3.1.53 \(\int (d x)^m (a+b \sec ^{-1}(c x)) \, dx\) [53]

Optimal. Leaf size=67 \[ \frac {(d x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{d (1+m)}-\frac {b (d x)^m \, _2F_1\left (\frac {1}{2},-\frac {m}{2};1-\frac {m}{2};\frac {1}{c^2 x^2}\right )}{c m (1+m)} \]

[Out]

(d*x)^(1+m)*(a+b*arcsec(c*x))/d/(1+m)-b*(d*x)^m*hypergeom([1/2, -1/2*m],[1-1/2*m],1/c^2/x^2)/c/m/(1+m)

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Rubi [A]
time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5328, 346, 371} \begin {gather*} \frac {(d x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{d (m+1)}-\frac {b (d x)^m \, _2F_1\left (\frac {1}{2},-\frac {m}{2};1-\frac {m}{2};\frac {1}{c^2 x^2}\right )}{c m (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(a + b*ArcSec[c*x]),x]

[Out]

((d*x)^(1 + m)*(a + b*ArcSec[c*x]))/(d*(1 + m)) - (b*(d*x)^m*Hypergeometric2F1[1/2, -1/2*m, 1 - m/2, 1/(c^2*x^
2)])/(c*m*(1 + m))

Rule 346

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(-c^(-1))*(c*x)^(m + 1)*(1/x)^(m + 1),
Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m
]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 5328

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

Rubi steps

\begin {align*} \int (d x)^m \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac {(d x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{d (1+m)}-\frac {(b d) \int \frac {(d x)^{-1+m}}{\sqrt {1-\frac {1}{c^2 x^2}}} \, dx}{c (1+m)}\\ &=\frac {(d x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{d (1+m)}+\frac {\left (b \left (\frac {1}{x}\right )^m (d x)^m\right ) \text {Subst}\left (\int \frac {x^{-1-m}}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c (1+m)}\\ &=\frac {(d x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{d (1+m)}-\frac {b (d x)^m \, _2F_1\left (\frac {1}{2},-\frac {m}{2};1-\frac {m}{2};\frac {1}{c^2 x^2}\right )}{c m (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 82, normalized size = 1.22 \begin {gather*} \frac {x (d x)^m \left ((1+m) \left (a+b \sec ^{-1}(c x)\right )+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}} x \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{\sqrt {1-c^2 x^2}}\right )}{(1+m)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m*(a + b*ArcSec[c*x]),x]

[Out]

(x*(d*x)^m*((1 + m)*(a + b*ArcSec[c*x]) + (b*c*Sqrt[1 - 1/(c^2*x^2)]*x*Hypergeometric2F1[1/2, (1 + m)/2, (3 +
m)/2, c^2*x^2])/Sqrt[1 - c^2*x^2]))/(1 + m)^2

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Maple [F]
time = 1.24, size = 0, normalized size = 0.00 \[\int \left (d x \right )^{m} \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a+b*arcsec(c*x)),x)

[Out]

int((d*x)^m*(a+b*arcsec(c*x)),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arcsec(c*x)),x, algorithm="maxima")

[Out]

(d^m*x*x^m*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - (c^2*d^m*m + c^2*d^m)*integrate(-sqrt(c*x + 1)*sqrt(c*x - 1)*
x^m/(c^2*m - (c^4*m + c^4)*x^2 + c^2), x))*b/(m + 1) + (d*x)^(m + 1)*a/(d*(m + 1))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arcsec(c*x)),x, algorithm="fricas")

[Out]

integral((b*arcsec(c*x) + a)*(d*x)^m, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{m} \left (a + b \operatorname {asec}{\left (c x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*(a+b*asec(c*x)),x)

[Out]

Integral((d*x)**m*(a + b*asec(c*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arcsec(c*x)),x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)*(d*x)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,x\right )}^m\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a + b*acos(1/(c*x))),x)

[Out]

int((d*x)^m*(a + b*acos(1/(c*x))), x)

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