3.187 \(\int x^3 (a+b \tanh ^{-1}(c \sqrt{x})) \, dx\)

Optimal. Leaf size=88 \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+\frac{b x^{5/2}}{20 c^3}+\frac{b x^{3/2}}{12 c^5}+\frac{b \sqrt{x}}{4 c^7}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{4 c^8}+\frac{b x^{7/2}}{28 c} \]

[Out]

(b*Sqrt[x])/(4*c^7) + (b*x^(3/2))/(12*c^5) + (b*x^(5/2))/(20*c^3) + (b*x^(7/2))/(28*c) - (b*ArcTanh[c*Sqrt[x]]
)/(4*c^8) + (x^4*(a + b*ArcTanh[c*Sqrt[x]]))/4

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Rubi [A]  time = 0.0415064, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6097, 50, 63, 206} \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+\frac{b x^{5/2}}{20 c^3}+\frac{b x^{3/2}}{12 c^5}+\frac{b \sqrt{x}}{4 c^7}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{4 c^8}+\frac{b x^{7/2}}{28 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*Sqrt[x]]),x]

[Out]

(b*Sqrt[x])/(4*c^7) + (b*x^(3/2))/(12*c^5) + (b*x^(5/2))/(20*c^3) + (b*x^(7/2))/(28*c) - (b*ArcTanh[c*Sqrt[x]]
)/(4*c^8) + (x^4*(a + b*ArcTanh[c*Sqrt[x]]))/4

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

\begin{align*} \int x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{1}{8} (b c) \int \frac{x^{7/2}}{1-c^2 x} \, dx\\ &=\frac{b x^{7/2}}{28 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \int \frac{x^{5/2}}{1-c^2 x} \, dx}{8 c}\\ &=\frac{b x^{5/2}}{20 c^3}+\frac{b x^{7/2}}{28 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \int \frac{x^{3/2}}{1-c^2 x} \, dx}{8 c^3}\\ &=\frac{b x^{3/2}}{12 c^5}+\frac{b x^{5/2}}{20 c^3}+\frac{b x^{7/2}}{28 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \int \frac{\sqrt{x}}{1-c^2 x} \, dx}{8 c^5}\\ &=\frac{b \sqrt{x}}{4 c^7}+\frac{b x^{3/2}}{12 c^5}+\frac{b x^{5/2}}{20 c^3}+\frac{b x^{7/2}}{28 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \int \frac{1}{\sqrt{x} \left (1-c^2 x\right )} \, dx}{8 c^7}\\ &=\frac{b \sqrt{x}}{4 c^7}+\frac{b x^{3/2}}{12 c^5}+\frac{b x^{5/2}}{20 c^3}+\frac{b x^{7/2}}{28 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )}{4 c^7}\\ &=\frac{b \sqrt{x}}{4 c^7}+\frac{b x^{3/2}}{12 c^5}+\frac{b x^{5/2}}{20 c^3}+\frac{b x^{7/2}}{28 c}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{4 c^8}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.031173, size = 114, normalized size = 1.3 \[ \frac{a x^4}{4}+\frac{b x^{5/2}}{20 c^3}+\frac{b x^{3/2}}{12 c^5}+\frac{b \sqrt{x}}{4 c^7}+\frac{b \log \left (1-c \sqrt{x}\right )}{8 c^8}-\frac{b \log \left (c \sqrt{x}+1\right )}{8 c^8}+\frac{b x^{7/2}}{28 c}+\frac{1}{4} b x^4 \tanh ^{-1}\left (c \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTanh[c*Sqrt[x]]),x]

[Out]

(b*Sqrt[x])/(4*c^7) + (b*x^(3/2))/(12*c^5) + (b*x^(5/2))/(20*c^3) + (b*x^(7/2))/(28*c) + (a*x^4)/4 + (b*x^4*Ar
cTanh[c*Sqrt[x]])/4 + (b*Log[1 - c*Sqrt[x]])/(8*c^8) - (b*Log[1 + c*Sqrt[x]])/(8*c^8)

