4
*f^2 - e^2*h^2])])/h - (b*p*q*ArcSin[(h*x)/2]*Log[1 + (2*E^(I*ArcSin[(h*x)/2])*f)/(I*e*h + Sqrt[4*f^2 - e^2*h^
2])])/h + (ArcSin[(h*x)/2]*(a + b*Log[c*(d*(e + f*x)^p)^q]))/h + (I*b*p*q*PolyLog[2, (-2*E^(I*ArcSin[(h*x)/2])
*f)/(I*e*h - Sqrt[4*f^2 - e^2*h^2])])/h + (I*b*p*q*PolyLog[2, (-2*E^(I*ArcSin[(h*x)/2])*f)/(I*e*h + Sqrt[4*f^2
- e^2*h^2])])/h

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 2405

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/(Sqrt[(f1_) + (g1_.)*(x_)]*Sqrt[(f2_) + (g2_.)*(x_)])
, x_Symbol] :> With[{u = IntHide[1/Sqrt[f1*f2 + g1*g2*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, f1, g1, f2, g2, n}, x] && EqQ[f
2*g1 + f1*g2, 0] && GtQ[f1, 0] && GtQ[f2, 0]

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 4521

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] + (Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(I*a - Rt[-a^2 + b^
2, 2] + b*E^(I*(c + d*x))), x], x] + Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(I*a + Rt[-a^2 + b^2, 2] + b*E^
(I*(c + d*x))), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NegQ[a^2 - b^2]

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 2445

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

\begin{align*} \int \frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt{2-h x} \sqrt{2+h x}} \, dx &=\operatorname{Subst}\left (\int \frac{a+b \log \left (c d^q (e+f x)^{p q}\right )}{\sqrt{2-h x} \sqrt{2+h x}} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{\sin ^{-1}\left (\frac{h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\operatorname{Subst}\left ((b f p q) \int \frac{\sin ^{-1}\left (\frac{h x}{2}\right )}{e h+f h x} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{\sin ^{-1}\left (\frac{h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\operatorname{Subst}\left ((b f p q) \operatorname{Subst}\left (\int \frac{x \cos (x)}{\frac{e h^2}{2}+f h \sin (x)} \, dx,x,\sin ^{-1}\left (\frac{h x}{2}\right )\right ),c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{i b p q \sin ^{-1}\left (\frac{h x}{2}\right )^2}{2 h}+\frac{\sin ^{-1}\left (\frac{h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\operatorname{Subst}\left ((i b f p q) \operatorname{Subst}\left (\int \frac{e^{i x} x}{e^{i x} f h+\frac{1}{2} i e h^2-\frac{1}{2} h \sqrt{4 f^2-e^2 h^2}} \, dx,x,\sin ^{-1}\left (\frac{h x}{2}\right )\right ),c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )-\operatorname{Subst}\left ((i b f p q) \operatorname{Subst}\left (\int \frac{e^{i x} x}{e^{i x} f h+\frac{1}{2} i e h^2+\frac{1}{2} h \sqrt{4 f^2-e^2 h^2}} \, dx,x,\sin ^{-1}\left (\frac{h x}{2}\right )\right ),c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{i b p q \sin ^{-1}\left (\frac{h x}{2}\right )^2}{2 h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 e^{i \sin ^{-1}\left (\frac{h x}{2}\right )} f}{i e h-\sqrt{4 f^2-e^2 h^2}}\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 e^{i \sin ^{-1}\left (\frac{h x}{2}\right )} f}{i e h+\sqrt{4 f^2-e^2 h^2}}\right )}{h}+\frac{\sin ^{-1}\left (\frac{h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}+\operatorname{Subst}\left (\frac{(b p q) \operatorname{Subst}\left (\int \log \left (1+\frac{e^{i x} f h}{\frac{1}{2} i e h^2-\frac{1}{2} h \sqrt{4 f^2-e^2 h^2}}\right ) \, dx,x,\sin ^{-1}\left (\frac{h x}{2}\right )\right )}{h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )+\operatorname{Subst}\left (\frac{(b p q) \operatorname{Subst}\left (\int \log \left (1+\frac{e^{i x} f h}{\frac{1}{2} i e h^2+\frac{1}{2} h \sqrt{4 f^2-e^2 h^2}}\right ) \, dx,x,\sin ^{-1}\left (\frac{h x}{2}\right )\right )}{h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{i b p q \sin ^{-1}\left (\frac{h x}{2}\right )^2}{2 h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 e^{i \sin ^{-1}\left (\frac{h x}{2}\right )} f}{i e h-\sqrt{4 f^2-e^2 h^2}}\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 e^{i \sin ^{-1}\left (\frac{h x}{2}\right )} f}{i e h+\sqrt{4 f^2-e^2 h^2}}\right )}{h}+\frac{\sin ^{-1}\left (\frac{h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\operatorname{Subst}\left (\frac{(i b p q) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{f h x}{\frac{1}{2} i e h^2-\frac{1}{2} h \sqrt{4 f^2-e^2 h^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}\right )}{h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )-\operatorname{Subst}\left (\frac{(i b p q) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{f h x}{\frac{1}{2} i e h^2+\frac{1}{2} h \sqrt{4 f^2-e^2 h^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}\right )}{h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{i b p q \sin ^{-1}\left (\frac{h x}{2}\right )^2}{2 h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 e^{i \sin ^{-1}\left (\frac{h x}{2}\right )} f}{i e h-\sqrt{4 f^2-e^2 h^2}}\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{2 e^{i \sin ^{-1}\left (\frac{h x}{2}\right )} f}{i e h+\sqrt{4 f^2-e^2 h^2}}\right )}{h}+\frac{\sin ^{-1}\left (\frac{h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}+\frac{i b p q \text{Li}_2\left (-\frac{2 e^{i \sin ^{-1}\left (\frac{h x}{2}\right )} f}{i e h-\sqrt{4 f^2-e^2 h^2}}\right )}{h}+\frac{i b p q \text{Li}_2\left (-\frac{2 e^{i \sin ^{-1}\left (\frac{h x}{2}\right )} f}{i e h+\sqrt{4 f^2-e^2 h^2}}\right )}{h}\\ \end{align*}

