3.6.21 \(\int \frac {a+b \log (c (d (e+f x)^p)^q)}{\sqrt {2-h x} \sqrt {2+h x}} \, dx\) [521]

Optimal. Leaf size=287 \[ \frac {i b p q \sin ^{-1}\left (\frac {h x}{2}\right )^2}{2 h}-\frac {b p q \sin ^{-1}\left (\frac {h x}{2}\right ) \log \left (1+\frac {2 e^{i \sin ^{-1}\left (\frac {h x}{2}\right )} f}{i e h-\sqrt {4 f^2-e^2 h^2}}\right )}{h}-\frac {b p q \sin ^{-1}\left (\frac {h x}{2}\right ) \log \left (1+\frac {2 e^{i \sin ^{-1}\left (\frac {h x}{2}\right )} f}{i e h+\sqrt {4 f^2-e^2 h^2}}\right )}{h}+\frac {\sin ^{-1}\left (\frac {h x}{2}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}+\frac {i b p q \text {Li}_2\left (-\frac {2 e^{i \sin ^{-1}\left (\frac {h x}{2}\right )} f}{i e h-\sqrt {4 f^2-e^2 h^2}}\right )}{h}+\frac {i b p q \text {Li}_2\left (-\frac {2 e^{i \sin ^{-1}\left (\frac {h x}{2}\right )} f}{i e h+\sqrt {4 f^2-e^2 h^2}}\right )}{h} \]

[Out]

1/2*I*b*p*q*arcsin(1/2*h*x)^2/h+arcsin(1/2*h*x)*(a+b*ln(c*(d*(f*x+e)^p)^q))/h-b*p*q*arcsin(1/2*h*x)*ln(1+2*(1/
2*I*h*x+1/2*(-h^2*x^2+4)^(1/2))*f/(I*e*h-(-e^2*h^2+4*f^2)^(1/2)))/h-b*p*q*arcsin(1/2*h*x)*ln(1+2*(1/2*I*h*x+1/
2*(-h^2*x^2+4)^(1/2))*f/(I*e*h+(-e^2*h^2+4*f^2)^(1/2)))/h+I*b*p*q*polylog(2,-2*(1/2*I*h*x+1/2*(-h^2*x^2+4)^(1/
2))*f/(I*e*h-(-e^2*h^2+4*f^2)^(1/2)))/h+I*b*p*q*polylog(2,-2*(1/2*I*h*x+1/2*(-h^2*x^2+4)^(1/2))*f/(I*e*h+(-e^2
*h^2+4*f^2)^(1/2)))/h

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Rubi [A]
time = 0.68, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {222, 2452, 4825, 4617, 2221, 2317, 2438, 2495} \begin {gather*} \frac {i b p q \text {PolyLog}\left (2,-\frac {2 f e^{i \text {ArcSin}\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 \text {ArcSin}\left (\frac {h x}{2}\right )}}{\sqrt {4 f^2-e^2 h^2}+i e h}\right )}{h}+\frac {\text {ArcSin}\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 \text {ArcSin}\left (\frac {h x}{2}\right ) \log \left (1+\frac {2 f e^{i \text {ArcSin}\left (\frac {h x}{2}\right )}}{-\sqrt {4 f^2-e^2 h^2}+i e h}\right )}{h}-\frac {b p q \text {ArcSin}\left (\frac {h x}{2}\right ) \log \left (1+\frac {2 f e^{i \text {ArcSin}\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 {ArcSin}\left (\frac {h x}{2}\right )^2}{2 h} \end {gather*}

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 222

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 2452

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 2495

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]]

Rule 4617

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 4825

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 \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {2-h x} \sqrt {2+h x}} \, dx &=\text {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}-\text {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}-\text {Subst}\left ((b f p q) \text {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}-\text {Subst}\left ((i b f p q) \text {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 )-\text {Subst}\left ((i b f p q) \text {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}+\text {Subst}\left (\frac {(b p q) \text {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 )+\text {Subst}\left (\frac {(b p q) \text {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}-\text {Subst}\left (\frac {(i b p q) \text {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 )-\text {Subst}\left (\frac {(i b p q) \text {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*}

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Mathematica [A]
time = 0.02, size = 316, normalized size = 1.10 \begin {gather*} \frac {a \sin ^{-1}\left (\frac {h x}{2}\right )}{h}+\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 {e^{i \sin ^{-1}\left (\frac {h x}{2}\right )} f h}{\frac {1}{2} i e h^2-\frac {1}{2} 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 {e^{i \sin ^{-1}\left (\frac {h x}{2}\right )} f h}{\frac {1}{2} i e h^2+\frac {1}{2} h \sqrt {4 f^2-e^2 h^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 {i b p q \text {Li}_2\left (\frac {2 i e^{i \sin ^{-1}\left (\frac {h x}{2}\right )} f}{e h-i \sqrt {4 f^2-e^2 h^2}}\right )}{h}+\frac {i b p q \text {Li}_2\left (\frac {2 i e^{i \sin ^{-1}\left (\frac {h x}{2}\right )} f}{e h+i \sqrt {4 f^2-e^2 h^2}}\right )}{h} \end {gather*}

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 = 0.34, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )}{\sqrt {-h x +2}\, \sqrt {h x +2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )}{\sqrt {2-h\,x}\,\sqrt {h\,x+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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