3.636 \(\int (a+a \sec (c+d x))^m (B-C+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=171 \[ \frac {\sqrt {2} (B-C) \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^m F_1\left (m+\frac {3}{2};\frac {1}{2},1;m+\frac {5}{2};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d (2 m+3) \sqrt {1-\sec (c+d x)}}+\frac {C 2^{m+\frac {3}{2}} \tan (c+d x) (\sec (c+d x)+1)^{-m-\frac {1}{2}} (a \sec (c+d x)+a)^m \, _2F_1\left (\frac {1}{2},-m-\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x))\right )}{d} \]
[Out]
2^(3/2+m)*C*hypergeom([1/2, -1/2-m],[3/2],1/2-1/2*sec(d*x+c))*(1+sec(d*x+c))^(-1/2-m)*(a+a*sec(d*x+c))^m*tan(d
*x+c)/d+(B-C)*AppellF1(3/2+m,1,1/2,5/2+m,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*(1+sec(d*x+c))*(a+a*sec(d*x+c))^m*2^
(1/2)*tan(d*x+c)/d/(3+2*m)/(1-sec(d*x+c))^(1/2)
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Rubi [A] time = 0.26, antiderivative size = 171, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used =
{4041, 3924, 3779, 3778, 136, 3828, 3827, 69} \[ \frac {\sqrt {2} (B-C) \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^m F_1\left (m+\frac {3}{2};\frac {1}{2},1;m+\frac {5}{2};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d (2 m+3) \sqrt {1-\sec (c+d x)}}+\frac {C 2^{m+\frac {3}{2}} \tan (c+d x) (\sec (c+d x)+1)^{-m-\frac {1}{2}} (a \sec (c+d x)+a)^m \, _2F_1\left (\frac {1}{2},-m-\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x))\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])^m*(B - C + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(Sqrt[2]*(B - C)*AppellF1[3/2 + m, 1/2, 1, 5/2 + m, (1 + Sec[c + d*x])/2, 1 + Sec[c + d*x]]*(1 + Sec[c + d*x])
*(a + a*Sec[c + d*x])^m*Tan[c + d*x])/(d*(3 + 2*m)*Sqrt[1 - Sec[c + d*x]]) + (2^(3/2 + m)*C*Hypergeometric2F1[
1/2, -1/2 - m, 3/2, (1 - Sec[c + d*x])/2]*(1 + Sec[c + d*x])^(-1/2 - m)*(a + a*Sec[c + d*x])^m*Tan[c + d*x])/d
Rule 69
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))
Rule 136
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && !IntegerQ[m] && !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])
Rule 3778
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^n*Cot[c + d*x])/(d*Sqrt[1 + Csc[c + d*x]
]*Sqrt[1 - Csc[c + d*x]]), Subst[Int[(1 + (b*x)/a)^(n - 1/2)/(x*Sqrt[1 - (b*x)/a]), x], x, Csc[c + d*x]], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0]
Rule 3779
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Csc[c + d*x])^FracPart
[n])/(1 + (b*Csc[c + d*x])/a)^FracPart[n], Int[(1 + (b*Csc[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
Rule 3827
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !In
tegerQ[m] && GtQ[a, 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 3924
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, I
nt[(a + b*Csc[e + f*x])^m, x], x] + Dist[d, Int[(a + b*Csc[e + f*x])^m*Csc[e + f*x], x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[2*m]
Rule 4041
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[b*B - a*C + b*C*Csc[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^m \left (B-C+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {\int (a+a \sec (c+d x))^{1+m} (a (B-C)+a C \sec (c+d x)) \, dx}{a^2}\\ &=\frac {(B-C) \int (a+a \sec (c+d x))^{1+m} \, dx}{a}+\frac {C \int \sec (c+d x) (a+a \sec (c+d x))^{1+m} \, dx}{a}\\ &=\left ((B-C) (1+\sec (c+d x))^{-m} (a+a \sec (c+d x))^m\right ) \int (1+\sec (c+d x))^{1+m} \, dx+\left (C (1+\sec (c+d x))^{-m} (a+a \sec (c+d x))^m\right ) \int \sec (c+d x) (1+\sec (c+d x))^{1+m} \, dx\\ &=-\frac {\left ((B-C) (1+\sec (c+d x))^{-\frac {1}{2}-m} (a+a \sec (c+d x))^m \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {(1+x)^{\frac {1}{2}+m}}{\sqrt {1-x} x} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}}-\frac {\left (C (1+\sec (c+d x))^{-\frac {1}{2}-m} (a+a \sec (c+d x))^m \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {(1+x)^{\frac {1}{2}+m}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}}\\ &=\frac {\sqrt {2} (B-C) F_1\left (\frac {3}{2}+m;\frac {1}{2},1;\frac {5}{2}+m;\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (1+\sec (c+d x)) (a+a \sec (c+d x))^m \tan (c+d x)}{d (3+2 m) \sqrt {1-\sec (c+d x)}}+\frac {2^{\frac {3}{2}+m} C \, _2F_1\left (\frac {1}{2},-\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sec (c+d x))\right ) (1+\sec (c+d x))^{-\frac {1}{2}-m} (a+a \sec (c+d x))^m \tan (c+d x)}{d}\\ \end {align*}
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Mathematica [B] time = 14.66, size = 1817, normalized size = 10.63 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[c + d*x])^m*(B - C + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
((C*(1 + Cos[c + d*x] + 4*m*Cos[c + d*x]*Hypergeometric2F1[1/2, 2 + m, 3/2, Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*
Sec[(c + d*x)/2]^2)^m)*(a*(1 + Sec[c + d*x]))^(1 + m)*(B - C + C*Sec[c + d*x])*Tan[(c + d*x)/2])/(d*(1 + 2*m)*