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Maple [A]  time = 0.026, size = 84, normalized size = 1. \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}}{4}{\it Artanh} \left ( c\sqrt{x} \right ) }+{\frac{b}{28\,c}{x}^{{\frac{7}{2}}}}+{\frac{b}{20\,{c}^{3}}{x}^{{\frac{5}{2}}}}+{\frac{b}{12\,{c}^{5}}{x}^{{\frac{3}{2}}}}+{\frac{b}{4\,{c}^{7}}\sqrt{x}}+{\frac{b}{8\,{c}^{8}}\ln \left ( c\sqrt{x}-1 \right ) }-{\frac{b}{8\,{c}^{8}}\ln \left ( 1+c\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x^(1/2))),x)

[Out]

1/4*x^4*a+1/4*b*x^4*arctanh(c*x^(1/2))+1/28*b*x^(7/2)/c+1/20*b*x^(5/2)/c^3+1/12*b*x^(3/2)/c^5+1/4*b*x^(1/2)/c^
7+1/8/c^8*b*ln(c*x^(1/2)-1)-1/8/c^8*b*ln(1+c*x^(1/2))

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Maxima [A]  time = 1.023, size = 116, normalized size = 1.32 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{840} \,{\left (210 \, x^{4} \operatorname{artanh}\left (c \sqrt{x}\right ) + c{\left (\frac{2 \,{\left (15 \, c^{6} x^{\frac{7}{2}} + 21 \, c^{4} x^{\frac{5}{2}} + 35 \, c^{2} x^{\frac{3}{2}} + 105 \, \sqrt{x}\right )}}{c^{8}} - \frac{105 \, \log \left (c \sqrt{x} + 1\right )}{c^{9}} + \frac{105 \, \log \left (c \sqrt{x} - 1\right )}{c^{9}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^(1/2))),x, algorithm="maxima")

[Out]

1/4*a*x^4 + 1/840*(210*x^4*arctanh(c*sqrt(x)) + c*(2*(15*c^6*x^(7/2) + 21*c^4*x^(5/2) + 35*c^2*x^(3/2) + 105*s
qrt(x))/c^8 - 105*log(c*sqrt(x) + 1)/c^9 + 105*log(c*sqrt(x) - 1)/c^9))*b

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Fricas [A]  time = 1.78655, size = 213, normalized size = 2.42 \begin{align*} \frac{210 \, a c^{8} x^{4} + 105 \,{\left (b c^{8} x^{4} - b\right )} \log \left (-\frac{c^{2} x + 2 \, c \sqrt{x} + 1}{c^{2} x - 1}\right ) + 2 \,{\left (15 \, b c^{7} x^{3} + 21 \, b c^{5} x^{2} + 35 \, b c^{3} x + 105 \, b c\right )} \sqrt{x}}{840 \, c^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^(1/2))),x, algorithm="fricas")

[Out]

1/840*(210*a*c^8*x^4 + 105*(b*c^8*x^4 - b)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) + 2*(15*b*c^7*x^3 + 21*
b*c^5*x^2 + 35*b*c^3*x + 105*b*c)*sqrt(x))/c^8

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (a + b \operatorname{atanh}{\left (c \sqrt{x} \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c*x**(1/2))),x)

[Out]

Integral(x**3*(a + b*atanh(c*sqrt(x))), x)

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Giac [A]  time = 1.20434, size = 142, normalized size = 1.61 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{840} \,{\left (105 \, x^{4} \log \left (-\frac{c \sqrt{x} + 1}{c \sqrt{x} - 1}\right ) - c{\left (\frac{105 \, \log \left ({\left | c \sqrt{x} + 1 \right |}\right )}{c^{9}} - \frac{105 \, \log \left ({\left | c \sqrt{x} - 1 \right |}\right )}{c^{9}} - \frac{2 \,{\left (15 \, c^{12} x^{\frac{7}{2}} + 21 \, c^{10} x^{\frac{5}{2}} + 35 \, c^{8} x^{\frac{3}{2}} + 105 \, c^{6} \sqrt{x}\right )}}{c^{14}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^(1/2))),x, algorithm="giac")

[Out]

1/4*a*x^4 + 1/840*(105*x^4*log(-(c*sqrt(x) + 1)/(c*sqrt(x) - 1)) - c*(105*log(abs(c*sqrt(x) + 1))/c^9 - 105*lo
g(abs(c*sqrt(x) - 1))/c^9 - 2*(15*c^12*x^(7/2) + 21*c^10*x^(5/2) + 35*c^8*x^(3/2) + 105*c^6*sqrt(x))/c^14))*b