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

Optimal. Leaf size=137 \[ \frac{b \left (8 a^2 C+9 a b B+2 b^2 C\right ) \tan (c+d x)}{3 d}+\frac{\left (6 a^2 b B+2 a^3 C+3 a b^2 C+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^3 B x+\frac{b^2 (5 a C+3 b B) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{b C \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

[Out]

a^3*B*x + ((6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (b*(9*a*b*B + 8*a^2*C + 2*
b^2*C)*Tan[c + d*x])/(3*d) + (b^2*(3*b*B + 5*a*C)*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (b*C*(a + b*Sec[c + d*x])
^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.240119, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4072, 3918, 4048, 3770, 3767, 8} \[ \frac{b \left (8 a^2 C+9 a b B+2 b^2 C\right ) \tan (c+d x)}{3 d}+\frac{\left (6 a^2 b B+2 a^3 C+3 a b^2 C+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^3 B x+\frac{b^2 (5 a C+3 b B) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{b C \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*B*x + ((6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (b*(9*a*b*B + 8*a^2*C + 2*
b^2*C)*Tan[c + d*x])/(3*d) + (b^2*(3*b*B + 5*a*C)*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (b*C*(a + b*Sec[c + d*x])
^2*Tan[c + d*x])/(3*d)

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3918

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 - 2)*Simp[a^2*c
*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

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 (c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{b C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+b \sec (c+d x)) \left (3 a^2 B+\left (6 a b B+3 a^2 C+2 b^2 C\right ) \sec (c+d x)+b (3 b B+5 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b^2 (3 b B+5 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{b C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{6} \int \left (6 a^3 B+3 \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 C\right ) \sec (c+d x)+2 b \left (9 a b B+8 a^2 C+2 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^3 B x+\frac{b^2 (3 b B+5 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{b C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} \left (b \left (9 a b B+8 a^2 C+2 b^2 C\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 C\right ) \int \sec (c+d x) \, dx\\ &=a^3 B x+\frac{\left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^2 (3 b B+5 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{b C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{\left (b \left (9 a b B+8 a^2 C+2 b^2 C\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=a^3 B x+\frac{\left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b \left (9 a b B+8 a^2 C+2 b^2 C\right ) \tan (c+d x)}{3 d}+\frac{b^2 (3 b B+5 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{b C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.585876, size = 108, normalized size = 0.79 \[ \frac{3 \left (6 a^2 b B+2 a^3 C+3 a b^2 C+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))+3 b \tan (c+d x) \left (6 a^2 C+b (3 a C+b B) \sec (c+d x)+6 a b B+2 b^2 C\right )+6 a^3 B d x+2 b^3 C \tan ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^3*B*d*x + 3*(6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*C)*ArcTanh[Sin[c + d*x]] + 3*b*(6*a*b*B + 6*a^2*C + 2*
b^2*C + b*(b*B + 3*a*C)*Sec[c + d*x])*Tan[c + d*x] + 2*b^3*C*Tan[c + d*x]^3)/(6*d)

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Maple [A]  time = 0.066, size = 223, normalized size = 1.6 \begin{align*}{a}^{3}Bx+{\frac{B{a}^{3}c}{d}}+{\frac{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{B{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{2}bC\tan \left ( dx+c \right ) }{d}}+3\,{\frac{Ba{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,Ca{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{B{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,C{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.984015, size = 292, normalized size = 2.13 \begin{align*} \frac{12 \,{\left (d x + c\right )} B a^{3} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{3} - 9 \, C a b^{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 )} - 3 \, B 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^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, B 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 )} + 36 \, C a^{2} b \tan \left (d x + c\right ) + 36 \, B a b^{2} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(d*x + c)*B*a^3 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b^3 - 9*C*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)
^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*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^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*B*a^
2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*C*a^2*b*tan(d*x + c) + 36*B*a*b^2*tan(d*x + c))/d

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Fricas [A]  time = 0.547865, size = 458, normalized size = 3.34 \begin{align*} \frac{12 \, B a^{3} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, C b^{3} + 2 \,{\left (9 \, C a^{2} b + 9 \, B a b^{2} + 2 \, C b^{3}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*B*a^3*d*x*cos(d*x + c)^3 + 3*(2*C*a^3 + 6*B*a^2*b + 3*C*a*b^2 + B*b^3)*cos(d*x + c)^3*log(sin(d*x + c
) + 1) - 3*(2*C*a^3 + 6*B*a^2*b + 3*C*a*b^2 + B*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*C*b^3 + 2*(9
*C*a^2*b + 9*B*a*b^2 + 2*C*b^3)*cos(d*x + c)^2 + 3*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x
+ c)^3)

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

[Out]

Timed out

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Giac [B]  time = 1.23836, size = 454, normalized size = 3.31 \begin{align*} \frac{6 \,{\left (d x + c\right )} B a^{3} + 3 \,{\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 36 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)*B*a^3 + 3*(2*C*a^3 + 6*B*a^2*b + 3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2
*C*a^3 + 6*B*a^2*b + 3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(18*C*a^2*b*tan(1/2*d*x + 1/2*c
)^5 + 18*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 9*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^3*tan(1/2*d*x + 1/2*c)^5 +
6*C*b^3*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*C*b
^3*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b*tan(1/2*d*x + 1/2*c) + 18*B*a*b^2*tan(1/2*d*x + 1/2*c) + 9*C*a*b^2*tan(
1/2*d*x + 1/2*c) + 3*B*b^3*tan(1/2*d*x + 1/2*c) + 6*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3
)/d