3.3.10 \(\int \sec ^m(c+d x) (b \sec (c+d x))^n \, dx\) [210]

Optimal. Leaf size=89 \[ -\frac {\, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-m-n);\frac {1}{2} (3-m-n);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1-m-n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-hypergeom([1/2, 1/2-1/2*m-1/2*n],[3/2-1/2*m-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*(b*sec(d*x+c))^n*sin(d*x+c
)/d/(1-m-n)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {20, 3857, 2722} \begin {gather*} -\frac {\sin (c+d x) \sec ^{m-1}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-m-n+1);\frac {1}{2} (-m-n+3);\cos ^2(c+d x)\right )}{d (-m-n+1) \sqrt {\sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^m*(b*Sec[c + d*x])^n,x]

[Out]

-((Hypergeometric2F1[1/2, (1 - m - n)/2, (3 - m - n)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*(b*Sec[c + d*x])
^n*Sin[c + d*x])/(d*(1 - m - n)*Sqrt[Sin[c + d*x]^2]))

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[b^IntPart[n]*((b*v)^FracPart[n]/(a^IntPart[n]
*(a*v)^FracPart[n])), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \sec ^m(c+d x) (b \sec (c+d x))^n \, dx &=\left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{m+n}(c+d x) \, dx\\ &=\left (\cos ^{m+n}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n\right ) \int \cos ^{-m-n}(c+d x) \, dx\\ &=-\frac {\, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-m-n);\frac {1}{2} (3-m-n);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1-m-n) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 76, normalized size = 0.85 \begin {gather*} \frac {\csc (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+n}{2};\frac {1}{2} (2+m+n);\sec ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (m+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^n,x]

[Out]

(Csc[c + d*x]*Hypergeometric2F1[1/2, (m + n)/2, (2 + m + n)/2, Sec[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*(b*Sec[c
+ d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*(m + n))

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Maple [F]
time = 0.48, size = 0, normalized size = 0.00 \[\int \left (\sec ^{m}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^m*(b*sec(d*x+c))^n,x)

[Out]

int(sec(d*x+c)^m*(b*sec(d*x+c))^n,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(sec(d*x+c)^m*(b*sec(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^n*sec(d*x + c)^m, x)

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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(sec(d*x+c)^m*(b*sec(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c))^n*sec(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**m*(b*sec(d*x+c))**n,x)

[Out]

Integral((b*sec(c + d*x))**n*sec(c + d*x)**m, 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(sec(d*x+c)^m*(b*sec(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^n*sec(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(c + d*x))^n*(1/cos(c + d*x))^m,x)

[Out]

int((b/cos(c + d*x))^n*(1/cos(c + d*x))^m, x)

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