3.3.32 \(\int (c (d \sec (e+f x))^p)^n (a+a \sec (e+f x))^2 \, dx\) [232]

Optimal. Leaf size=205 \[ \frac {2 a^2 \, _2F_1\left (\frac {1}{2},-\frac {n p}{2};\frac {1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f n p \sqrt {\sin ^2(e+f x)}}-\frac {a^2 (1+2 n p) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f \left (1-n^2 p^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {a^2 \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p)} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4033, 3873, 3857, 2722, 4131} \begin {gather*} -\frac {a^2 (2 n p+1) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f \left (1-n^2 p^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {2 a^2 \sin (e+f x) \, _2F_1\left (\frac {1}{2},-\frac {n p}{2};\frac {1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f n p \sqrt {\sin ^2(e+f x)}}+\frac {a^2 \tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*(d*Sec[e + f*x])^p)^n*(a + a*Sec[e + f*x])^2,x]

[Out]

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

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]

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4033

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

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 2.82, size = 299, normalized size = 1.46 \begin {gather*} -\frac {i 2^{-2+n p} a^2 e^{-i (e+f x)} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{1+n p} \left (4 e^{2 i (e+f x)} n p (1+n p) \, _2F_1\left (1,-\frac {n p}{2};2+\frac {n p}{2};-e^{2 i (e+f x)}\right )+\left (1+e^{2 i (e+f x)}\right ) (2+n p) \left (4 e^{i (e+f x)} n p \, _2F_1\left (1,\frac {1}{2} (1-n p);\frac {1}{2} (3+n p);-e^{2 i (e+f x)}\right )+\left (1+e^{2 i (e+f x)}\right ) (1+n p) \, _2F_1\left (1,1-\frac {n p}{2};1+\frac {n p}{2};-e^{2 i (e+f x)}\right )\right )\right ) \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sec ^{-2-n p}(e+f x) \left (c (d \sec (e+f x))^p\right )^n (1+\sec (e+f x))^2}{f n p (1+n p) (2+n p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*(d*Sec[e + f*x])^p)^n*(a + a*Sec[e + f*x])^2,x]

[Out]

((-I)*2^(-2 + n*p)*a^2*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(1 + n*p)*(4*E^((2*I)*(e + f*x))*n*p*(1 + n
*p)*Hypergeometric2F1[1, -1/2*(n*p), 2 + (n*p)/2, -E^((2*I)*(e + f*x))] + (1 + E^((2*I)*(e + f*x)))*(2 + n*p)*
(4*E^(I*(e + f*x))*n*p*Hypergeometric2F1[1, (1 - n*p)/2, (3 + n*p)/2, -E^((2*I)*(e + f*x))] + (1 + E^((2*I)*(e
+ f*x)))*(1 + n*p)*Hypergeometric2F1[1, 1 - (n*p)/2, 1 + (n*p)/2, -E^((2*I)*(e + f*x))]))*Sec[(e + f*x)/2]^4*
Sec[e + f*x]^(-2 - n*p)*(c*(d*Sec[e + f*x])^p)^n*(1 + Sec[e + f*x])^2)/(E^(I*(e + f*x))*f*n*p*(1 + n*p)*(2 + n
*p))

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \left (c \left (d \sec \left (f x +e \right )\right )^{p}\right )^{n} \left (a +a \sec \left (f x +e \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^2,x)

[Out]

int((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^2,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((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^2*((d*sec(f*x + e))^p*c)^n, 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((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

integral((a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2)*((d*sec(f*x + e))^p*c)^n, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n}\, dx + \int 2 \left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n} \sec {\left (e + f x \right )}\, dx + \int \left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n} \sec ^{2}{\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sec(f*x+e))**p)**n*(a+a*sec(f*x+e))**2,x)

[Out]

a**2*(Integral((c*(d*sec(e + f*x))**p)**n, x) + Integral(2*(c*(d*sec(e + f*x))**p)**n*sec(e + f*x), x) + Integ
ral((c*(d*sec(e + f*x))**p)**n*sec(e + f*x)**2, 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((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((a*sec(f*x + e) + a)^2*((d*sec(f*x + e))^p*c)^n, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (c\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^p\right )}^n\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d/cos(e + f*x))^p)^n*(a + a/cos(e + f*x))^2,x)

[Out]

int((c*(d/cos(e + f*x))^p)^n*(a + a/cos(e + f*x))^2, x)

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