(C + B*Cos[c + d*x] - C*Cos[c + d*x])*(1 + Sec[c + d*x])) + (2^(1 + m)*B*Cos[c + d*x]*Hypergeometric2F1[1/2, 1
+ m, 3/2, Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^m*(1 + Se
c[c + d*x])^(-1 - m)*(a*(1 + Sec[c + d*x]))^(1 + m)*(B - C + C*Sec[c + d*x])*Tan[(c + d*x)/2])/(d*(C + B*Cos[c
+ d*x] - C*Cos[c + d*x])) + (30*B*AppellF1[1/2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c +
d*x)/2]^2*Cos[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(1 + m)*(B - C + C*Sec[c + d*x])*Sin[c + d*x]*(3*AppellF1[1/2,
m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(AppellF1[3/2, m, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c
+ d*x)/2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))/
(d*(C + B*Cos[c + d*x] - C*Cos[c + d*x])*(1 + Sec[c + d*x])*(45*AppellF1[1/2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -
Tan[(c + d*x)/2]^2]^2*Cos[(c + d*x)/2]^2*(1 + 2*m - 2*m*Cos[c + d*x] + Cos[2*(c + d*x)]) + 6*AppellF1[1/2, m,
1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sin[(c + d*x)/2]^2*(-5*AppellF1[3/2, m, 2, 5/2, Tan[(c + d*x)
/2]^2, -Tan[(c + d*x)/2]^2]*(1 + 2*m - 2*(2 + m)*Cos[c + d*x] + Cos[2*(c + d*x)]) + 5*m*AppellF1[3/2, 1 + m, 1
, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + 2*m - 2*(2 + m)*Cos[c + d*x] + Cos[2*(c + d*x)]) - 48*(2*
AppellF1[5/2, m, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*m*AppellF1[5/2, 1 + m, 2, 7/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2] + m*(1 + m)*AppellF1[5/2, 2 + m, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2])*Cot[c + d*x]*Csc[c + d*x]*Sin[(c + d*x)/2]^4) + 40*(AppellF1[3/2, m, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c
+ d*x)/2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])^2*Cos[c + d*x]*Sin[(c
+ d*x)/2]^2*Tan[(c + d*x)/2]^2)) - (30*C*AppellF1[1/2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos
[(c + d*x)/2]^2*Cos[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(1 + m)*(B - C + C*Sec[c + d*x])*Sin[c + d*x]*(3*AppellF
1[1/2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(AppellF1[3/2, m, 2, 5/2, Tan[(c + d*x)/2]^2, -
Tan[(c + d*x)/2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2
]^2))/(d*(C + B*Cos[c + d*x] - C*Cos[c + d*x])*(1 + Sec[c + d*x])*(45*AppellF1[1/2, m, 1, 3/2, Tan[(c + d*x)/2
]^2, -Tan[(c + d*x)/2]^2]^2*Cos[(c + d*x)/2]^2*(1 + 2*m - 2*m*Cos[c + d*x] + Cos[2*(c + d*x)]) + 6*AppellF1[1/
2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sin[(c + d*x)/2]^2*(-5*AppellF1[3/2, m, 2, 5/2, Tan[(c
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + 2*m - 2*(2 + m)*Cos[c + d*x] + Cos[2*(c + d*x)]) + 5*m*AppellF1[3/2, 1
+ m, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + 2*m - 2*(2 + m)*Cos[c + d*x] + Cos[2*(c + d*x)]) -
48*(2*AppellF1[5/2, m, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*m*AppellF1[5/2, 1 + m, 2, 7/2, Tan
[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + m*(1 + m)*AppellF1[5/2, 2 + m, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d
*x)/2]^2])*Cot[c + d*x]*Csc[c + d*x]*Sin[(c + d*x)/2]^4) + 40*(AppellF1[3/2, m, 2, 5/2, Tan[(c + d*x)/2]^2, -T
an[(c + d*x)/2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])^2*Cos[c + d*x]*S
in[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2)))/a
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + B - C\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^m*(B-C+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 - C)*(a*sec(d*x + c) + a)^m, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + B - C\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^m*(B-C+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 - C)*(a*sec(d*x + c) + a)^m, x)
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maple [F] time = 2.26, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{m} \left (B -C +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))^m*(B-C+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
int((a+a*sec(d*x+c))^m*(B-C+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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + B - C\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^m*(B-C+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 - C)*(a*sec(d*x + c) + a)^m, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^m\,\left (B-C+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a/cos(c + d*x))^m*(B - C + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
int((a + a/cos(c + d*x))^m*(B - C + B/cos(c + d*x) + C/cos(c + d*x)^2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{m} \left (\sec {\left (c + d x \right )} + 1\right ) \left (B + C \sec {\left (c + d x \right )} - C\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))**m*(B-C+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Integral((a*(sec(c + d*x) + 1))**m*(sec(c + d*x) + 1)*(B + C*sec(c + d*x) - C), x)
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