3.864 \(\int (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=69 \[ \frac {(2 a B+2 A b+b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+a A x+\frac {(a C+b B) \tan (c+d x)}{d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

a*A*x+1/2*(2*A*b+2*B*a+C*b)*arctanh(sin(d*x+c))/d+(B*b+C*a)*tan(d*x+c)/d+1/2*b*C*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4048, 3770, 3767, 8} \[ \frac {(2 a B+2 A b+b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+a A x+\frac {(a C+b B) \tan (c+d x)}{d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*x + ((2*A*b + 2*a*B + b*C)*ArcTanh[Sin[c + d*x]])/(2*d) + ((b*B + a*C)*Tan[c + d*x])/d + (b*C*Sec[c + d*x]
*Tan[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 92, normalized size = 1.33 \[ a A x+\frac {a B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \tan (c+d x)}{d}+\frac {A b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b B \tan (c+d x)}{d}+\frac {b C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*x + (A*b*ArcTanh[Sin[c + d*x]])/d + (a*B*ArcTanh[Sin[c + d*x]])/d + (b*C*ArcTanh[Sin[c + d*x]])/(2*d) + (b
*B*Tan[c + d*x])/d + (a*C*Tan[c + d*x])/d + (b*C*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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fricas [A]  time = 0.57, size = 118, normalized size = 1.71 \[ \frac {4 \, A a d x \cos \left (d x + c\right )^{2} + {\left (2 \, B a + {\left (2 \, A + C\right )} b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, B a + {\left (2 \, A + C\right )} b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b + 2 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*A*a*d*x*cos(d*x + c)^2 + (2*B*a + (2*A + C)*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*B*a + (2*A + C
)*b)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(C*b + 2*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c
)^2)

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giac [B]  time = 0.26, size = 170, normalized size = 2.46 \[ \frac {2 \, {\left (d x + c\right )} A a + {\left (2 \, B a + 2 \, A b + C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, B a + 2 \, A b + C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*x + c)*A*a + (2*B*a + 2*A*b + C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*B*a + 2*A*b + C*b)*log(ab
s(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2*C*a*tan(1/2*d*x + 1/2*c)^3 + 2*B*b*tan(1/2*d*x + 1/2*c)^3 - C*b*tan(1/2*d*
x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x + 1/2*c) - 2*B*b*tan(1/2*d*x + 1/2*c) - C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*
x + 1/2*c)^2 - 1)^2)/d

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maple [A]  time = 0.95, size = 117, normalized size = 1.70 \[ a A x +\frac {A a c}{d}+\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a C \tan \left (d x +c \right )}{d}+\frac {A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b B \tan \left (d x +c \right )}{d}+\frac {b C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*A*x+1/d*A*a*c+1/d*a*B*ln(sec(d*x+c)+tan(d*x+c))+a*C*tan(d*x+c)/d+1/d*A*b*ln(sec(d*x+c)+tan(d*x+c))+b*B*tan(d
*x+c)/d+1/2*b*C*sec(d*x+c)*tan(d*x+c)/d+1/2/d*C*b*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.34, size = 116, normalized size = 1.68 \[ \frac {4 \, {\left (d x + c\right )} A a - C b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, A b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, C a \tan \left (d x + c\right ) + 4 \, B b \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(d*x + c)*A*a - C*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1
)) + 4*B*a*log(sec(d*x + c) + tan(d*x + c)) + 4*A*b*log(sec(d*x + c) + tan(d*x + c)) + 4*C*a*tan(d*x + c) + 4*
B*b*tan(d*x + c))/d

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mupad [B]  time = 4.59, size = 164, normalized size = 2.38 \[ \frac {2\,\left (A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+A\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}\right )}{d}+\frac {\frac {C\,b\,\sin \left (c+d\,x\right )}{2}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(A*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + A*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + B*a*a
tanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (C*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2))/d + ((C*b
*sin(c + d*x))/2 + (B*b*sin(2*c + 2*d*x))/2 + (C*a*sin(2*c + 2*d*x))/2)/(d*(cos(2*c + 2*d*x)/2 + 1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(c + d*x))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2), x)

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