3.40 \(\int \sec ^m(c+d x) (b \sec (c+d x))^n (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=167 \[ \frac {B \sin (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-m-n);\frac {1}{2} (-m-n+2);\cos ^2(c+d x)\right )}{d (m+n) \sqrt {\sin ^2(c+d x)}}+\frac {C \sin (c+d x) \sec ^{m+1}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-m-n-1);\frac {1}{2} (-m-n+1);\cos ^2(c+d x)\right )}{d (m+n+1) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {20, 4047, 3772, 2643, 12} \[ \frac {B \sin (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-m-n);\frac {1}{2} (-m-n+2);\cos ^2(c+d x)\right )}{d (m+n) \sqrt {\sin ^2(c+d x)}}+\frac {C \sin (c+d x) \sec ^{m+1}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-m-n-1);\frac {1}{2} (-m-n+1);\cos ^2(c+d x)\right )}{d (m+n+1) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^m*(b*Sec[c + d*x])^n*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(B*Hypergeometric2F1[1/2, (-m - n)/2, (2 - m - n)/2, Cos[c + d*x]^2]*Sec[c + d*x]^m*(b*Sec[c + d*x])^n*Sin[c +
d*x])/(d*(m + n)*Sqrt[Sin[c + d*x]^2]) + (C*Hypergeometric2F1[1/2, (-1 - m - n)/2, (1 - 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 4047

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 \sec ^m(c+d x) (b \sec (c+d x))^n \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{m+n}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\\ &=\left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int C \sec ^{2+m+n}(c+d x) \, dx+\left (B \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{1+m+n}(c+d x) \, dx\\ &=\left (B \cos ^{m+n}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n\right ) \int \cos ^{-1-m-n}(c+d x) \, dx+\left (C \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{2+m+n}(c+d x) \, dx\\ &=\frac {B \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-m-n);\frac {1}{2} (2-m-n);\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt {\sin ^2(c+d x)}}+\left (C \cos ^{m+n}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n\right ) \int \cos ^{-2-m-n}(c+d x) \, dx\\ &=\frac {B \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-m-n);\frac {1}{2} (2-m-n);\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt {\sin ^2(c+d x)}}+\frac {C \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1-m-n);\frac {1}{2} (1-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 = 0.29, size = 129, normalized size = 0.77 \[ \frac {\sqrt {-\tan ^2(c+d x)} \csc (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \left (B (m+n+2) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+1);\frac {1}{2} (m+n+3);\sec ^2(c+d x)\right )+C (m+n+1) \sec (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+2);\frac {1}{2} (m+n+4);\sec ^2(c+d x)\right )\right )}{d (m+n+1) (m+n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^n*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(Csc[c + d*x]*Sec[c + d*x]^m*(b*Sec[c + d*x])^n*(B*(2 + m + n)*Hypergeometric2F1[1/2, (1 + m + n)/2, (3 + m +
n)/2, Sec[c + d*x]^2] + C*(1 + m + n)*Hypergeometric2F1[1/2, (2 + m + n)/2, (4 + m + n)/2, Sec[c + d*x]^2]*Sec
[c + d*x])*Sqrt[-Tan[c + d*x]^2])/(d*(1 + m + n)*(2 + m + n))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c))^n*sec(d*x + c)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c))^n*sec(d*x + c)^m, x)

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maple [F]  time = 4.77, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^m*(b*sec(d*x+c))^n*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)^m*(b*sec(d*x+c))^n*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*(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 \[ \int \left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n*(1/cos(c + d*x))^m,x)

[Out]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n*(1/cos(c + d*x))^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**m*(b*sec(d*x+c))**n*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((b*sec(c + d*x))**n*(B + C*sec(c + d*x))*sec(c + d*x)*sec(c + d*x)**m, x)

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