3.75 \(\int (a+b \cos ^{-1}(1+d x^2))^2 \, dx\)

Optimal. Leaf size=63 \[ -\frac {4 b \sqrt {-d^2 x^4-2 d x^2} \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )}{d x}+x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^2-8 b^2 x \]

[Out]

-8*b^2*x+x*(a+b*arccos(d*x^2+1))^2-4*b*(a+b*arccos(d*x^2+1))*(-d^2*x^4-2*d*x^2)^(1/2)/d/x

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Rubi [A]  time = 0.01, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4815, 8} \[ -\frac {4 b \sqrt {-d^2 x^4-2 d x^2} \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )}{d x}+x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^2-8 b^2 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[1 + d*x^2])^2,x]

[Out]

-8*b^2*x - (4*b*Sqrt[-2*d*x^2 - d^2*x^4]*(a + b*ArcCos[1 + d*x^2]))/(d*x) + x*(a + b*ArcCos[1 + d*x^2])^2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4815

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcCos[c + d*x^2])^n, x] + (-
Dist[4*b^2*n*(n - 1), Int[(a + b*ArcCos[c + d*x^2])^(n - 2), x], x] - Simp[(2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*(
a + b*ArcCos[c + d*x^2])^(n - 1))/(d*x), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^2 \, dx &=-\frac {4 b \sqrt {-2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )}{d x}+x \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^2-\left (8 b^2\right ) \int 1 \, dx\\ &=-8 b^2 x-\frac {4 b \sqrt {-2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )}{d x}+x \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^2\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 98, normalized size = 1.56 \[ x \left (a^2-8 b^2\right )-\frac {4 a b \sqrt {-d x^2 \left (d x^2+2\right )}}{d x}+\frac {2 b \cos ^{-1}\left (d x^2+1\right ) \left (a d x^2-2 b \sqrt {-d x^2 \left (d x^2+2\right )}\right )}{d x}+b^2 x \cos ^{-1}\left (d x^2+1\right )^2 \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCos[1 + d*x^2])^2,x]

[Out]

(a^2 - 8*b^2)*x - (4*a*b*Sqrt[-(d*x^2*(2 + d*x^2))])/(d*x) + (2*b*(a*d*x^2 - 2*b*Sqrt[-(d*x^2*(2 + d*x^2))])*A
rcCos[1 + d*x^2])/(d*x) + b^2*x*ArcCos[1 + d*x^2]^2

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fricas [A]  time = 0.45, size = 91, normalized size = 1.44 \[ \frac {b^{2} d x^{2} \arccos \left (d x^{2} + 1\right )^{2} + 2 \, a b d x^{2} \arccos \left (d x^{2} + 1\right ) + {\left (a^{2} - 8 \, b^{2}\right )} d x^{2} - 4 \, \sqrt {-d^{2} x^{4} - 2 \, d x^{2}} {\left (b^{2} \arccos \left (d x^{2} + 1\right ) + a b\right )}}{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(d*x^2+1))^2,x, algorithm="fricas")

[Out]

(b^2*d*x^2*arccos(d*x^2 + 1)^2 + 2*a*b*d*x^2*arccos(d*x^2 + 1) + (a^2 - 8*b^2)*d*x^2 - 4*sqrt(-d^2*x^4 - 2*d*x
^2)*(b^2*arccos(d*x^2 + 1) + a*b))/(d*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(d*x^2+1))^2,x, algorithm="giac")

[Out]

integrate((b*arccos(d*x^2 + 1) + a)^2, x)

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \left (a +b \arccos \left (d \,x^{2}+1\right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccos(d*x^2+1))^2,x)

[Out]

int((a+b*arccos(d*x^2+1))^2,x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(d*x^2+1))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (a+b\,\mathrm {acos}\left (d\,x^2+1\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(d*x^2 + 1))^2,x)

[Out]

int((a + b*acos(d*x^2 + 1))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {acos}{\left (d x^{2} + 1 \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acos(d*x**2+1))**2,x)

[Out]

Integral((a + b*acos(d*x**2 + 1))**2, x)

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