Mathematica [A]  time = 0.0292965, size = 316, normalized size = 1.1 \[ \frac{i b p q \text{PolyLog}\left (2,\frac{2 i f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{e h-i \sqrt{4 f^2-e^2 h^2}}\right )}{h}+\frac{i b p q \text{PolyLog}\left (2,\frac{2 i f e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{e h+i \sqrt{4 f^2-e^2 h^2}}\right )}{h}+\frac{a \sin ^{-1}\left (\frac{h x}{2}\right )}{h}+\frac{b \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (c \left (d (e+f x)^p\right )^q\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{f h e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{-\frac{1}{2} h \sqrt{4 f^2-e^2 h^2}+\frac{1}{2} i e h^2}\right )}{h}-\frac{b p q \sin ^{-1}\left (\frac{h x}{2}\right ) \log \left (1+\frac{f h e^{i \sin ^{-1}\left (\frac{h x}{2}\right )}}{\frac{1}{2} h \sqrt{4 f^2-e^2 h^2}+\frac{1}{2} i e h^2}\right )}{h}+\frac{i b p q \sin ^{-1}\left (\frac{h x}{2}\right )^2}{2 h} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])/(Sqrt[2 - h*x]*Sqrt[2 + h*x]),x]

[Out]

(a*ArcSin[(h*x)/2])/h + ((I/2)*b*p*q*ArcSin[(h*x)/2]^2)/h - (b*p*q*ArcSin[(h*x)/2]*Log[1 + (E^(I*ArcSin[(h*x)/
2])*f*h)/((I/2)*e*h^2 - (h*Sqrt[4*f^2 - e^2*h^2])/2)])/h - (b*p*q*ArcSin[(h*x)/2]*Log[1 + (E^(I*ArcSin[(h*x)/2
])*f*h)/((I/2)*e*h^2 + (h*Sqrt[4*f^2 - e^2*h^2])/2)])/h + (b*ArcSin[(h*x)/2]*Log[c*(d*(e + f*x)^p)^q])/h + (I*
b*p*q*PolyLog[2, ((2*I)*E^(I*ArcSin[(h*x)/2])*f)/(e*h - I*Sqrt[4*f^2 - e^2*h^2])])/h + (I*b*p*q*PolyLog[2, ((2
*I)*E^(I*ArcSin[(h*x)/2])*f)/(e*h + I*Sqrt[4*f^2 - e^2*h^2])])/h

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Maple [F]  time = 0.962, size = 0, normalized size = 0. \begin{align*} \int{(a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) ){\frac{1}{\sqrt{-hx+2}}}{\frac{1}{\sqrt{hx+2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(d*(f*x+e)^p)^q))/(-h*x+2)^(1/2)/(h*x+2)^(1/2),x)

[Out]

int((a+b*ln(c*(d*(f*x+e)^p)^q))/(-h*x+2)^(1/2)/(h*x+2)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right ) + \log \left (c\right ) + \log \left (d^{q}\right )}{\sqrt{h x + 2} \sqrt{-h x + 2}}\,{d x} + \frac{a \arcsin \left (\frac{h^{2} x}{2 \, \sqrt{h^{2}}}\right )}{\sqrt{h^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(-h*x+2)^(1/2)/(h*x+2)^(1/2),x, algorithm="maxima")

[Out]

b*integrate((log(((f*x + e)^p)^q) + log(c) + log(d^q))/(sqrt(h*x + 2)*sqrt(-h*x + 2)), x) + a*arcsin(1/2*h^2*x
/sqrt(h^2))/sqrt(h^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{h x + 2} \sqrt{-h x + 2} b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + \sqrt{h x + 2} \sqrt{-h x + 2} a}{h^{2} x^{2} - 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(-h*x+2)^(1/2)/(h*x+2)^(1/2),x, algorithm="fricas")

[Out]

integral(-(sqrt(h*x + 2)*sqrt(-h*x + 2)*b*log(((f*x + e)^p*d)^q*c) + sqrt(h*x + 2)*sqrt(-h*x + 2)*a)/(h^2*x^2
- 4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \log{\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}}{\sqrt{- h x + 2} \sqrt{h x + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d*(f*x+e)**p)**q))/(-h*x+2)**(1/2)/(h*x+2)**(1/2),x)

[Out]

Integral((a + b*log(c*(d*(e + f*x)**p)**q))/(sqrt(-h*x + 2)*sqrt(h*x + 2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a}{\sqrt{h x + 2} \sqrt{-h x + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(-h*x+2)^(1/2)/(h*x+2)^(1/2),x, algorithm="giac")

[Out]

integrate((b*log(((f*x + e)^p*d)^q*c) + a)/(sqrt(h*x + 2)*sqrt(-h*x + 2)), x)