3.9.61 \(\int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [861]

Optimal. Leaf size=165 \[ \frac {(4 a A+3 b B+3 a C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(5 A b+5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {(4 a A+3 b B+3 a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A b+5 a B+4 b C) \tan ^3(c+d x)}{15 d} \]

[Out]

1/8*(4*A*a+3*B*b+3*C*a)*arctanh(sin(d*x+c))/d+1/5*(5*A*b+5*B*a+4*C*b)*tan(d*x+c)/d+1/8*(4*A*a+3*B*b+3*C*a)*sec
(d*x+c)*tan(d*x+c)/d+1/4*(B*b+C*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*b*C*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(5*A*b+5*B
*a+4*C*b)*tan(d*x+c)^3/d

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Rubi [A]
time = 0.17, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4161, 4132, 3852, 4131, 3853, 3855} \begin {gather*} \frac {\tan ^3(c+d x) (5 a B+5 A b+4 b C)}{15 d}+\frac {\tan (c+d x) (5 a B+5 A b+4 b C)}{5 d}+\frac {(4 a A+3 a C+3 b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec (c+d x) (4 a A+3 a C+3 b B)}{8 d}+\frac {(a C+b B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((4*a*A + 3*b*B + 3*a*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((5*A*b + 5*a*B + 4*b*C)*Tan[c + d*x])/(5*d) + ((4*a*A
+ 3*b*B + 3*a*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + ((b*B + a*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (b*C*Se
c[c + d*x]^4*Tan[c + d*x])/(5*d) + ((5*A*b + 5*a*B + 4*b*C)*Tan[c + d*x]^3)/(15*d)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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 ^3(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 ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (5 a A+(5 A b+5 a B+4 b C) \sec (c+d x)+5 (b B+a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (5 a A+5 (b B+a C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} (5 A b+5 a B+4 b C) \int \sec ^4(c+d x) \, dx\\ &=\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (4 a A+3 b B+3 a C) \int \sec ^3(c+d x) \, dx-\frac {(5 A b+5 a B+4 b C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {(5 A b+5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {(4 a A+3 b B+3 a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A b+5 a B+4 b C) \tan ^3(c+d x)}{15 d}+\frac {1}{8} (4 a A+3 b B+3 a C) \int \sec (c+d x) \, dx\\ &=\frac {(4 a A+3 b B+3 a C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(5 A b+5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {(4 a A+3 b B+3 a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A b+5 a B+4 b C) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]
time = 1.26, size = 124, normalized size = 0.75 \begin {gather*} \frac {15 (4 a A+3 b B+3 a C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (15 (4 a A+3 b B+3 a C) \sec (c+d x)+30 (b B+a C) \sec ^3(c+d x)+8 \left (15 (A b+a B+b C)+5 (A b+a B+2 b C) \tan ^2(c+d x)+3 b C \tan ^4(c+d x)\right )\right )}{120 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(15*(4*a*A + 3*b*B + 3*a*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(4*a*A + 3*b*B + 3*a*C)*Sec[c + d*x] + 30
*(b*B + a*C)*Sec[c + d*x]^3 + 8*(15*(A*b + a*B + b*C) + 5*(A*b + a*B + 2*b*C)*Tan[c + d*x]^2 + 3*b*C*Tan[c + d
*x]^4)))/(120*d)

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Maple [A]
time = 0.10, size = 210, normalized size = 1.27

method	result	size
derivativedivides	\(\frac {-A b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C b \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(210\)
default	\(\frac {-A b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C b \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(210\)
norman	\(\frac {-\frac {4 \left (25 A b +25 B a +29 C b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {\left (4 a A -8 A b -8 B a +5 b B +5 a C -8 C b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (4 a A +8 A b +8 B a +5 b B +5 a C +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a A -32 A b -32 B a +3 b B +3 a C -16 C b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (12 a A +32 A b +32 B a +3 b B +3 a C +16 C b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {\left (4 a A +3 b B +3 a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 a A +3 b B +3 a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(270\)
risch	\(-\frac {i \left (60 a A \,{\mathrm e}^{9 i \left (d x +c \right )}+45 B b \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C a \,{\mathrm e}^{9 i \left (d x +c \right )}+120 A a \,{\mathrm e}^{7 i \left (d x +c \right )}+210 B b \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C a \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A b \,{\mathrm e}^{6 i \left (d x +c \right )}-240 B a \,{\mathrm e}^{6 i \left (d x +c \right )}-560 A b \,{\mathrm e}^{4 i \left (d x +c \right )}-560 B a \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C b \,{\mathrm e}^{4 i \left (d x +c \right )}-120 a A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 B b \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C a \,{\mathrm e}^{3 i \left (d x +c \right )}-400 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-400 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-320 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-60 A a \,{\mathrm e}^{i \left (d x +c \right )}-45 B b \,{\mathrm e}^{i \left (d x +c \right )}-45 C a \,{\mathrm e}^{i \left (d x +c \right )}-80 A b -80 B a -64 C b \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b B}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b B}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(414\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(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*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b*B*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*
x+c)+tan(d*x+c)))-C*b*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+a*A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2
*ln(sec(d*x+c)+tan(d*x+c)))-B*a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a*C*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*ta
n(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [A]
time = 0.29, size = 266, normalized size = 1.61 \begin {gather*} \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A 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 )}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b + 16*(3*tan(d*x + c
)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*b - 15*C*a*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4
- 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 15*B*b*(2*(3*sin(d*x + c)^3 -
5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) -
60*A*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]
time = 2.17, size = 182, normalized size = 1.10 \begin {gather*} \frac {15 \, {\left ({\left (4 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (4 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, B a + {\left (5 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left ({\left (4 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, B a + {\left (5 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 24 \, C b + 30 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*((4*A + 3*C)*a + 3*B*b)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*((4*A + 3*C)*a + 3*B*b)*cos(d*x +
c)^5*log(-sin(d*x + c) + 1) + 2*(16*(5*B*a + (5*A + 4*C)*b)*cos(d*x + c)^4 + 15*((4*A + 3*C)*a + 3*B*b)*cos(d*
x + c)^3 + 8*(5*B*a + (5*A + 4*C)*b)*cos(d*x + c)^2 + 24*C*b + 30*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*c
os(d*x + c)^5)

