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

Optimal. Leaf size=145 \[ \frac{C \sin (c+d x) \sec ^{m+1}(c+d x) (b \sec (c+d x))^n}{d (m+n+1)}-\frac{(A (m+n+1)+C (m+n)) \sin (c+d x) \sec ^{m-1}(c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-m-n+1),\frac{1}{2} (-m-n+3),\cos ^2(c+d x)\right )}{d (-m-n+1) (m+n+1) \sqrt{\sin ^2(c+d x)}} \]

[Out]

(C*Sec[c + d*x]^(1 + m)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + m + n)) - ((C*(m + n) + A*(1 + m + n))*Hyperg
eometric2F1[1/2, (1 - m - n)/2, (3 - 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)*(1 + m + n)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.110575, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {20, 4046, 3772, 2643} \[ \frac{C \sin (c+d x) \sec ^{m+1}(c+d x) (b \sec (c+d x))^n}{d (m+n+1)}-\frac{(A (m+n+1)+C (m+n)) \sin (c+d x) \sec ^{m-1}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-m-n+1);\frac{1}{2} (-m-n+3);\cos ^2(c+d x)\right )}{d (-m-n+1) (m+n+1) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*Sec[c + d*x]^(1 + m)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + m + n)) - ((C*(m + n) + A*(1 + m + n))*Hyperg
eometric2F1[1/2, (1 - m - n)/2, (3 - 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)*(1 + m + n)*Sqrt[Sin[c + d*x]^2])

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 4046

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 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 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]

Rubi steps

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

Mathematica [C]  time = 7.79207, size = 289, normalized size = 1.99 \[ -\frac{i 2^{m+n+1} e^{-i (m+n+1) (c+d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+n+1} \sec ^{-n-2}(c+d x) \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \left (\frac{2 (A+2 C) e^{i (m+n+2) (c+d x)} \text{Hypergeometric2F1}\left (1,\frac{1}{2} (-m-n),\frac{1}{2} (m+n+4),-e^{2 i (c+d x)}\right )}{m+n+2}+\frac{A e^{i (m+n) (c+d x)} \text{Hypergeometric2F1}\left (1,\frac{1}{2} (-m-n-2),\frac{1}{2} (m+n+2),-e^{2 i (c+d x)}\right )}{m+n}+\frac{A e^{i (m+n+4) (c+d x)} \text{Hypergeometric2F1}\left (1,\frac{1}{2} (-m-n+2),\frac{1}{2} (m+n+6),-e^{2 i (c+d x)}\right )}{m+n+4}\right )}{d (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(1 + m + n)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(1 + m + n)*((A*E^(I*(m + n)*(c + d*x))*Hyperg
eometric2F1[1, (-2 - m - n)/2, (2 + m + n)/2, -E^((2*I)*(c + d*x))])/(m + n) + (2*(A + 2*C)*E^(I*(2 + m + n)*(
c + d*x))*Hypergeometric2F1[1, (-m - n)/2, (4 + m + n)/2, -E^((2*I)*(c + d*x))])/(2 + m + n) + (A*E^(I*(4 + m
+ n)*(c + d*x))*Hypergeometric2F1[1, (2 - m - n)/2, (6 + m + n)/2, -E^((2*I)*(c + d*x))])/(4 + m + n))*Sec[c +
d*x]^(-2 - n)*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2))/(d*E^(I*(1 + m + n)*(c + d*x))*(A + 2*C + A*Cos[2*c
+ 2*d*x]))

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Maple [F]  time = 1.011, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{m} \left ( b\sec \left ( dx+c \right ) \right ) ^{n} \left ( A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec{\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sec(c + d*x))**n*(A + C*sec(c + d*x)**2)*sec(c + d*x)**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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