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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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])

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]

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*}

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Mathematica [A]  time = 0.03, size = 114, normalized size = 1.30 \[ \frac {a x^4}{4}+\frac {b \log \left (1-c \sqrt {x}\right )}{8 c^8}-\frac {b \log \left (c \sqrt {x}+1\right )}{8 c^8}+\frac {b \sqrt {x}}{4 c^7}+\frac {b x^{3/2}}{12 c^5}+\frac {b x^{5/2}}{20 c^3}+\frac {b x^{7/2}}{28 c}+\frac {1}{4} b x^4 \tanh ^{-1}\left (c \sqrt {x}\right ) \]

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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fricas [A]  time = 0.64, size = 89, normalized size = 1.01 \[ \frac {210 \, a c^{8} x^{4} + 105 \, {\left (b c^{8} x^{4} - b\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 2 \, {\left (15 \, b c^{7} x^{3} + 21 \, b c^{5} x^{2} + 35 \, b c^{3} x + 105 \, b c\right )} \sqrt {x}}{840 \, c^{8}} \]

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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giac [B]  time = 0.21, size = 359, normalized size = 4.08 \[ \frac {1}{4} \, a x^{4} + \frac {2}{105} \, b c {\left (\frac {\frac {105 \, {\left (c \sqrt {x} + 1\right )}^{6}}{{\left (c \sqrt {x} - 1\right )}^{6}} - \frac {315 \, {\left (c \sqrt {x} + 1\right )}^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {770 \, {\left (c \sqrt {x} + 1\right )}^{4}}{{\left (c \sqrt {x} - 1\right )}^{4}} - \frac {770 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {609 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} - \frac {203 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 44}{c^{9} {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} - 1\right )}^{7}} + \frac {105 \, {\left (\frac {{\left (c \sqrt {x} + 1\right )}^{7}}{{\left (c \sqrt {x} - 1\right )}^{7}} + \frac {7 \, {\left (c \sqrt {x} + 1\right )}^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {7 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {c \sqrt {x} + 1}{c \sqrt {x} - 1}\right )} \log \left (-\frac {\frac {c {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} + 1\right )}}{\frac {{\left (c \sqrt {x} + 1\right )} c}{c \sqrt {x} - 1} - c} + 1}{\frac {c {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} + 1\right )}}{\frac {{\left (c \sqrt {x} + 1\right )} c}{c \sqrt {x} - 1} - c} - 1}\right )}{c^{9} {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} - 1\right )}^{8}}\right )} \]

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 + 2/105*b*c*((105*(c*sqrt(x) + 1)^6/(c*sqrt(x) - 1)^6 - 315*(c*sqrt(x) + 1)^5/(c*sqrt(x) - 1)^5 + 77
0*(c*sqrt(x) + 1)^4/(c*sqrt(x) - 1)^4 - 770*(c*sqrt(x) + 1)^3/(c*sqrt(x) - 1)^3 + 609*(c*sqrt(x) + 1)^2/(c*sqr
t(x) - 1)^2 - 203*(c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 44)/(c^9*((c*sqrt(x) + 1)/(c*sqrt(x) - 1) - 1)^7) + 105*((
c*sqrt(x) + 1)^7/(c*sqrt(x) - 1)^7 + 7*(c*sqrt(x) + 1)^5/(c*sqrt(x) - 1)^5 + 7*(c*sqrt(x) + 1)^3/(c*sqrt(x) -
1)^3 + (c*sqrt(x) + 1)/(c*sqrt(x) - 1))*log(-(c*((c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 1)/((c*sqrt(x) + 1)*c/(c*sq
rt(x) - 1) - c) + 1)/(c*((c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 1)/((c*sqrt(x) + 1)*c/(c*sqrt(x) - 1) - c) - 1))/(c
^9*((c*sqrt(x) + 1)/(c*sqrt(x) - 1) - 1)^8))

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maple [A]  time = 0.03, size = 84, normalized size = 0.95 \[ \frac {x^{4} a}{4}+\frac {b \,x^{4} \arctanh \left (c \sqrt {x}\right )}{4}+\frac {b \,x^{\frac {7}{2}}}{28 c}+\frac {b \,x^{\frac {5}{2}}}{20 c^{3}}+\frac {b \,x^{\frac {3}{2}}}{12 c^{5}}+\frac {b \sqrt {x}}{4 c^{7}}+\frac {b \ln \left (c \sqrt {x}-1\right )}{8 c^{8}}-\frac {b \ln \left (1+c \sqrt {x}\right )}{8 c^{8}} \]

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 = 0.32, size = 86, normalized size = 0.98 \[ \frac {1}{4} \, a x^{4} + \frac {1}{840} \, {\left (210 \, x^{4} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (15 \, c^{6} x^{\frac {7}{2}} + 21 \, c^{4} x^{\frac {5}{2}} + 35 \, c^{2} x^{\frac {3}{2}} + 105 \, \sqrt {x}\right )}}{c^{8}} - \frac {105 \, \log \left (c \sqrt {x} + 1\right )}{c^{9}} + \frac {105 \, \log \left (c \sqrt {x} - 1\right )}{c^{9}}\right )}\right )} b \]

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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mupad [B]  time = 1.43, size = 86, normalized size = 0.98 \[ \frac {\frac {b\,c^3\,x^{3/2}}{12}-\frac {b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{4}+\frac {b\,c^5\,x^{5/2}}{20}+\frac {b\,c^7\,x^{7/2}}{28}+\frac {b\,c\,\sqrt {x}}{4}}{c^8}+\frac {b\,\left (105\,x^4\,\ln \left (c\,\sqrt {x}+1\right )-105\,x^4\,\ln \left (1-c\,\sqrt {x}\right )\right )}{840}+\frac {a\,x^4}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x^(1/2))),x)

[Out]

((b*c^3*x^(3/2))/12 - (b*atanh(c*x^(1/2)))/4 + (b*c^5*x^(5/2))/20 + (b*c^7*x^(7/2))/28 + (b*c*x^(1/2))/4)/c^8
+ (b*(105*x^4*log(c*x^(1/2) + 1) - 105*x^4*log(1 - c*x^(1/2))))/840 + (a*x^4)/4

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )\, dx \]

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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