3.1.94 \(\int \sec (c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [94]

Optimal. Leaf size=132 \[ \frac {a^2 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (12 A+7 C) \tan (c+d x)}{6 d}+\frac {a^2 (12 A+7 C) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d} \]

[Out]

1/8*a^2*(12*A+7*C)*arctanh(sin(d*x+c))/d+1/6*a^2*(12*A+7*C)*tan(d*x+c)/d+1/24*a^2*(12*A+7*C)*sec(d*x+c)*tan(d*
x+c)/d-1/12*C*(a+a*sec(d*x+c))^2*tan(d*x+c)/d+1/4*C*(a+a*sec(d*x+c))^3*tan(d*x+c)/a/d

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Rubi [A]
time = 0.15, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4168, 4086, 3873, 3852, 8, 4131, 3855} \begin {gather*} \frac {a^2 (12 A+7 C) \tan (c+d x)}{6 d}+\frac {a^2 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (12 A+7 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}-\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A + 7*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(12*A + 7*C)*Tan[c + d*x])/(6*d) + (a^2*(12*A + 7*C)*Sec
[c + d*x]*Tan[c + d*x])/(24*d) - (C*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + (C*(a + a*Sec[c + d*x])^3*Ta
n[c + d*x])/(4*a*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 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4086

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

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 4168

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(
m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2))
, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C*Csc[e + f*x], x], x], x] /; Fre
eQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac {\int \sec (c+d x) (a+a \sec (c+d x))^2 (a (4 A+3 C)-a C \sec (c+d x)) \, dx}{4 a}\\ &=-\frac {C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac {1}{12} (12 A+7 C) \int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx\\ &=-\frac {C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac {1}{12} (12 A+7 C) \int \sec (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac {1}{6} \left (a^2 (12 A+7 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^2 (12 A+7 C) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac {1}{8} \left (a^2 (12 A+7 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (12 A+7 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac {a^2 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (12 A+7 C) \tan (c+d x)}{6 d}+\frac {a^2 (12 A+7 C) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(291\) vs. \(2(132)=264\).
time = 1.51, size = 291, normalized size = 2.20 \begin {gather*} -\frac {a^2 (1+\cos (c+d x))^2 \left (C+A \cos ^2(c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (24 (12 A+7 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-48 (3 A+2 C) \sin (c)+3 (4 A+15 C) \sin (d x)+12 A \sin (2 c+d x)+45 C \sin (2 c+d x)+144 A \sin (c+2 d x)+128 C \sin (c+2 d x)-48 A \sin (3 c+2 d x)+12 A \sin (2 c+3 d x)+21 C \sin (2 c+3 d x)+12 A \sin (4 c+3 d x)+21 C \sin (4 c+3 d x)+48 A \sin (3 c+4 d x)+32 C \sin (3 c+4 d x))\right )}{384 d (A+2 C+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/384*(a^2*(1 + Cos[c + d*x])^2*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^4*Sec[c + d*x]^4*(24*(12*A + 7*C)*Cos
[c + d*x]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-4
8*(3*A + 2*C)*Sin[c] + 3*(4*A + 15*C)*Sin[d*x] + 12*A*Sin[2*c + d*x] + 45*C*Sin[2*c + d*x] + 144*A*Sin[c + 2*d
*x] + 128*C*Sin[c + 2*d*x] - 48*A*Sin[3*c + 2*d*x] + 12*A*Sin[2*c + 3*d*x] + 21*C*Sin[2*c + 3*d*x] + 12*A*Sin[
4*c + 3*d*x] + 21*C*Sin[4*c + 3*d*x] + 48*A*Sin[3*c + 4*d*x] + 32*C*Sin[3*c + 4*d*x])))/(d*(A + 2*C + A*Cos[2*
(c + d*x)]))

