3.5.81 \(\int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [481]

Optimal. Leaf size=261 \[ \frac {b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}-\frac {\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d \left (1-m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\left (b^2 B (1+m)+a (2 A b+a B) (2+m)\right ) \, _2F_1\left (\frac {1}{2},-\frac {m}{2};\frac {2-m}{2};\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m (2+m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

b*(A*b*(2+m)+a*B*(3+m))*sec(d*x+c)^(1+m)*sin(d*x+c)/d/(1+m)/(2+m)+b*B*sec(d*x+c)^(1+m)*(a+b*sec(d*x+c))*sin(d*
x+c)/d/(2+m)-(A*b^2*m+2*a*b*B*m+a^2*A*(1+m))*hypergeom([1/2, 1/2-1/2*m],[3/2-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(
-1+m)*sin(d*x+c)/d/(-m^2+1)/(sin(d*x+c)^2)^(1/2)+(b^2*B*(1+m)+a*(2*A*b+B*a)*(2+m))*hypergeom([1/2, -1/2*m],[1-
1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*sin(d*x+c)/d/m/(2+m)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4111, 4132, 3857, 2722, 4131} \begin {gather*} -\frac {\sin (c+d x) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(c+d x)\right )}{d \left (1-m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},-\frac {m}{2};\frac {2-m}{2};\cos ^2(c+d x)\right )}{d m (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(A*b*(2 + m) + a*B*(3 + m))*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(1 + m)*(2 + m)) + (b*B*Sec[c + d*x]^(1 +
m)*(a + b*Sec[c + d*x])*Sin[c + d*x])/(d*(2 + m)) - ((A*b^2*m + 2*a*b*B*m + a^2*A*(1 + m))*Hypergeometric2F1[
1/2, (1 - m)/2, (3 - m)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(1 - m^2)*Sqrt[Sin[c + d*x]^
2]) + ((b^2*B*(1 + m) + a*(2*A*b + a*B)*(2 + m))*Hypergeometric2F1[1/2, -1/2*m, (2 - m)/2, Cos[c + d*x]^2]*Sec
[c + d*x]^m*Sin[c + d*x])/(d*m*(2 + m)*Sqrt[Sin[c + d*x]^2])

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 4111

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

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

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Mathematica [A]
time = 1.04, size = 239, normalized size = 0.92 \begin {gather*} \frac {\csc (c+d x) \left (a^2 A \left (6+11 m+6 m^2+m^3\right ) \cos ^3(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {2+m}{2};\sec ^2(c+d x)\right )+a (2 A b+a B) m \left (6+5 m+m^2\right ) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sec ^2(c+d x)\right )+b m (1+m) \left ((A b+2 a B) (3+m) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\sec ^2(c+d x)\right )+b B (2+m) \, _2F_1\left (\frac {1}{2},\frac {3+m}{2};\frac {5+m}{2};\sec ^2(c+d x)\right )\right )\right ) \sec ^{2+m}(c+d x) \sqrt {-\tan ^2(c+d x)}}{d m (1+m) (2+m) (3+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*(a^2*A*(6 + 11*m + 6*m^2 + m^3)*Cos[c + d*x]^3*Hypergeometric2F1[1/2, m/2, (2 + m)/2, Sec[c + d*
x]^2] + a*(2*A*b + a*B)*m*(6 + 5*m + m^2)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sec[c +
d*x]^2] + b*m*(1 + m)*((A*b + 2*a*B)*(3 + m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sec[c +
d*x]^2] + b*B*(2 + m)*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Sec[c + d*x]^2]))*Sec[c + d*x]^(2 + m)*Sqr
t[-Tan[c + d*x]^2])/(d*m*(1 + m)*(2 + m)*(3 + m))

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Maple [F]
time = 0.32, size = 0, normalized size = 0.00 \[\int \left (\sec ^{m}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right )^{2} \left (A +B \sec \left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^m*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)

[Out]

int(sec(d*x+c)^m*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),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(sec(d*x+c)^m*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

[Out]

integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*se
c(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**m*(a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**2*sec(c + d*x)**m, 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(sec(d*x+c)^m*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*sec(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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