3.196 \(\int x^2 (a+b \tanh ^{-1}(c \sqrt {x}))^2 \, dx\)

Optimal. Leaf size=173 \[ -\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{3 c^6}+\frac {2 a b \sqrt {x}}{3 c^5}+\frac {2 b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{9 c^3}+\frac {2 b x^{5/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{15 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+\frac {2 b^2 \sqrt {x} \tanh ^{-1}\left (c \sqrt {x}\right )}{3 c^5}+\frac {8 b^2 x}{45 c^4}+\frac {b^2 x^2}{30 c^2}+\frac {23 b^2 \log \left (1-c^2 x\right )}{45 c^6} \]

[Out]

8/45*b^2*x/c^4+1/30*b^2*x^2/c^2+2/9*b*x^(3/2)*(a+b*arctanh(c*x^(1/2)))/c^3+2/15*b*x^(5/2)*(a+b*arctanh(c*x^(1/
2)))/c-1/3*(a+b*arctanh(c*x^(1/2)))^2/c^6+1/3*x^3*(a+b*arctanh(c*x^(1/2)))^2+23/45*b^2*ln(-c^2*x+1)/c^6+2/3*a*
b*x^(1/2)/c^5+2/3*b^2*arctanh(c*x^(1/2))*x^(1/2)/c^5

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Rubi [F]  time = 0.02, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx \]

Verification is Not applicable to the result.

[In]

Int[x^2*(a + b*ArcTanh[c*Sqrt[x]])^2,x]

[Out]

Defer[Int][x^2*(a + b*ArcTanh[c*Sqrt[x]])^2, x]

Rubi steps

\begin {align*} \int x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx &=\int x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 194, normalized size = 1.12 \[ \frac {30 a^2 c^6 x^3+12 a b c^5 x^{5/2}+20 a b c^3 x^{3/2}+4 b c \sqrt {x} \tanh ^{-1}\left (c \sqrt {x}\right ) \left (15 a c^5 x^{5/2}+b \left (3 c^4 x^2+5 c^2 x+15\right )\right )+60 a b c \sqrt {x}+2 b (15 a+23 b) \log \left (1-c \sqrt {x}\right )-30 a b \log \left (c \sqrt {x}+1\right )+30 b^2 \left (c^6 x^3-1\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^2+3 b^2 c^4 x^2+16 b^2 c^2 x+46 b^2 \log \left (c \sqrt {x}+1\right )}{90 c^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTanh[c*Sqrt[x]])^2,x]

[Out]

(60*a*b*c*Sqrt[x] + 16*b^2*c^2*x + 20*a*b*c^3*x^(3/2) + 3*b^2*c^4*x^2 + 12*a*b*c^5*x^(5/2) + 30*a^2*c^6*x^3 +
4*b*c*Sqrt[x]*(15*a*c^5*x^(5/2) + b*(15 + 5*c^2*x + 3*c^4*x^2))*ArcTanh[c*Sqrt[x]] + 30*b^2*(-1 + c^6*x^3)*Arc
Tanh[c*Sqrt[x]]^2 + 2*b*(15*a + 23*b)*Log[1 - c*Sqrt[x]] - 30*a*b*Log[1 + c*Sqrt[x]] + 46*b^2*Log[1 + c*Sqrt[x
]])/(90*c^6)

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fricas [A]  time = 1.21, size = 241, normalized size = 1.39 \[ \frac {60 \, a^{2} c^{6} x^{3} + 6 \, b^{2} c^{4} x^{2} + 32 \, b^{2} c^{2} x + 15 \, {\left (b^{2} c^{6} x^{3} - b^{2}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right )^{2} + 4 \, {\left (15 \, a b c^{6} - 15 \, a b + 23 \, b^{2}\right )} \log \left (c \sqrt {x} + 1\right ) - 4 \, {\left (15 \, a b c^{6} - 15 \, a b - 23 \, b^{2}\right )} \log \left (c \sqrt {x} - 1\right ) + 4 \, {\left (15 \, a b c^{6} x^{3} - 15 \, a b c^{6} + {\left (3 \, b^{2} c^{5} x^{2} + 5 \, b^{2} c^{3} x + 15 \, b^{2} c\right )} \sqrt {x}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 8 \, {\left (3 \, a b c^{5} x^{2} + 5 \, a b c^{3} x + 15 \, a b c\right )} \sqrt {x}}{180 \, c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/180*(60*a^2*c^6*x^3 + 6*b^2*c^4*x^2 + 32*b^2*c^2*x + 15*(b^2*c^6*x^3 - b^2)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(
c^2*x - 1))^2 + 4*(15*a*b*c^6 - 15*a*b + 23*b^2)*log(c*sqrt(x) + 1) - 4*(15*a*b*c^6 - 15*a*b - 23*b^2)*log(c*s
qrt(x) - 1) + 4*(15*a*b*c^6*x^3 - 15*a*b*c^6 + (3*b^2*c^5*x^2 + 5*b^2*c^3*x + 15*b^2*c)*sqrt(x))*log(-(c^2*x +
2*c*sqrt(x) + 1)/(c^2*x - 1)) + 8*(3*a*b*c^5*x^2 + 5*a*b*c^3*x + 15*a*b*c)*sqrt(x))/c^6

