3.69 \(\int (a (b \sec (c+d x))^p)^n \, dx\)

Optimal. Leaf size=83 \[ -\frac {\sin (c+d x) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {1}{2} (3-n p);\cos ^2(c+d x)\right ) \left (a (b \sec (c+d x))^p\right )^n}{d (1-n p) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 83, 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 = {4123, 3772, 2643} \[ -\frac {\sin (c+d x) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {1}{2} (3-n p);\cos ^2(c+d x)\right ) \left (a (b \sec (c+d x))^p\right )^n}{d (1-n p) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*(b*Sec[c + d*x])^p)^n,x]

[Out]

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

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3772

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]

Rule 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 71, normalized size = 0.86 \[ \frac {\sqrt {-\tan ^2(c+d x)} \cot (c+d x) \, _2F_1\left (\frac {1}{2},\frac {n p}{2};\frac {1}{2} (n p+2);\sec ^2(c+d x)\right ) \left (a (b \sec (c+d x))^p\right )^n}{d n p} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*(b*Sec[c + d*x])^p)^n,x]

[Out]

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

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\left (b \sec \left (d x + c\right )\right )^{p} a\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((b*sec(d*x + c))^p*a)^n, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\left (b \sec \left (d x + c\right )\right )^{p} a\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*(b*sec(d*x+c))^p)^n,x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.66, size = 0, normalized size = 0.00 \[ \int \left (a \left (b \sec \left (d x +c \right )\right )^{p}\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\left (b \sec \left (d x + c\right )\right )^{p} a\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*(b*sec(d*x+c))^p)^n,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^p\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*(b/cos(c + d*x))^p)^n,x)

[Out]

int((a*(b/cos(c + d*x))^p)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(b*sec(c + d*x))**p)**n, x)

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