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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 ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(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)**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (153) = 306\).
time = 0.52, size = 473, normalized size = 2.87 \begin {gather*} \frac {15 \, {\left (4 \, A a + 3 \, C a + 3 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a + 3 \, C a + 3 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 400 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 400 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 464 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, 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 )}^{5}}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(4*A*a + 3*C*a + 3*B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a + 3*C*a + 3*B*b)*log(abs(tan(
1/2*d*x + 1/2*c) - 1)) + 2*(60*A*a*tan(1/2*d*x + 1/2*c)^9 - 120*B*a*tan(1/2*d*x + 1/2*c)^9 + 75*C*a*tan(1/2*d*
x + 1/2*c)^9 - 120*A*b*tan(1/2*d*x + 1/2*c)^9 + 75*B*b*tan(1/2*d*x + 1/2*c)^9 - 120*C*b*tan(1/2*d*x + 1/2*c)^9
- 120*A*a*tan(1/2*d*x + 1/2*c)^7 + 320*B*a*tan(1/2*d*x + 1/2*c)^7 - 30*C*a*tan(1/2*d*x + 1/2*c)^7 + 320*A*b*t
an(1/2*d*x + 1/2*c)^7 - 30*B*b*tan(1/2*d*x + 1/2*c)^7 + 160*C*b*tan(1/2*d*x + 1/2*c)^7 - 400*B*a*tan(1/2*d*x +
1/2*c)^5 - 400*A*b*tan(1/2*d*x + 1/2*c)^5 - 464*C*b*tan(1/2*d*x + 1/2*c)^5 + 120*A*a*tan(1/2*d*x + 1/2*c)^3 +
320*B*a*tan(1/2*d*x + 1/2*c)^3 + 30*C*a*tan(1/2*d*x + 1/2*c)^3 + 320*A*b*tan(1/2*d*x + 1/2*c)^3 + 30*B*b*tan(
1/2*d*x + 1/2*c)^3 + 160*C*b*tan(1/2*d*x + 1/2*c)^3 - 60*A*a*tan(1/2*d*x + 1/2*c) - 120*B*a*tan(1/2*d*x + 1/2*
c) - 75*C*a*tan(1/2*d*x + 1/2*c) - 120*A*b*tan(1/2*d*x + 1/2*c) - 75*B*b*tan(1/2*d*x + 1/2*c) - 120*C*b*tan(1/
2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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Mupad [B]
time = 7.67, size = 301, normalized size = 1.82 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A\,a}{2}+\frac {3\,B\,b}{8}+\frac {3\,C\,a}{8}\right )}{2\,A\,a+\frac {3\,B\,b}{2}+\frac {3\,C\,a}{2}}\right )\,\left (A\,a+\frac {3\,B\,b}{4}+\frac {3\,C\,a}{4}\right )}{d}-\frac {\left (2\,A\,b-A\,a+2\,B\,a-\frac {5\,B\,b}{4}-\frac {5\,C\,a}{4}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,a-\frac {16\,A\,b}{3}-\frac {16\,B\,a}{3}+\frac {B\,b}{2}+\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,b}{3}+\frac {20\,B\,a}{3}+\frac {116\,C\,b}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-2\,A\,a-\frac {16\,A\,b}{3}-\frac {16\,B\,a}{3}-\frac {B\,b}{2}-\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a+2\,A\,b+2\,B\,a+\frac {5\,B\,b}{4}+\frac {5\,C\,a}{4}+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 )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\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)^3,x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((A*a)/2 + (3*B*b)/8 + (3*C*a)/8))/(2*A*a + (3*B*b)/2 + (3*C*a)/2))*(A*a + (3*B*b
)/4 + (3*C*a)/4))/d - (tan(c/2 + (d*x)/2)^7*(2*A*a - (16*A*b)/3 - (16*B*a)/3 + (B*b)/2 + (C*a)/2 - (8*C*b)/3)
- tan(c/2 + (d*x)/2)^3*(2*A*a + (16*A*b)/3 + (16*B*a)/3 + (B*b)/2 + (C*a)/2 + (8*C*b)/3) - tan(c/2 + (d*x)/2)^
9*(A*a - 2*A*b - 2*B*a + (5*B*b)/4 + (5*C*a)/4 - 2*C*b) + tan(c/2 + (d*x)/2)*(A*a + 2*A*b + 2*B*a + (5*B*b)/4
+ (5*C*a)/4 + 2*C*b) + tan(c/2 + (d*x)/2)^5*((20*A*b)/3 + (20*B*a)/3 + (116*C*b)/15))/(d*(5*tan(c/2 + (d*x)/2)
^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))

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