3.9.63 \(\int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [863]

Optimal. Leaf size=101 \[ \frac {(b B+a (2 A+C)) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(3 A b+3 a B+2 b C) \tan (c+d x)}{3 d}+\frac {(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {b C \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

1/2*(b*B+a*(2*A+C))*arctanh(sin(d*x+c))/d+1/3*(3*A*b+3*B*a+2*C*b)*tan(d*x+c)/d+1/2*(B*b+C*a)*sec(d*x+c)*tan(d*
x+c)/d+1/3*b*C*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]
time = 0.10, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4161, 4132, 3852, 8, 4131, 3855} \begin {gather*} \frac {\tan (c+d x) (3 a B+3 A b+2 b C)}{3 d}+\frac {(a (2 A+C)+b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(a C+b B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {b C \tan (c+d x) \sec ^2(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B + a*(2*A + C))*ArcTanh[Sin[c + d*x]])/(2*d) + ((3*A*b + 3*a*B + 2*b*C)*Tan[c + d*x])/(3*d) + ((b*B + a*C
)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (b*C*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 8

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

Rule 3852

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 3855

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

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]

Rule 4161

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_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f
*x])^n/(f*(n + 2))), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1
) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C
, n}, x] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \sec (c+d x) (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 ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int \sec (c+d x) \left (3 a A+(3 A b+3 a B+2 b C) \sec (c+d x)+3 (b B+a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int \sec (c+d x) \left (3 a A+3 (b B+a C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{3} (3 A b+3 a B+2 b C) \int \sec ^2(c+d x) \, dx\\ &=\frac {(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {b C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} (b B+a (2 A+C)) \int \sec (c+d x) \, dx-\frac {(3 A b+3 a B+2 b C) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {(b B+a (2 A+C)) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(3 A b+3 a B+2 b C) \tan (c+d x)}{3 d}+\frac {(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {b C \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 75, normalized size = 0.74 \begin {gather*} \frac {3 (b B+a (2 A+C)) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (6 A b+6 a B+6 b C+3 (b B+a C) \sec (c+d x)+2 b C \tan ^2(c+d x)\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*B + a*(2*A + C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*A*b + 6*a*B + 6*b*C + 3*(b*B + a*C)*Sec[c + d*x
] + 2*b*C*Tan[c + d*x]^2))/(6*d)

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Maple [A]
time = 0.07, size = 131, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {A b \tan \left (d x +c \right )+b B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \tan \left (d x +c \right )+a C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(131\)
default	\(\frac {A b \tan \left (d x +c \right )+b B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \tan \left (d x +c \right )+a C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(131\)
norman	\(\frac {\frac {4 \left (3 A b +3 B a +C b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (2 A b +2 B a -b B -a C +2 C b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 A b +2 B a +b B +a C +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {\left (2 a A +b B +a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 a A +b B +a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(173\)
risch	\(-\frac {i \left (3 B b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 B a \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B b \,{\mathrm e}^{i \left (d x +c \right )}-3 C a \,{\mathrm e}^{i \left (d x +c \right )}-6 A b -6 B a -4 C b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b B}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b B}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(270\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(A*b*tan(d*x+c)+b*B*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-C*b*(-2/3-1/3*sec(d*x+c)^2)*
tan(d*x+c)+a*A*ln(sec(d*x+c)+tan(d*x+c))+B*a*tan(d*x+c)+a*C*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d
*x+c))))

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Maxima [A]
time = 0.28, size = 155, normalized size = 1.53 \begin {gather*} \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b - 3 \, C a {\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 )} - 3 \, B 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 )} + 12 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, B a \tan \left (d x + c\right ) + 12 \, A b \tan \left (d x + c\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b - 3*C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) - 3*B*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d
*x + c) - 1)) + 12*A*a*log(sec(d*x + c) + tan(d*x + c)) + 12*B*a*tan(d*x + c) + 12*A*b*tan(d*x + c))/d

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Fricas [A]
time = 2.93, size = 128, normalized size = 1.27 \begin {gather*} \frac {3 \, {\left ({\left (2 \, A + C\right )} a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A + C\right )} a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, B a + {\left (3 \, A + 2 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 2 \, C b + 3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*((2*A + C)*a + B*b)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*((2*A + C)*a + B*b)*cos(d*x + c)^3*log(-s
in(d*x + c) + 1) + 2*(2*(3*B*a + (3*A + 2*C)*b)*cos(d*x + c)^2 + 2*C*b + 3*(C*a + B*b)*cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c)^3)

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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 )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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)*sec(c + d*x), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (93) = 186\).
time = 0.48, size = 261, normalized size = 2.58 \begin {gather*} \frac {3 \, {\left (2 \, A a + C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a + C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, 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 )}^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*A*a + C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a + C*a + B*b)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) - 2*(6*B*a*tan(1/2*d*x + 1/2*c)^5 - 3*C*a*tan(1/2*d*x + 1/2*c)^5 + 6*A*b*tan(1/2*d*x + 1/2*c)^5 - 3
*B*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 12*A*b*tan(1/2*d*
x + 1/2*c)^3 - 4*C*b*tan(1/2*d*x + 1/2*c)^3 + 6*B*a*tan(1/2*d*x + 1/2*c) + 3*C*a*tan(1/2*d*x + 1/2*c) + 6*A*b*
tan(1/2*d*x + 1/2*c) + 3*B*b*tan(1/2*d*x + 1/2*c) + 6*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3
)/d

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Mupad [B]
time = 7.66, size = 190, normalized size = 1.88 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a+\frac {B\,b}{2}+\frac {C\,a}{2}\right )}{4\,A\,a+2\,B\,b+2\,C\,a}\right )\,\left (2\,A\,a+B\,b+C\,a\right )}{d}-\frac {\left (2\,A\,b+2\,B\,a-B\,b-C\,a+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,b-4\,B\,a-\frac {4\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,b+2\,B\,a+B\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \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) + C/cos(c + d*x)^2))/cos(c + d*x),x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*(A*a + (B*b)/2 + (C*a)/2))/(4*A*a + 2*B*b + 2*C*a))*(2*A*a + B*b + C*a))/d - (tan
(c/2 + (d*x)/2)^5*(2*A*b + 2*B*a - B*b - C*a + 2*C*b) + tan(c/2 + (d*x)/2)*(2*A*b + 2*B*a + B*b + C*a + 2*C*b)
- tan(c/2 + (d*x)/2)^3*(4*A*b + 4*B*a + (4*C*b)/3))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan
(c/2 + (d*x)/2)^6 - 1))

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