3.633 \(\int \sec ^m(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=259 \[ \frac {2^{n+\frac {1}{2}} \tan (c+d x) (A (m+n+1)-B (m+n+1)+C (m-n)) (\sec (c+d x)+1)^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {1}{2};1-m,\frac {1}{2}-n;\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (m+n+1)}+\frac {2^{n+\frac {3}{2}} (B (m+n+1)+C n) \tan (c+d x) (\sec (c+d x)+1)^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {1}{2};1-m,-n-\frac {1}{2};\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (m+n+1)}+\frac {C \sin (c+d x) \sec ^{m+1}(c+d x) (a \sec (c+d x)+a)^n}{d (m+n+1)} \]

[Out]

C*sec(d*x+c)^(1+m)*(a+a*sec(d*x+c))^n*sin(d*x+c)/d/(1+n+m)+2^(3/2+n)*(C*n+B*(1+n+m))*AppellF1(1/2,1-m,-1/2-n,3
/2,1-sec(d*x+c),1/2-1/2*sec(d*x+c))*(1+sec(d*x+c))^(-1/2-n)*(a+a*sec(d*x+c))^n*tan(d*x+c)/d/(1+n+m)+2^(1/2+n)*
(C*(m-n)+A*(1+n+m)-B*(1+n+m))*AppellF1(1/2,1-m,1/2-n,3/2,1-sec(d*x+c),1/2-1/2*sec(d*x+c))*(1+sec(d*x+c))^(-1/2
-n)*(a+a*sec(d*x+c))^n*tan(d*x+c)/d/(1+n+m)

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Rubi [A]  time = 0.61, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4088, 4023, 3828, 3825, 133} \[ \frac {2^{n+\frac {1}{2}} \tan (c+d x) (A (m+n+1)-B (m+n+1)+C (m-n)) (\sec (c+d x)+1)^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {1}{2};1-m,\frac {1}{2}-n;\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (m+n+1)}+\frac {2^{n+\frac {3}{2}} (B (m+n+1)+C n) \tan (c+d x) (\sec (c+d x)+1)^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {1}{2};1-m,-n-\frac {1}{2};\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (m+n+1)}+\frac {C \sin (c+d x) \sec ^{m+1}(c+d x) (a \sec (c+d x)+a)^n}{d (m+n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*Sec[c + d*x]^(1 + m)*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + m + n)) + (2^(3/2 + n)*(C*n + B*(1 + m +
n))*AppellF1[1/2, 1 - m, -1/2 - n, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + d*x])/2]*(1 + Sec[c + d*x])^(-1/2 - n)*
(a + a*Sec[c + d*x])^n*Tan[c + d*x])/(d*(1 + m + n)) + (2^(1/2 + n)*(C*(m - n) + A*(1 + m + n) - B*(1 + m + n)
)*AppellF1[1/2, 1 - m, 1/2 - n, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + d*x])/2]*(1 + Sec[c + d*x])^(-1/2 - n)*(a
+ a*Sec[c + d*x])^n*Tan[c + d*x])/(d*(1 + m + n))

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 3825

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

Rule 3828

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

Rule 4023

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

Rule 4088

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

Rubi steps

\begin {align*} \int \sec ^m(c+d x) (a+a \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec ^{1+m}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}+\frac {\int \sec ^m(c+d x) (a+a \sec (c+d x))^n (a (C m+A (1+m+n))+a (C n+B (1+m+n)) \sec (c+d x)) \, dx}{a (1+m+n)}\\ &=\frac {C \sec ^{1+m}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}+\left (A-B+\frac {C (m-n)}{1+m+n}\right ) \int \sec ^m(c+d x) (a+a \sec (c+d x))^n \, dx+\frac {(C n+B (1+m+n)) \int \sec ^m(c+d x) (a+a \sec (c+d x))^{1+n} \, dx}{a (1+m+n)}\\ &=\frac {C \sec ^{1+m}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}+\left (\left (A-B+\frac {C (m-n)}{1+m+n}\right ) (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec ^m(c+d x) (1+\sec (c+d x))^n \, dx+\frac {\left ((C n+B (1+m+n)) (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec ^m(c+d x) (1+\sec (c+d x))^{1+n} \, dx}{1+m+n}\\ &=\frac {C \sec ^{1+m}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}+\frac {\left (\left (A-B+\frac {C (m-n)}{1+m+n}\right ) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {(1-x)^{-1+m} (2-x)^{-\frac {1}{2}+n}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}}+\frac {\left ((C n+B (1+m+n)) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {(1-x)^{-1+m} (2-x)^{\frac {1}{2}+n}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d (1+m+n) \sqrt {1-\sec (c+d x)}}\\ &=\frac {C \sec ^{1+m}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}+\frac {2^{\frac {3}{2}+n} (C n+B (1+m+n)) F_1\left (\frac {1}{2};1-m,-\frac {1}{2}-n;\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)}{d (1+m+n)}+\frac {2^{\frac {1}{2}+n} \left (A-B+\frac {C (m-n)}{1+m+n}\right ) F_1\left (\frac {1}{2};1-m,\frac {1}{2}-n;\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [F]  time = 4.32, size = 0, normalized size = 0.00 \[ \int \sec ^m(c+d x) (a+a \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(a+a*sec(d*x+c))^n*(A+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) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(a+a*sec(d*x+c))^n*(A+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) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^m, x)

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maple [F]  time = 3.59, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sec(d*x+c)^m*(a+a*sec(d*x+c))^n*(A+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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(a+a*sec(d*x+c))^n*(A+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) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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