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

Optimal. Leaf size=335 \[ \frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}-\frac {b p q \text {Li}_2\left (-\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}-\sqrt {h e^2+2 f^2}}\right )}{\sqrt {h}}-\frac {b p q \text {Li}_2\left (-\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{\sqrt {h} e+\sqrt {h e^2+2 f^2}}\right )}{\sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (\frac {\sqrt {2} f e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}}{e \sqrt {h}-\sqrt {e^2 h+2 f^2}}+1\right )}{\sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (\frac {\sqrt {2} f e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}}{\sqrt {e^2 h+2 f^2}+e \sqrt {h}}+1\right )}{\sqrt {h}}+\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )^2}{2 \sqrt {h}} \]

[Out]

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

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Rubi [A]  time = 0.83, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {215, 2404, 12, 5799, 5561, 2190, 2279, 2391, 2445} \[ -\frac {b p q \text {PolyLog}\left (2,-\frac {\sqrt {2} f e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}}{e \sqrt {h}-\sqrt {e^2 h+2 f^2}}\right )}{\sqrt {h}}-\frac {b p q \text {PolyLog}\left (2,-\frac {\sqrt {2} f e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}}{\sqrt {e^2 h+2 f^2}+e \sqrt {h}}\right )}{\sqrt {h}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (\frac {\sqrt {2} f e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}}{e \sqrt {h}-\sqrt {e^2 h+2 f^2}}+1\right )}{\sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (\frac {\sqrt {2} f e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}}{\sqrt {e^2 h+2 f^2}+e \sqrt {h}}+1\right )}{\sqrt {h}}+\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )^2}{2 \sqrt {h}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*p*q*ArcSinh[(Sqrt[h]*x)/Sqrt[2]]^2)/(2*Sqrt[h]) - (b*p*q*ArcSinh[(Sqrt[h]*x)/Sqrt[2]]*Log[1 + (Sqrt[2]*E^Ar
cSinh[(Sqrt[h]*x)/Sqrt[2]]*f)/(e*Sqrt[h] - Sqrt[2*f^2 + e^2*h])])/Sqrt[h] - (b*p*q*ArcSinh[(Sqrt[h]*x)/Sqrt[2]
]*Log[1 + (Sqrt[2]*E^ArcSinh[(Sqrt[h]*x)/Sqrt[2]]*f)/(e*Sqrt[h] + Sqrt[2*f^2 + e^2*h])])/Sqrt[h] + (ArcSinh[(S
qrt[h]*x)/Sqrt[2]]*(a + b*Log[c*(d*(e + f*x)^p)^q]))/Sqrt[h] - (b*p*q*PolyLog[2, -((Sqrt[2]*E^ArcSinh[(Sqrt[h]
*x)/Sqrt[2]]*f)/(e*Sqrt[h] - Sqrt[2*f^2 + e^2*h]))])/Sqrt[h] - (b*p*q*PolyLog[2, -((Sqrt[2]*E^ArcSinh[(Sqrt[h]
*x)/Sqrt[2]]*f)/(e*Sqrt[h] + Sqrt[2*f^2 + e^2*h]))])/Sqrt[h]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 215

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

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 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]]

