3.882 \(\int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=204 \[ -\frac{b \tan (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{2 d}+\frac{b \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{1}{2} a x \left (a^2 (A+2 C)+6 a b B+6 A b^2\right )-\frac{b^2 \tan (c+d x) \sec (c+d x) (2 a B+4 A b-b C)}{2 d}+\frac{(2 a B+3 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d} \]

[Out]

(a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*x)/2 + (b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/
(2*d) + ((3*A*b + 2*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (A*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*S
in[c + d*x])/(2*d) - (b*(9*a*A*b + 4*a^2*B - 2*b^2*B - 6*a*b*C)*Tan[c + d*x])/(2*d) - (b^2*(4*A*b + 2*a*B - b*
C)*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.455644, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4094, 4048, 3770, 3767, 8} \[ -\frac{b \tan (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{2 d}+\frac{b \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{1}{2} a x \left (a^2 (A+2 C)+6 a b B+6 A b^2\right )-\frac{b^2 \tan (c+d x) \sec (c+d x) (2 a B+4 A b-b C)}{2 d}+\frac{(2 a B+3 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*x)/2 + (b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/
(2*d) + ((3*A*b + 2*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (A*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*S
in[c + d*x])/(2*d) - (b*(9*a*A*b + 4*a^2*B - 2*b^2*B - 6*a*b*C)*Tan[c + d*x])/(2*d) - (b^2*(4*A*b + 2*a*B - b*
C)*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 4094

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_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4048

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 3770

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

Rule 3767

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 8

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

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (3 A b+2 a B+(2 b B+a (A+2 C)) \sec (c+d x)-2 b (A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(3 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}+\frac{1}{2} \int (a+b \sec (c+d x)) \left (6 A b^2+6 a b B+a^2 (A+2 C)-b (a A-2 b B-4 a C) \sec (c+d x)-2 b (4 A b+2 a B-b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(3 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 (4 A b+2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{4} \int \left (2 a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right )+2 b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \sec (c+d x)-2 b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{1}{2} a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) x+\frac{(3 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 (4 A b+2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \left (b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) x+\frac{b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(3 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 (4 A b+2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\left (b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=\frac{1}{2} a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) x+\frac{b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(3 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \tan (c+d x)}{2 d}-\frac{b^2 (4 A b+2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 3.03943, size = 320, normalized size = 1.57 \[ \frac{2 a (c+d x) \left (a^2 (A+2 C)+6 a b B+6 A b^2\right )-2 b \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 b \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a^2 (a B+3 A b) \sin (c+d x)+a^3 A \sin (2 (c+d x))+\frac{4 b^2 (3 a C+b B) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b^2 (3 a C+b B) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{b^3 C}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^3 C}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*(c + d*x) - 2*b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]] + 2*b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]
+ (b^3*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*b^2*(b*B + 3*a*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2]
- Sin[(c + d*x)/2]) - (b^3*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*b^2*(b*B + 3*a*C)*Sin[(c + d*x)/2])
/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*a^2*(3*A*b + a*B)*Sin[c + d*x] + a^3*A*Sin[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.081, size = 267, normalized size = 1.3 \begin{align*}{\frac{A{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}Ax}{2}}+{\frac{A{a}^{3}c}{2\,d}}+{\frac{B{a}^{3}\sin \left ( dx+c \right ) }{d}}+{a}^{3}Cx+{\frac{C{a}^{3}c}{d}}+3\,{\frac{A{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,B{a}^{2}bx+3\,{\frac{B{a}^{2}bc}{d}}+3\,{\frac{{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,Aa{b}^{2}x+3\,{\frac{Aa{b}^{2}c}{d}}+3\,{\frac{Ba{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{Ca{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{A{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{C{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{C{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0552, size = 328, normalized size = 1.61 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 4 \,{\left (d x + c\right )} C a^{3} + 12 \,{\left (d x + c\right )} B a^{2} b + 12 \,{\left (d x + c\right )} A a b^{2} - C 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 )} + 6 \, C a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, A a^{2} b \sin \left (d x + c\right ) + 12 \, C a b^{2} \tan \left (d x + c\right ) + 4 \, B b^{3} \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.57597, size = 500, normalized size = 2.45 \begin{align*} \frac{2 \,{\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} d x \cos \left (d x + c\right )^{2} +{\left (6 \, C a^{2} b + 6 \, B a b^{2} +{\left (2 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (6 \, C a^{2} b + 6 \, B a b^{2} +{\left (2 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a^{3} \cos \left (d x + c\right )^{3} + C b^{3} + 2 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*((A + 2*C)*a^3 + 6*B*a^2*b + 6*A*a*b^2)*d*x*cos(d*x + c)^2 + (6*C*a^2*b + 6*B*a*b^2 + (2*A + C)*b^3)*co
s(d*x + c)^2*log(sin(d*x + c) + 1) - (6*C*a^2*b + 6*B*a*b^2 + (2*A + C)*b^3)*cos(d*x + c)^2*log(-sin(d*x + c)
+ 1) + 2*(A*a^3*cos(d*x + c)^3 + C*b^3 + 2*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)^2 + 2*(3*C*a*b^2 + B*b^3)*cos(d*x
+ c))*sin(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.33571, size = 729, normalized size = 3.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((A*a^3 + 2*C*a^3 + 6*B*a^2*b + 6*A*a*b^2)*(d*x + c) + (6*C*a^2*b + 6*B*a*b^2 + 2*A*b^3 + C*b^3)*log(abs(t
an(1/2*d*x + 1/2*c) + 1)) - (6*C*a^2*b + 6*B*a*b^2 + 2*A*b^3 + C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(
A*a^3*tan(1/2*d*x + 1/2*c)^7 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 6*C*a*b^2*t
an(1/2*d*x + 1/2*c)^7 + 2*B*b^3*tan(1/2*d*x + 1/2*c)^7 - C*b^3*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^3*tan(1/2*d*x +
1/2*c)^5 + 2*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^
5 + 2*B*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3
*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 2*B*b^3*tan(1/
2*d*x + 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 - A*a^3*tan(1/2*d*x + 1/2*c) - 2*B*a^3*tan(1/2*d*x + 1/2*c)
- 6*A*a^2*b*tan(1/2*d*x + 1/2*c) - 6*C*a*b^2*tan(1/2*d*x + 1/2*c) - 2*B*b^3*tan(1/2*d*x + 1/2*c) - C*b^3*tan(1
/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d