3.7.56 \(\int (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [656]

Optimal. Leaf size=167 \[ a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \tan (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4142, 4141, 4133, 3855, 3852, 8} \begin {gather*} a^3 A x+\frac {a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \tan (c+d x)}{2 d}+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*A*x + (b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(6*A*b^2 + (a^2 + 4*b^2)*C
)*Tan[c + d*x])/(2*d) + (b*(2*a^2*C + b^2*(4*A + 3*C))*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*C*(a + b*Sec[c +
d*x])^2*Tan[c + d*x])/(4*d) + (C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*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 4133

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]

Rule 4141

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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/(m + 1), I
nt[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) +
a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

Rule 4142

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/(m + 1), Int[(a + b*Csc[e + f*x])^(m - 1)*
Simp[a*A*(m + 1) + (A*b*(m + 1) + b*C*m)*Csc[e + f*x] + a*C*m*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f,
A, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+b \sec (c+d x))^2 \left (4 a A+b (4 A+3 C) \sec (c+d x)+3 a C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+b \sec (c+d x)) \left (12 a^2 A+3 a b (8 A+5 C) \sec (c+d x)+3 \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{24} \int \left (24 a^3 A+3 b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \sec (c+d x)+12 a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^3 A x+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{2} \left (a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \tan (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1241\) vs. \(2(167)=334\).
time = 6.43, size = 1241, normalized size = 7.43 \begin {gather*} \frac {2 a^3 A (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x))}+\frac {\left (-24 a^2 A b-4 A b^3-12 a^2 b C-3 b^3 C\right ) \cos ^5(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{4 d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x))}+\frac {\left (24 a^2 A b+4 A b^3+12 a^2 b C+3 b^3 C\right ) \cos ^5(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{4 d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x))}+\frac {b^3 C \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{8 d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {\left (4 A b^3+12 a^2 b C+4 a b^2 C+3 b^3 C\right ) \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{8 d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {a b^2 C \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {b^3 C \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{8 d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {a b^2 C \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\left (-4 A b^3-12 a^2 b C-4 a b^2 C-3 b^3 C\right ) \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{8 d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \left (3 a A b^2 \sin \left (\frac {1}{2} (c+d x)\right )+a^3 C \sin \left (\frac {1}{2} (c+d x)\right )+2 a b^2 C \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \left (3 a A b^2 \sin \left (\frac {1}{2} (c+d x)\right )+a^3 C \sin \left (\frac {1}{2} (c+d x)\right )+2 a b^2 C \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*A*(c + d*x)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/(d*(b + a*Cos[c + d*x])^3*(A
+ 2*C + A*Cos[2*c + 2*d*x])) + ((-24*a^2*A*b - 4*A*b^3 - 12*a^2*b*C - 3*b^3*C)*Cos[c + d*x]^5*Log[Cos[(c + d*x
)/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/(4*d*(b + a*Cos[c + d*x])^3*(A + 2*C +
A*Cos[2*c + 2*d*x])) + ((24*a^2*A*b + 4*A*b^3 + 12*a^2*b*C + 3*b^3*C)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/(4*d*(b + a*Cos[c + d*x])^3*(A + 2*C + A*Cos[2
*c + 2*d*x])) + (b^3*C*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/(8*d*(b + a*Cos[c + d*x])
^3*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + ((4*A*b^3 + 12*a^2*b*C + 4*a*b^2*
C + 3*b^3*C)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/(8*d*(b + a*Cos[c + d*x])^3*(A + 2*
C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (a*b^2*C*Cos[c + d*x]^5*(a + b*Sec[c + d*x]
)^3*(A + C*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(d*(b + a*Cos[c + d*x])^3*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2])^3) - (b^3*C*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/(8*d*(
b + a*Cos[c + d*x])^3*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + (a*b^2*C*Cos[c
+ d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(d*(b + a*Cos[c + d*x])^3*(A + 2*C +
A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((-4*A*b^3 - 12*a^2*b*C - 4*a*b^2*C - 3*b^3*C)
*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/(8*d*(b + a*Cos[c + d*x])^3*(A + 2*C + A*Cos[2*
c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (2*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c +
d*x]^2)*(3*a*A*b^2*Sin[(c + d*x)/2] + a^3*C*Sin[(c + d*x)/2] + 2*a*b^2*C*Sin[(c + d*x)/2]))/(d*(b + a*Cos[c +
d*x])^3*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Cos[c + d*x]^5*(a + b*Sec[
c + d*x])^3*(A + C*Sec[c + d*x]^2)*(3*a*A*b^2*Sin[(c + d*x)/2] + a^3*C*Sin[(c + d*x)/2] + 2*a*b^2*C*Sin[(c + d
*x)/2]))/(d*(b + a*Cos[c + d*x])^3*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]
time = 0.11, size = 209, normalized size = 1.25 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 254, normalized size = 1.52 \begin {gather*} \frac {16 \, {\left (d x + c\right )} A a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b^{2} - C b^{3} {\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 \, C a^{2} 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 \, A b^{3} {\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} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 16 \, C a^{3} \tan \left (d x + c\right ) + 48 \, A a b^{2} \tan \left (d x + c\right )}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*(d*x + c)*A*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a*b^2 - C*b^3*(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*C*a
^2*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*A*b^3*(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*b*log(sec(d*x + c) + t
an(d*x + c)) + 16*C*a^3*tan(d*x + c) + 48*A*a*b^2*tan(d*x + c))/d

