3.3.37 \(\int (c (d \sec (e+f x))^p)^n (a+b \sec (e+f x))^3 \, dx\) [237]

Optimal. Leaf size=296 \[ \frac {b \left (b^2 (1+n p)+3 a^2 (2+n p)\right ) \, _2F_1\left (\frac {1}{2},-\frac {n p}{2};\frac {1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f n p (2+n p) \sqrt {\sin ^2(e+f x)}}-\frac {a \left (3 b^2 n p+a^2 (1+n p)\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f \left (1-n^2 p^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {a b^2 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac {b^2 \left (c (d \sec (e+f x))^p\right )^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n p)} \]

[Out]

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

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Rubi [A]
time = 0.35, antiderivative size = 296, 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, 3927, 4132, 3857, 2722, 4131} \begin {gather*} -\frac {a \left (a^2 (n p+1)+3 b^2 n p\right ) \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 {b \left (3 a^2 (n p+2)+b^2 (n p+1)\right ) \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 b^2 (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 {b^2 \tan (e+f x) (a+b \sec (e+f x)) \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 + b*Sec[e + f*x])^3,x]

[Out]

(b*(b^2*(1 + n*p) + 3*a^2*(2 + 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*b^2*n*p + a^2*(1 + 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*b^2*(5 + 2*n*p)*(c*(d*Sec[e + f*x])^p)^n*Tan[e + f*x])/(f*(1 + n*p)
*(2 + n*p)) + (b^2*(c*(d*Sec[e + f*x])^p)^n*(a + b*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 3927

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[1/(d*(m + n - 1)),
Int[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2)
+ 3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e,
f, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Int
egerQ[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 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]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

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

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Mathematica [A]
time = 1.09, size = 278, normalized size = 0.94 \begin {gather*} -\frac {\csc ^3(e+f x) \left (a^3 \left (6+11 n p+6 n^2 p^2+n^3 p^3\right ) \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2},\frac {n p}{2};1+\frac {n p}{2};\sec ^2(e+f x)\right )+b n p \left (3 a b \left (3+4 n p+n^2 p^2\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},1+\frac {n p}{2};2+\frac {n p}{2};\sec ^2(e+f x)\right )+(2+n p) \left (3 a^2 (3+n p) \cos ^2(e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);\sec ^2(e+f x)\right )+b^2 (1+n p) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (3+n p);\frac {1}{2} (5+n p);\sec ^2(e+f x)\right )\right )\right )\right ) \left (c (d \sec (e+f x))^p\right )^n \left (-\tan ^2(e+f x)\right )^{3/2}}{f n p (1+n p) (2+n p) (3+n p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Csc[e + f*x]^3*(a^3*(6 + 11*n*p + 6*n^2*p^2 + n^3*p^3)*Cos[e + f*x]^3*Hypergeometric2F1[1/2, (n*p)/2, 1 + (
n*p)/2, Sec[e + f*x]^2] + b*n*p*(3*a*b*(3 + 4*n*p + n^2*p^2)*Cos[e + f*x]*Hypergeometric2F1[1/2, 1 + (n*p)/2,
2 + (n*p)/2, Sec[e + f*x]^2] + (2 + n*p)*(3*a^2*(3 + n*p)*Cos[e + f*x]^2*Hypergeometric2F1[1/2, (1 + n*p)/2, (
3 + n*p)/2, Sec[e + f*x]^2] + b^2*(1 + n*p)*Hypergeometric2F1[1/2, (3 + n*p)/2, (5 + n*p)/2, Sec[e + f*x]^2]))
)*(c*(d*Sec[e + f*x])^p)^n*(-Tan[e + f*x]^2)^(3/2))/(f*n*p*(1 + n*p)*(2 + n*p)*(3 + n*p)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sec(f*x+e))**p)**n*(a+b*sec(f*x+e))**3,x)

[Out]

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

[Out]

integrate((b*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 {b}{\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 + b/cos(e + f*x))^3,x)

[Out]

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

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