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Maple [A]
time = 0.63, size = 182, normalized size = 1.38

method	result	size
norman	\(\frac {\frac {5 a^{2} \left (4 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 a^{2} \left (12 A +7 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {a^{2} \left (12 A +7 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a^{2} \left (156 A +83 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a^{2} \left (12 A +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{2} \left (12 A +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(175\)
derivativedivides	\(\frac {a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} A \tan \left (d x +c \right )-2 a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(182\)
default	\(\frac {a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} A \tan \left (d x +c \right )-2 a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(182\)
risch	\(-\frac {i a^{2} \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+21 C \,{\mathrm e}^{7 i \left (d x +c \right )}-48 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+45 C \,{\mathrm e}^{5 i \left (d x +c \right )}-144 A \,{\mathrm e}^{4 i \left (d x +c \right )}-96 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-45 C \,{\mathrm e}^{3 i \left (d x +c \right )}-144 A \,{\mathrm e}^{2 i \left (d x +c \right )}-128 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 \,{\mathrm e}^{i \left (d x +c \right )} A -21 C \,{\mathrm e}^{i \left (d x +c \right )}-48 A -32 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a^2*C*(-(-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)))+2*a^2*A*tan(d*x+c)-2*a^2*C*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a^2*
A*ln(sec(d*x+c)+tan(d*x+c))+a^2*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 = 0.29, size = 227, normalized size = 1.72 \begin {gather*} \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 3 \, C a^{2} {\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 )} - 12 \, A a^{2} {\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 \, C a^{2} {\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 )} + 48 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, A a^{2} \tan \left (d x + c\right )}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 3*C*a^2*(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)) - 12*A*a^2*(2*sin(d*x + c)/(s
in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2
- 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*A*a^2*log(sec(d*x + c) + tan(d*x + c)) + 96*A*a^2*
tan(d*x + c))/d

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Fricas [A]
time = 2.94, size = 141, normalized size = 1.07 \begin {gather*} \frac {3 \, {\left (12 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (12 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 16 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(12*A + 7*C)*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(12*A + 7*C)*a^2*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(16*(3*A + 2*C)*a^2*cos(d*x + c)^3 + 3*(4*A + 7*C)*a^2*cos(d*x + c)^2 + 16*C*a^2*cos(d*x + c) +
6*C*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 2 A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*sec(c + d*x), x) + Integral(2*A*sec(c + d*x)**2, x) + Integral(A*sec(c + d*x)**3, x) + Integr
al(C*sec(c + d*x)**3, x) + Integral(2*C*sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**5, x))

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Giac [A]
time = 0.49, size = 212, normalized size = 1.61 \begin {gather*} \frac {3 \, {\left (12 \, A a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (12 \, A a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (36 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 132 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 77 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 156 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 83 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 75 \, C a^{2} \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 )}^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(12*A*a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*A*a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(36*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 21*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 132*A*a^2*tan(1/2*d*x +
1/2*c)^5 - 77*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 156*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 83*C*a^2*tan(1/2*d*x + 1/2*c)
^3 - 60*A*a^2*tan(1/2*d*x + 1/2*c) - 75*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 5.00, size = 185, normalized size = 1.40 \begin {gather*} \frac {\left (-3\,A\,a^2-\frac {7\,C\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (11\,A\,a^2+\frac {77\,C\,a^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-13\,A\,a^2-\frac {83\,C\,a^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (5\,A\,a^2+\frac {25\,C\,a^2}{4}\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 )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (12\,A+7\,C\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(5*A*a^2 + (25*C*a^2)/4) - tan(c/2 + (d*x)/2)^7*(3*A*a^2 + (7*C*a^2)/4) + tan(c/2 + (d*x)/
2)^5*(11*A*a^2 + (77*C*a^2)/12) - tan(c/2 + (d*x)/2)^3*(13*A*a^2 + (83*C*a^2)/12))/(d*(6*tan(c/2 + (d*x)/2)^4
- 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (a^2*atanh(tan(c/2 + (d*x)/2)
)*(12*A + 7*C))/(4*d)

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