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Fricas [A]
time = 2.68, size = 199, normalized size = 1.19 \begin {gather*} \frac {16 \, A a^{3} d x \cos \left (d x + c\right )^{4} + {\left (12 \, {\left (2 \, A + C\right )} a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (12 \, {\left (2 \, A + C\right )} a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, C a b^{2} \cos \left (d x + c\right ) + 2 \, C b^{3} + 8 \, {\left (C a^{3} + {\left (3 \, A + 2 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (12 \, C a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*A*a^3*d*x*cos(d*x + c)^4 + (12*(2*A + C)*a^2*b + (4*A + 3*C)*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1
) - (12*(2*A + C)*a^2*b + (4*A + 3*C)*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*C*a*b^2*cos(d*x + c) +
2*C*b^3 + 8*(C*a^3 + (3*A + 2*C)*a*b^2)*cos(d*x + c)^3 + (12*C*a^2*b + (4*A + 3*C)*b^3)*cos(d*x + c)^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*} \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*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. 526 vs. \(2 (157) = 314\).
time = 0.50, size = 526, normalized size = 3.15 \begin {gather*} \frac {8 \, {\left (d x + c\right )} A a^{3} + {\left (24 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 3 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (24 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 3 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, C b^{3} \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}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*(d*x + c)*A*a^3 + (24*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 3*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2
4*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 3*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(8*C*a^3*tan(1/2*d*x + 1/2*
c)^7 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b^2*tan(1/2*d*x + 1/2*c)
^7 - 4*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 5*C*b^3*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*C*
a^2*b*tan(1/2*d*x + 1/2*c)^5 - 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*b^3
*tan(1/2*d*x + 1/2*c)^5 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 24*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*b*tan(1/
2*d*x + 1/2*c)^3 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^3*tan(1/2*d*x
+ 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 - 8*C*a^3*tan(1/2*d*x + 1/2*c) - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)
- 24*A*a*b^2*tan(1/2*d*x + 1/2*c) - 24*C*a*b^2*tan(1/2*d*x + 1/2*c) - 4*A*b^3*tan(1/2*d*x + 1/2*c) - 5*C*b^3*t
an(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 = 6.03, size = 1547, normalized size = 9.26 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^3,x)

[Out]

((3*A*a^3*atan((64*A^2*a^6*sin(c/2 + (d*x)/2) + 16*A^2*b^6*sin(c/2 + (d*x)/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) +
192*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 72*C^2*a^2*b^4*sin(c/2 + (d*x)/2) +
144*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*b^6*sin(c/2 + (d*x)/2) + 240*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 576
*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(64*A^2*a^6 + 16*A^2*b^6 + 9*C^2*b^6 + 192*A^2*a^2*b^4 +
576*A^2*a^4*b^2 + 72*C^2*a^2*b^4 + 144*C^2*a^4*b^2 + 24*A*C*b^6 + 240*A*C*a^2*b^4 + 576*A*C*a^4*b^2))))/4 + (3
*A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + (9*C*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
)/32 + (A*b^3*sin(3*c + 3*d*x))/8 + (C*a^3*sin(2*c + 2*d*x))/4 + (C*a^3*sin(4*c + 4*d*x))/8 + (3*C*b^3*sin(3*c
+ 3*d*x))/32 + (A*b^3*sin(c + d*x))/8 + (11*C*b^3*sin(c + d*x))/32 + (3*C*a^2*b*sin(c + d*x))/8 + (9*A*a^2*b*
atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (9*C*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 +
(3*A*a*b^2*sin(2*c + 2*d*x))/4 + (3*A*a*b^2*sin(4*c + 4*d*x))/8 + C*a*b^2*sin(2*c + 2*d*x) + (3*C*a^2*b*sin(3
*c + 3*d*x))/8 + (C*a*b^2*sin(4*c + 4*d*x))/4 + A*a^3*atan((64*A^2*a^6*sin(c/2 + (d*x)/2) + 16*A^2*b^6*sin(c/2
+ (d*x)/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) + 192*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A^2*a^4*b^2*sin(c/2 + (d
*x)/2) + 72*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 144*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*b^6*sin(c/2 + (d*x)/2
) + 240*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(64*A^2*a^6 +
16*A^2*b^6 + 9*C^2*b^6 + 192*A^2*a^2*b^4 + 576*A^2*a^4*b^2 + 72*C^2*a^2*b^4 + 144*C^2*a^4*b^2 + 24*A*C*b^6 +
240*A*C*a^2*b^4 + 576*A*C*a^4*b^2)))*cos(2*c + 2*d*x) + (A*a^3*atan((64*A^2*a^6*sin(c/2 + (d*x)/2) + 16*A^2*b^
6*sin(c/2 + (d*x)/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) + 192*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A^2*a^4*b^2*sin
(c/2 + (d*x)/2) + 72*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 144*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*b^6*sin(c/2
+ (d*x)/2) + 240*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(64*
A^2*a^6 + 16*A^2*b^6 + 9*C^2*b^6 + 192*A^2*a^2*b^4 + 576*A^2*a^4*b^2 + 72*C^2*a^2*b^4 + 144*C^2*a^4*b^2 + 24*A
*C*b^6 + 240*A*C*a^2*b^4 + 576*A*C*a^4*b^2)))*cos(4*c + 4*d*x))/4 + (A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2))*cos(2*c + 2*d*x))/2 + (A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/8 + (3*C
*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/8 + (3*C*b^3*atanh(sin(c/2 + (d*x)/2)/cos(
c/2 + (d*x)/2))*cos(4*c + 4*d*x))/32 + 3*A*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)
+ (3*A*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/4 + (3*C*a^2*b*atanh(sin(c/2 + (d
*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2 + (3*C*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4
*c + 4*d*x))/8)/(d*(cos(2*c + 2*d*x)/2 + cos(4*c + 4*d*x)/8 + 3/8))

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