3.3.31 \(\int (c (d \sec (e+f x))^p)^n (a+a \sec (e+f x))^3 \, dx\) [231]

Optimal. Leaf size=275 \[ \frac {a^3 (7+4 n p) \, _2F_1\left (\frac {1}{2},-\frac {n p}{2};\frac {1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f n p (2+n p) \sqrt {\sin ^2(e+f x)}}-\frac {a^3 (1+4 n p) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f \left (1-n^2 p^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {a^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac {\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)} \]

[Out]

a^3*(4*n*p+7)*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/(n*
p+2)/(sin(f*x+e)^2)^(1/2)-a^3*(4*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^3*(2*n*p+5)*(c*(d*sec(f*x+e))^p)^n*tan
(f*x+e)/f/(n*p+1)/(n*p+2)+(c*(d*sec(f*x+e))^p)^n*(a^3+a^3*sec(f*x+e))*tan(f*x+e)/f/(n*p+2)

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Rubi [A]
time = 0.30, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4033, 3899, 4082, 3872, 3857, 2722} \begin {gather*} -\frac {a^3 (4 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 {a^3 (4 n p+7) \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 (n p+2) \sqrt {\sin ^2(e+f x)}}+\frac {a^3 (2 n p+5) \tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+1) (n p+2)}+\frac {\tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right ) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*(d*Sec[e + f*x])^p)^n*(a + a*Sec[e + f*x])^3,x]

[Out]

(a^3*(7 + 4*n*p)*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*(2 + n*p)*Sqrt[Sin[e + f*x]^2]) - (a^3*(1 + 4*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^3*(5 + 2*n*p)*(c*(d*Sec[e + f*x])^p)^n*Tan[e + f*x])/(f*(1 + n*p)*(2 + n*p)) + ((c*(d*Sec[e + f*x])^p
)^n*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(f*(2 + 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 3872

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

Rule 3899

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-b^2)
*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Dist[b/(m + n - 1), Int[
(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

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 4082

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Dist[1/(n + 1), Int[(d
*Csc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e,
f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rubi steps

\begin {align*} \int \left (c (d \sec (e+f x))^p\right )^n (a+a \sec (e+f x))^3 \, 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))^3 \, dx\\ &=\frac {\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}+\frac {\left (a (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)) (a (2+2 n p)+a (5+2 n p) \sec (e+f x)) \, dx}{2+n p}\\ &=\frac {a^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac {\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}+\frac {\left (a (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 (2+n p) (1+4 n p)+a^2 (1+n p) (7+4 n p) \sec (e+f x)\right ) \, dx}{(1+n p) (2+n p)}\\ &=\frac {a^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac {\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}+\frac {\left (a^3 (1+4 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 {\left (a^3 (7+4 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))^{1+n p} \, dx}{d (2+n p)}\\ &=\frac {a^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac {\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}+\frac {\left (a^3 (1+4 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 {\left (a^3 (7+4 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 )^{-1-n p} \, dx}{d (2+n p)}\\ &=\frac {a^3 (7+4 n p) \, _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 (2+n p) \sqrt {\sin ^2(e+f x)}}-\frac {a^3 (1+4 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^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac {\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.08, size = 343, normalized size = 1.25 \begin {gather*} -i 2^{-3+n p} a^3 \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{n p} \left (\frac {12 e^{2 i (e+f x)} \, _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 ) f (2+n p)}+\frac {8 e^{3 i (e+f x)} \, _2F_1\left (1,\frac {1}{2} (-1-n p);\frac {1}{2} (5+n p);-e^{2 i (e+f x)}\right )}{\left (1+e^{2 i (e+f x)}\right )^2 f (3+n p)}+\frac {6 e^{i (e+f x)} \, _2F_1\left (1,\frac {1}{2} (1-n p);\frac {1}{2} (3+n p);-e^{2 i (e+f x)}\right )}{f+f n p}+\frac {\left (1+e^{2 i (e+f x)}\right ) \, _2F_1\left (1,1-\frac {n p}{2};1+\frac {n p}{2};-e^{2 i (e+f x)}\right )}{f n p}\right ) \sec ^6\left (\frac {1}{2} (e+f x)\right ) \sec ^{-3-n p}(e+f x) \left (c (d \sec (e+f x))^p\right )^n (1+\sec (e+f x))^3 \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*(d*Sec[e + f*x])^p)^n*(a + a*Sec[e + f*x])^3,x]

[Out]

(-I)*2^(-3 + n*p)*a^3*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(n*p)*((12*E^((2*I)*(e + f*x))*Hypergeometri
c2F1[1, -1/2*(n*p), 2 + (n*p)/2, -E^((2*I)*(e + f*x))])/((1 + E^((2*I)*(e + f*x)))*f*(2 + n*p)) + (8*E^((3*I)*
(e + f*x))*Hypergeometric2F1[1, (-1 - n*p)/2, (5 + n*p)/2, -E^((2*I)*(e + f*x))])/((1 + E^((2*I)*(e + f*x)))^2
*f*(3 + n*p)) + (6*E^(I*(e + f*x))*Hypergeometric2F1[1, (1 - n*p)/2, (3 + n*p)/2, -E^((2*I)*(e + f*x))])/(f +
f*n*p) + ((1 + E^((2*I)*(e + f*x)))*Hypergeometric2F1[1, 1 - (n*p)/2, 1 + (n*p)/2, -E^((2*I)*(e + f*x))])/(f*n
*p))*Sec[(e + f*x)/2]^6*Sec[e + f*x]^(-3 - n*p)*(c*(d*Sec[e + f*x])^p)^n*(1 + Sec[e + f*x])^3

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \left (c \left (d \sec \left (f x +e \right )\right )^{p}\right )^{n} \left (a +a \sec \left (f x +e \right )\right )^{3}\, dx\]

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))^3,x)

[Out]

int((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^3,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))^3,x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^3*((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))^3,x, algorithm="fricas")

[Out]

integral((a^3*sec(f*x + e)^3 + 3*a^3*sec(f*x + e)^2 + 3*a^3*sec(f*x + e) + a^3)*((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^{3} \left (\int \left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n}\, dx + \int 3 \left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n} \sec {\left (e + f x \right )}\, dx + \int 3 \left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n} \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n} \sec ^{3}{\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))**3,x)

[Out]

a**3*(Integral((c*(d*sec(e + f*x))**p)**n, x) + Integral(3*(c*(d*sec(e + f*x))**p)**n*sec(e + f*x), x) + Integ
ral(3*(c*(d*sec(e + f*x))**p)**n*sec(e + f*x)**2, x) + Integral((c*(d*sec(e + f*x))**p)**n*sec(e + f*x)**3, 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))^3,x, algorithm="giac")

[Out]

integrate((a*sec(f*x + e) + a)^3*((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 )}^3 \,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))^3,x)

[Out]

int((c*(d/cos(e + f*x))^p)^n*(a + a/cos(e + f*x))^3, x)

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