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {artanh}\left (c \sqrt {x}\right ) + a\right )}^{2} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctanh(c*sqrt(x)) + a)^2*x^2, x)

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maple [B]  time = 0.06, size = 358, normalized size = 2.07 \[ \frac {x^{3} a^{2}}{3}+\frac {b^{2} x^{3} \arctanh \left (c \sqrt {x}\right )^{2}}{3}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) x^{\frac {5}{2}}}{15 c}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) x^{\frac {3}{2}}}{9 c^{3}}+\frac {2 b^{2} \arctanh \left (c \sqrt {x}\right ) \sqrt {x}}{3 c^{5}}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{3 c^{6}}-\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{3 c^{6}}+\frac {b^{2} \ln \left (c \sqrt {x}-1\right )^{2}}{12 c^{6}}-\frac {b^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{6 c^{6}}+\frac {b^{2} \ln \left (1+c \sqrt {x}\right )^{2}}{12 c^{6}}-\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{6 c^{6}}+\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{6 c^{6}}+\frac {b^{2} x^{2}}{30 c^{2}}+\frac {8 b^{2} x}{45 c^{4}}+\frac {23 b^{2} \ln \left (c \sqrt {x}-1\right )}{45 c^{6}}+\frac {23 b^{2} \ln \left (1+c \sqrt {x}\right )}{45 c^{6}}+\frac {2 a b \,x^{3} \arctanh \left (c \sqrt {x}\right )}{3}+\frac {2 x^{\frac {5}{2}} a b}{15 c}+\frac {2 a b \,x^{\frac {3}{2}}}{9 c^{3}}+\frac {2 a b \sqrt {x}}{3 c^{5}}+\frac {a b \ln \left (c \sqrt {x}-1\right )}{3 c^{6}}-\frac {a b \ln \left (1+c \sqrt {x}\right )}{3 c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*a^2+1/3*b^2*x^3*arctanh(c*x^(1/2))^2+2/15/c*b^2*arctanh(c*x^(1/2))*x^(5/2)+2/9/c^3*b^2*arctanh(c*x^(1/
2))*x^(3/2)+2/3*b^2*arctanh(c*x^(1/2))*x^(1/2)/c^5+1/3/c^6*b^2*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)-1/3/c^6*b^2*
arctanh(c*x^(1/2))*ln(1+c*x^(1/2))+1/12/c^6*b^2*ln(c*x^(1/2)-1)^2-1/6/c^6*b^2*ln(c*x^(1/2)-1)*ln(1/2+1/2*c*x^(
1/2))+1/12/c^6*b^2*ln(1+c*x^(1/2))^2-1/6/c^6*b^2*ln(-1/2*c*x^(1/2)+1/2)*ln(1+c*x^(1/2))+1/6/c^6*b^2*ln(-1/2*c*
x^(1/2)+1/2)*ln(1/2+1/2*c*x^(1/2))+1/30*b^2*x^2/c^2+8/45*b^2*x/c^4+23/45/c^6*b^2*ln(c*x^(1/2)-1)+23/45/c^6*b^2
*ln(1+c*x^(1/2))+2/3*a*b*x^3*arctanh(c*x^(1/2))+2/15/c*x^(5/2)*a*b+2/9/c^3*a*b*x^(3/2)+2/3*a*b*x^(1/2)/c^5+1/3
/c^6*a*b*ln(c*x^(1/2)-1)-1/3/c^6*a*b*ln(1+c*x^(1/2))