Rule 5561

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

Rule 5799

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Subst[Int[((a + b*x)^n*Cosh[x
])/(c*d + e*Sinh[x]), x], x, ArcSinh[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^2}} \, dx &=\operatorname {Subst}\left (\int \frac {a+b \log \left (c d^q (e+f x)^{p q}\right )}{\sqrt {2+h x^2}} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}-\operatorname {Subst}\left ((b f p q) \int \frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}{\sqrt {h} (e+f x)} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}-\operatorname {Subst}\left (\frac {(b f p q) \int \frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}{e+f x} \, dx}{\sqrt {h}},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}-\operatorname {Subst}\left (\frac {(b f p q) \operatorname {Subst}\left (\int \frac {x \cosh (x)}{\frac {e \sqrt {h}}{\sqrt {2}}+f \sinh (x)} \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )\right )}{\sqrt {h}},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )^2}{2 \sqrt {h}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}-\operatorname {Subst}\left (\frac {(b f p q) \operatorname {Subst}\left (\int \frac {e^x x}{e^x f+\frac {e \sqrt {h}}{\sqrt {2}}-\frac {\sqrt {2 f^2+e^2 h}}{\sqrt {2}}} \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )\right )}{\sqrt {h}},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )-\operatorname {Subst}\left (\frac {(b f p q) \operatorname {Subst}\left (\int \frac {e^x x}{e^x f+\frac {e \sqrt {h}}{\sqrt {2}}+\frac {\sqrt {2 f^2+e^2 h}}{\sqrt {2}}} \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )\right )}{\sqrt {h}},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )^2}{2 \sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}-\sqrt {2 f^2+e^2 h}}\right )}{\sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}+\sqrt {2 f^2+e^2 h}}\right )}{\sqrt {h}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}+\operatorname {Subst}\left (\frac {(b p q) \operatorname {Subst}\left (\int \log \left (1+\frac {e^x f}{\frac {e \sqrt {h}}{\sqrt {2}}-\frac {\sqrt {2 f^2+e^2 h}}{\sqrt {2}}}\right ) \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )\right )}{\sqrt {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^x f}{\frac {e \sqrt {h}}{\sqrt {2}}+\frac {\sqrt {2 f^2+e^2 h}}{\sqrt {2}}}\right ) \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )\right )}{\sqrt {h}},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )^2}{2 \sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}-\sqrt {2 f^2+e^2 h}}\right )}{\sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}+\sqrt {2 f^2+e^2 h}}\right )}{\sqrt {h}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}+\operatorname {Subst}\left (\frac {(b p q) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {f x}{\frac {e \sqrt {h}}{\sqrt {2}}-\frac {\sqrt {2 f^2+e^2 h}}{\sqrt {2}}}\right )}{x} \, dx,x,e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}\right )}{\sqrt {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 \frac {\log \left (1+\frac {f x}{\frac {e \sqrt {h}}{\sqrt {2}}+\frac {\sqrt {2 f^2+e^2 h}}{\sqrt {2}}}\right )}{x} \, dx,x,e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}\right )}{\sqrt {h}},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )^2}{2 \sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}-\sqrt {2 f^2+e^2 h}}\right )}{\sqrt {h}}-\frac {b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}+\sqrt {2 f^2+e^2 h}}\right )}{\sqrt {h}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {h}}-\frac {b p q \text {Li}_2\left (-\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}-\sqrt {2 f^2+e^2 h}}\right )}{\sqrt {h}}-\frac {b p q \text {Li}_2\left (-\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{e \sqrt {h}+\sqrt {2 f^2+e^2 h}}\right )}{\sqrt {h}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 284, normalized size = 0.85 \[ \frac {\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right ) \left (2 a+2 b \log \left (c \left (d (e+f x)^p\right )^q\right )-2 b p q \log \left (\frac {\sqrt {2} f e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}}{e \sqrt {h}-\sqrt {e^2 h+2 f^2}}+1\right )-2 b p q \log \left (\frac {\sqrt {2} f e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )}}{\sqrt {e^2 h+2 f^2}+e \sqrt {h}}+1\right )+b p q \sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )\right )-2 b p q \text {Li}_2\left (\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{\sqrt {h e^2+2 f^2}-e \sqrt {h}}\right )-2 b p q \text {Li}_2\left (-\frac {\sqrt {2} e^{\sinh ^{-1}\left (\frac {\sqrt {h} x}{\sqrt {2}}\right )} f}{\sqrt {h} e+\sqrt {h e^2+2 f^2}}\right )}{2 \sqrt {h}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcSinh[(Sqrt[h]*x)/Sqrt[2]]*(2*a + b*p*q*ArcSinh[(Sqrt[h]*x)/Sqrt[2]] - 2*b*p*q*Log[1 + (Sqrt[2]*E^ArcSinh[(
Sqrt[h]*x)/Sqrt[2]]*f)/(e*Sqrt[h] - Sqrt[2*f^2 + e^2*h])] - 2*b*p*q*Log[1 + (Sqrt[2]*E^ArcSinh[(Sqrt[h]*x)/Sqr
t[2]]*f)/(e*Sqrt[h] + Sqrt[2*f^2 + e^2*h])] + 2*b*Log[c*(d*(e + f*x)^p)^q]) - 2*b*p*q*PolyLog[2, (Sqrt[2]*E^Ar
cSinh[(Sqrt[h]*x)/Sqrt[2]]*f)/(-(e*Sqrt[h]) + Sqrt[2*f^2 + e^2*h])] - 2*b*p*q*PolyLog[2, -((Sqrt[2]*E^ArcSinh[
(Sqrt[h]*x)/Sqrt[2]]*f)/(e*Sqrt[h] + Sqrt[2*f^2 + e^2*h]))])/(2*Sqrt[h])

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {h x^{2} + 2} b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + \sqrt {h x^{2} + 2} a}{h x^{2} + 2}, x\right ) \]

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+2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a}{\sqrt {h x^{2} + 2}}\,{d x} \]

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+2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )+a}{\sqrt {h \,x^{2}+2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {q \log \relax (d) + \log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right ) + \log \relax (c)}{\sqrt {h x^{2} + 2}}\,{d x} + \frac {a \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {2} \sqrt {h} x\right )}{\sqrt {h}} \]

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+2)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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+2)**(1/2),x)

[Out]

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

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