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

Optimal. Leaf size=108 \[ \frac {a^3 (6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(B+2 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{d}+a^3 x (3 B+C)-\frac {5 a^3 C \sin (c+d x)}{2 d}+\frac {a C \sin (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

[Out]

a^3*(3*B+C)*x+1/2*a^3*(6*B+7*C)*arctanh(sin(d*x+c))/d-5/2*a^3*C*sin(d*x+c)/d+1/2*a*C*(a+a*sec(d*x+c))^2*sin(d*
x+c)/d+(B+2*C)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.31, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4072, 4018, 3996, 3770} \[ \frac {a^3 (6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(B+2 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{d}+a^3 x (3 B+C)-\frac {5 a^3 C \sin (c+d x)}{2 d}+\frac {a C \sin (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*(3*B + C)*x + (a^3*(6*B + 7*C)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^3*C*Sin[c + d*x])/(2*d) + (a*C*(a + a*S
ec[c + d*x])^2*Sin[c + d*x])/(2*d) + ((B + 2*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/d

Rule 3770

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

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4018

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

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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos (c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac {a C (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) (a+a \sec (c+d x))^2 (a (2 B-C)+2 a (B+2 C) \sec (c+d x)) \, dx\\ &=\frac {a C (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {(B+2 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{d}+\frac {1}{2} \int \cos (c+d x) (a+a \sec (c+d x)) \left (-5 a^2 C+a^2 (6 B+7 C) \sec (c+d x)\right ) \, dx\\ &=-\frac {5 a^3 C \sin (c+d x)}{2 d}+\frac {a C (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {(B+2 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{d}-\frac {1}{2} \int \left (-2 a^3 (3 B+C)-a^3 (6 B+7 C) \sec (c+d x)\right ) \, dx\\ &=a^3 (3 B+C) x-\frac {5 a^3 C \sin (c+d x)}{2 d}+\frac {a C (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {(B+2 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{d}+\frac {1}{2} \left (a^3 (6 B+7 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 (3 B+C) x+\frac {a^3 (6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^3 C \sin (c+d x)}{2 d}+\frac {a C (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {(B+2 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 1.81, size = 208, normalized size = 1.93 \[ \frac {a^3 \left (4 (B+3 C) \tan (c+d x)+4 B \sin (c+d x)-12 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 B c+12 B d x+\frac {C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {C}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-14 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+14 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 c C+4 C d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(12*B*c + 4*c*C + 12*B*d*x + 4*C*d*x - 12*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 14*C*Log[Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]] + 12*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 14*C*Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]] + C/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - C/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 4*B*Si
n[c + d*x] + 4*(B + 3*C)*Tan[c + d*x]))/(4*d)

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fricas [A]  time = 0.49, size = 137, normalized size = 1.27 \[ \frac {4 \, {\left (3 \, B + C\right )} a^{3} d x \cos \left (d x + c\right )^{2} + {\left (6 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (6 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + C a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(3*B + C)*a^3*d*x*cos(d*x + c)^2 + (6*B + 7*C)*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (6*B + 7*C)*a
^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*B*a^3*cos(d*x + c)^2 + 2*(B + 3*C)*a^3*cos(d*x + c) + C*a^3)*s
in(d*x + c))/(d*cos(d*x + c)^2)

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giac [A]  time = 1.10, size = 192, normalized size = 1.78 \[ \frac {\frac {4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (3 \, B a^{3} + C a^{3}\right )} {\left (d x + c\right )} + {\left (6 \, B a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (6 \, B a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.18, size = 144, normalized size = 1.33 \[ \frac {a^{3} B \sin \left (d x +c \right )}{d}+a^{3} C x +\frac {C \,a^{3} c}{d}+3 a^{3} B x +\frac {3 a^{3} B c}{d}+\frac {7 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{3} C \tan \left (d x +c \right )}{d}+\frac {a^{3} B \tan \left (d x +c \right )}{d}+\frac {C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*B*sin(d*x+c)/d+a^3*C*x+1/d*C*a^3*c+3*a^3*B*x+3/d*a^3*B*c+7/2/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^3*B*l
n(sec(d*x+c)+tan(d*x+c))+3*a^3*C*tan(d*x+c)/d+1/d*a^3*B*tan(d*x+c)+1/2/d*C*a^3*sec(d*x+c)*tan(d*x+c)

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maxima [A]  time = 0.37, size = 165, normalized size = 1.53 \[ \frac {12 \, {\left (d x + c\right )} B a^{3} + 4 \, {\left (d x + c\right )} C a^{3} - C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right ) + 12 \, C a^{3} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(12*(d*x + c)*B*a^3 + 4*(d*x + c)*C*a^3 - C*a^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*B*a^3*(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)) + 4*B*a^3*sin(d*x + c) + 4*B*a^3*tan(d*x + c) + 12*C*a^3*tan(d*x + c))/d

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mupad [B]  time = 3.06, size = 207, normalized size = 1.92 \[ \frac {B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {6\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*a^3*sin(c + d*x))/d + (6*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*B*a^3*atanh(sin(c/2 + (d
*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*C*a^3*atanh(sin(c
/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (B*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (3*C*a^3*sin(c + d*x))/(d*cos(c
+ d*x)) + (C*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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