3.1.95 \(\int (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [95]

Optimal. Leaf size=96 \[ a^2 A x+\frac {a^2 (2 A+C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {C \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4140, 4002, 3999, 3852, 8, 3855} \begin {gather*} \frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {a^2 (2 A+C) \tanh ^{-1}(\sin (c+d x))}{d}+a^2 A x+\frac {C \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*A*x + (a^2*(2*A + C)*ArcTanh[Sin[c + d*x]])/d + (a^2*(A + C)*Tan[c + d*x])/d + (C*(a + a*Sec[c + d*x])^2*T
an[c + d*x])/(3*d) + (C*(a^2 + a^2*Sec[c + d*x])*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 3999

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 4002

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-b)
*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c
*m + (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &&
GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4140

Int[((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/(f*(m + 1))), x] + Dist[1/(b*(m + 1)), Int[(a + b*Csc[e + f*x])^m*Si
mp[A*b*(m + 1) + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !L
tQ[m, -2^(-1)]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1090\) vs. \(2(96)=192\).
time = 6.57, size = 1090, normalized size = 11.35 \begin {gather*} \frac {A x \cos ^4(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{2 (A+2 C+A \cos (2 c+2 d x))}+\frac {(-2 A-C) \cos ^4(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{2 d (A+2 C+A \cos (2 c+2 d x))}+\frac {(2 A+C) \cos ^4(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{2 d (A+2 C+A \cos (2 c+2 d x))}+\frac {C \cos ^4(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {d x}{2}\right )}{12 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\cos ^4(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (7 C \cos \left (\frac {c}{2}\right )-5 C \sin \left (\frac {c}{2}\right )\right )}{24 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^4(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (3 A \sin \left (\frac {d x}{2}\right )+5 C \sin \left (\frac {d x}{2}\right )\right )}{6 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {C \cos ^4(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {d x}{2}\right )}{12 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\cos ^4(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (-7 C \cos \left (\frac {c}{2}\right )-5 C \sin \left (\frac {c}{2}\right )\right )}{24 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^4(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (3 A \sin \left (\frac {d x}{2}\right )+5 C \sin \left (\frac {d x}{2}\right )\right )}{6 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(A*x*Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/(2*(A + 2*C + A*Cos[2*
c + 2*d*x])) + ((-2*A - C)*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a
+ a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((2*A + C)*Cos[c + d*x]^4*
Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2
))/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C
*Sec[c + d*x]^2)*Sin[(d*x)/2])/(12*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2]
- Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*
(7*C*Cos[c/2] - 5*C*Sin[c/2]))/(24*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2]
- Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*
(3*A*Sin[(d*x)/2] + 5*C*Sin[(d*x)/2]))/(6*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d
*x)/2] - Sin[c/2 + (d*x)/2])) + (C*Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d
*x]^2)*Sin[(d*x)/2])/(12*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2
+ (d*x)/2])^3) + (Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*(-7*C*Cos[
c/2] - 5*C*Sin[c/2]))/(24*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2
+ (d*x)/2])^2) + (Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*(3*A*Sin[
(d*x)/2] + 5*C*Sin[(d*x)/2]))/(6*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] +
Sin[c/2 + (d*x)/2]))

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Maple [A]
time = 0.54, size = 119, normalized size = 1.24

method	result	size
derivativedivides	\(\frac {a^{2} A \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} A \left (d x +c \right )+a^{2} C \tan \left (d x +c \right )}{d}\)	\(119\)
default	\(\frac {a^{2} A \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} A \left (d x +c \right )+a^{2} C \tan \left (d x +c \right )}{d}\)	\(119\)
risch	\(a^{2} A x -\frac {2 i a^{2} \left (3 C \,{\mathrm e}^{5 i \left (d x +c \right )}-3 A \,{\mathrm e}^{4 i \left (d x +c \right )}-3 C \,{\mathrm e}^{4 i \left (d x +c \right )}-6 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}-3 A -5 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\)	\(196\)
norman	\(\frac {a^{2} A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{2} A x +3 a^{2} A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{2} A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a^{2} \left (A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a^{2} \left (3 A +4 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (A +C \right ) a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {a^{2} \left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(198\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 131, normalized size = 1.36 \begin {gather*} \frac {6 \, {\left (d x + c\right )} A a^{2} + 2 \, {\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 \, \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^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 6 \, A a^{2} \tan \left (d x + c\right ) + 6 \, C a^{2} \tan \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)*A*a^2 + 2*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 3*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)) + 12*A*a^2*log(sec(d*x + c) + tan(d*x + c)) + 6*A*a^2*tan
(d*x + c) + 6*C*a^2*tan(d*x + c))/d

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Fricas [A]
time = 3.19, size = 131, normalized size = 1.36 \begin {gather*} \frac {6 \, A a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A + C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (3 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, C a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*A*a^2*d*x*cos(d*x + c)^3 + 3*(2*A + C)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*A + C)*a^2*cos(d
*x + c)^3*log(-sin(d*x + c) + 1) + 2*((3*A + 5*C)*a^2*cos(d*x + c)^2 + 3*C*a^2*cos(d*x + c) + C*a^2)*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*} a^{2} \left (\int A\, dx + \int 2 A \sec {\left (c + d x \right )}\, dx + \int A \sec ^{2}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (92) = 184\).
time = 0.50, size = 187, normalized size = 1.95 \begin {gather*} \frac {3 \, {\left (d x + c\right )} A a^{2} + 3 \, {\left (2 \, A a^{2} + C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a^{2} + 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 (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, 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 )}^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*A*a^2 + 3*(2*A*a^2 + C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a^2 + C*a^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*tan(
1/2*d*x + 1/2*c)^3 - 8*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^2*tan(1/2*d*x + 1/2*c) + 9*C*a^2*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 = 2.58, size = 184, normalized size = 1.92 \begin {gather*} \frac {2\,A\,a^2\,\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 )}{d}+\frac {4\,A\,a^2\,\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 )}{d}+\frac {2\,C\,a^2\,\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 )}{d}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {5\,C\,a^2\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \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,x)

[Out]

(2*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*A*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (2*C*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^2*sin(c + d*x))/(d*cos(c + d*x)) + (5*C
*a^2*sin(c + d*x))/(3*d*cos(c + d*x)) + (C*a^2*sin(c + d*x))/(d*cos(c + d*x)^2) + (C*a^2*sin(c + d*x))/(3*d*co
s(c + d*x)^3)

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