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

Optimal. Leaf size=131 \[ -\frac {b \left (2 a^2 B-3 a b C-b^2 B\right ) \tan (c+d x)}{d}+\frac {b \left (6 a^2 C+6 a b B+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (a C+3 b B)-\frac {b^2 (2 a B-b C) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4072, 4025, 4048, 3770, 3767, 8} \[ -\frac {b \left (2 a^2 B-3 a b C-b^2 B\right ) \tan (c+d x)}{d}+\frac {b \left (6 a^2 C+6 a b B+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (a C+3 b B)-\frac {b^2 (2 a B-b C) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3770

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

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && 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 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+b \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+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac {a B (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\int (a+b \sec (c+d x)) \left (-a (3 b B+a C)-b (b B+2 a C) \sec (c+d x)+b (2 a B-b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a B (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b^2 (2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-2 a^2 (3 b B+a C)-b \left (6 a b B+6 a^2 C+b^2 C\right ) \sec (c+d x)+2 b \left (2 a^2 B-b^2 B-3 a b C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^2 (3 b B+a C) x+\frac {a B (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b^2 (2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d}-\left (b \left (2 a^2 B-b^2 B-3 a b C\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (b \left (6 a b B+6 a^2 C+b^2 C\right )\right ) \int \sec (c+d x) \, dx\\ &=a^2 (3 b B+a C) x+\frac {b \left (6 a b B+6 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b^2 (2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\left (b \left (2 a^2 B-b^2 B-3 a b C\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=a^2 (3 b B+a C) x+\frac {b \left (6 a b B+6 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b \left (2 a^2 B-b^2 B-3 a b C\right ) \tan (c+d x)}{d}-\frac {b^2 (2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [B]  time = 4.45, size = 277, normalized size = 2.11 \[ \frac {4 a^3 B \sin (c+d x)-2 b \left (6 a^2 C+6 a b B+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 b \left (6 a^2 C+6 a b B+b^2 C\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 (c+d x) (a C+3 b B)+\frac {4 b^2 (3 a C+b B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 b^2 (3 a C+b B) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+\frac {b^3 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^3 C}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 167, normalized size = 1.27 \[ \frac {4 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} d x \cos \left (d x + c\right )^{2} + {\left (6 \, C a^{2} b + 6 \, B a b^{2} + C b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (6 \, C a^{2} b + 6 \, B a b^{2} + C b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + C b^{3} + 2 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(C*a^3 + 3*B*a^2*b)*d*x*cos(d*x + c)^2 + (6*C*a^2*b + 6*B*a*b^2 + C*b^3)*cos(d*x + c)^2*log(sin(d*x + c
) + 1) - (6*C*a^2*b + 6*B*a*b^2 + C*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*B*a^3*cos(d*x + c)^2 + C
*b^3 + 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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giac [A]  time = 0.34, size = 241, normalized size = 1.84 \[ \frac {\frac {4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} {\left (d x + c\right )} + {\left (6 \, C a^{2} b + 6 \, B a b^{2} + C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (6 \, C a^{2} b + 6 \, B a b^{2} + C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*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*(C*a^3 + 3*B*a^2*b)*(d*x + c) + (6*C*a^2*b
+ 6*B*a*b^2 + C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (6*C*a^2*b + 6*B*a*b^2 + C*b^3)*log(abs(tan(1/2*d*x
+ 1/2*c) - 1)) - 2*(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 - C*b^3*tan(1/2*d*x + 1/
2*c)^3 - 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)^2 - 1)^2)/d

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maple [A]  time = 0.94, size = 172, normalized size = 1.31 \[ \frac {a^{3} B \sin \left (d x +c \right )}{d}+a^{3} C x +\frac {C \,a^{3} c}{d}+3 B x \,a^{2} b +\frac {3 B \,a^{2} b c}{d}+\frac {3 C \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 C a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {b^{3} B \tan \left (d x +c \right )}{d}+\frac {b^{3} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {b^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.34, size = 169, normalized size = 1.29 \[ \frac {4 \, {\left (d x + c\right )} C a^{3} + 12 \, {\left (d x + c\right )} B a^{2} b - C b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, C a b^{2} \tan \left (d x + c\right ) + 4 \, B b^{3} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(d*x + c)*C*a^3 + 12*(d*x + c)*B*a^2*b - 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)) + 4*B*a^3*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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mupad [B]  time = 5.07, size = 249, normalized size = 1.90 \[ \frac {\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {2\,\left (-C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{2}-3\,B\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}+C\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}\right )}{d} \]

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 + b/cos(c + d*x))^3,x)

[Out]

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

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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+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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