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maxima [A]  time = 0.34, size = 241, normalized size = 1.39 \[ \frac {1}{3} \, b^{2} x^{3} \operatorname {artanh}\left (c \sqrt {x}\right )^{2} + \frac {1}{3} \, a^{2} x^{3} + \frac {1}{45} \, {\left (30 \, x^{3} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{\frac {5}{2}} + 5 \, c^{2} x^{\frac {3}{2}} + 15 \, \sqrt {x}\right )}}{c^{6}} - \frac {15 \, \log \left (c \sqrt {x} + 1\right )}{c^{7}} + \frac {15 \, \log \left (c \sqrt {x} - 1\right )}{c^{7}}\right )}\right )} a b + \frac {1}{180} \, {\left (4 \, c {\left (\frac {2 \, {\left (3 \, c^{4} x^{\frac {5}{2}} + 5 \, c^{2} x^{\frac {3}{2}} + 15 \, \sqrt {x}\right )}}{c^{6}} - \frac {15 \, \log \left (c \sqrt {x} + 1\right )}{c^{7}} + \frac {15 \, \log \left (c \sqrt {x} - 1\right )}{c^{7}}\right )} \operatorname {artanh}\left (c \sqrt {x}\right ) + \frac {6 \, c^{4} x^{2} + 32 \, c^{2} x - 2 \, {\left (15 \, \log \left (c \sqrt {x} - 1\right ) - 46\right )} \log \left (c \sqrt {x} + 1\right ) + 15 \, \log \left (c \sqrt {x} + 1\right )^{2} + 15 \, \log \left (c \sqrt {x} - 1\right )^{2} + 92 \, \log \left (c \sqrt {x} - 1\right )}{c^{6}}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3*arctanh(c*sqrt(x))^2 + 1/3*a^2*x^3 + 1/45*(30*x^3*arctanh(c*sqrt(x)) + c*(2*(3*c^4*x^(5/2) + 5*c^2
*x^(3/2) + 15*sqrt(x))/c^6 - 15*log(c*sqrt(x) + 1)/c^7 + 15*log(c*sqrt(x) - 1)/c^7))*a*b + 1/180*(4*c*(2*(3*c^
4*x^(5/2) + 5*c^2*x^(3/2) + 15*sqrt(x))/c^6 - 15*log(c*sqrt(x) + 1)/c^7 + 15*log(c*sqrt(x) - 1)/c^7)*arctanh(c
*sqrt(x)) + (6*c^4*x^2 + 32*c^2*x - 2*(15*log(c*sqrt(x) - 1) - 46)*log(c*sqrt(x) + 1) + 15*log(c*sqrt(x) + 1)^
2 + 15*log(c*sqrt(x) - 1)^2 + 92*log(c*sqrt(x) - 1))/c^6)*b^2

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mupad [B]  time = 1.57, size = 185, normalized size = 1.07 \[ \frac {46\,b^2\,\ln \left (c^2\,x-1\right )-30\,b^2\,{\mathrm {atanh}\left (c\,\sqrt {x}\right )}^2-60\,a\,b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+16\,b^2\,c^2\,x+30\,a^2\,c^6\,x^3+3\,b^2\,c^4\,x^2+30\,b^2\,c^6\,x^3\,{\mathrm {atanh}\left (c\,\sqrt {x}\right )}^2+60\,b^2\,c\,\sqrt {x}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+60\,a\,b\,c\,\sqrt {x}+20\,b^2\,c^3\,x^{3/2}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+12\,b^2\,c^5\,x^{5/2}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+20\,a\,b\,c^3\,x^{3/2}+12\,a\,b\,c^5\,x^{5/2}+60\,a\,b\,c^6\,x^3\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{90\,c^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atanh(c*x^(1/2)))^2,x)

[Out]

(46*b^2*log(c^2*x - 1) - 30*b^2*atanh(c*x^(1/2))^2 - 60*a*b*atanh(c*x^(1/2)) + 16*b^2*c^2*x + 30*a^2*c^6*x^3 +
3*b^2*c^4*x^2 + 30*b^2*c^6*x^3*atanh(c*x^(1/2))^2 + 60*b^2*c*x^(1/2)*atanh(c*x^(1/2)) + 60*a*b*c*x^(1/2) + 20
*b^2*c^3*x^(3/2)*atanh(c*x^(1/2)) + 12*b^2*c^5*x^(5/2)*atanh(c*x^(1/2)) + 20*a*b*c^3*x^(3/2) + 12*a*b*c^5*x^(5
/2) + 60*a*b*c^6*x^3*atanh(c*x^(1/2)))/(90*c^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*atanh(c*sqrt(x)))**2, x)

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