3.8.87 \(\int \cos (c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [787]

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

[Out]

a^3*B*x+1/2*(6*B*a^2*b+B*b^3+2*C*a^3+3*C*a*b^2)*arctanh(sin(d*x+c))/d+1/3*b*(9*B*a*b+8*C*a^2+2*C*b^2)*tan(d*x+
c)/d+1/6*b^2*(3*B*b+5*C*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*b*C*(a+b*sec(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]
time = 0.17, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 = {4157, 4003, 4133, 3855, 3852, 8} \begin {gather*} a^3 B x+\frac {b \left (8 a^2 C+9 a b B+2 b^2 C\right ) \tan (c+d x)}{3 d}+\frac {\left (2 a^3 C+6 a^2 b B+3 a b^2 C+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\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} \end {gather*}

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 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 4003

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 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 4157

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 (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 ) \text {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*}

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Mathematica [A]
time = 0.64, size = 108, normalized size = 0.79 \begin {gather*} \frac {6 a^3 B d x+3 \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))+3 b \left (6 a b B+6 a^2 C+2 b^2 C+b (b B+3 a C) \sec (c+d x)\right ) \tan (c+d x)+2 b^3 C \tan ^3(c+d x)}{6 d} \end {gather*}

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 = 0.12, size = 180, normalized size = 1.31

method	result	size
derivativedivides	\(\frac {b^{3} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,b^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a \,b^{2} B \tan \left (d x +c \right )+3 C \,b^{2} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{2} b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b C \tan \left (d x +c \right )+a^{3} B \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(180\)
default	\(\frac {b^{3} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,b^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a \,b^{2} B \tan \left (d x +c \right )+3 C \,b^{2} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{2} b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b C \tan \left (d x +c \right )+a^{3} B \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(180\)
risch	\(a^{3} B x -\frac {i b \left (3 B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+9 C a b \,{\mathrm e}^{5 i \left (d x +c \right )}-18 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-18 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-36 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-36 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-9 C b a \,{\mathrm e}^{i \left (d x +c \right )}-18 a b B -18 a^{2} C -4 b^{2} C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b B}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3} B}{2 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} C}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b B}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3} B}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} C}{2 d}\)	\(356\)
norman	\(\frac {a^{3} B x +a^{3} B x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (6 a b B +b^{2} B +6 a^{2} C +3 a b C +2 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-3 a^{3} B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a^{3} B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a^{3} B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} B x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 b \left (9 a b B +9 a^{2} C +b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 b \left (9 a b B +9 a^{2} C +b^{2} C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {b \left (6 a b B -b^{2} B +6 a^{2} C -3 a b C +2 b^{2} C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b^{2} \left (b B +3 a C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (6 a^{2} b B +b^{3} B +2 a^{3} C +3 C \,b^{2} a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (6 a^{2} b B +b^{3} B +2 a^{3} C +3 C \,b^{2} a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(398\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 216, normalized size = 1.58 \begin {gather*} \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 {gather*}

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 = 2.21, size = 189, normalized size = 1.38 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3} \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx \end {gather*}

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]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (129) = 258\).
time = 0.52, size = 336, normalized size = 2.45 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 5.58, size = 526, normalized size = 3.84 \begin {gather*} \frac {\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{6}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,C\,a^2\,b\,\sin \left (c+d\,x\right )}{4}+\frac {3\,B\,a^3\,\cos \left (c+d\,x\right )\,\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 )}{2}-\frac {B\,b^3\,\cos \left (c+d\,x\right )\,\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}}{4}-\frac {C\,a^3\,\cos \left (c+d\,x\right )\,\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}}{2}+\frac {3\,B\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {3\,C\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}-\frac {B\,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 )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{4}-\frac {C\,a^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 )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{2}-\frac {B\,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 )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{2}-\frac {C\,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 )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{4}-\frac {B\,a^2\,b\,\cos \left (c+d\,x\right )\,\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 )\,9{}\mathrm {i}}{2}-\frac {C\,a\,b^2\,\cos \left (c+d\,x\right )\,\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 )\,9{}\mathrm {i}}{4}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*b^3*sin(2*c + 2*d*x))/4 + (C*b^3*sin(3*c + 3*d*x))/6 + (C*b^3*sin(c + d*x))/2 + (3*B*a*b^2*sin(c + d*x))/4
+ (3*C*a^2*b*sin(c + d*x))/4 + (3*B*a^3*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (B*b^3*
cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/4 - (C*a^3*cos(c + d*x)*atan((sin(c/2 + (d*x
)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/2 + (3*B*a*b^2*sin(3*c + 3*d*x))/4 + (3*C*a*b^2*sin(2*c + 2*d*x))/4 + (3*C*a^
2*b*sin(3*c + 3*d*x))/4 + (B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 - (B*b^3*atan
((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/4 - (C*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(
c/2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/2 - (B*a^2*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3
*d*x)*3i)/2 - (C*a*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*3i)/4 - (B*a^2*b*cos(
c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*9i)/2 - (C*a*b^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/
2)*1i)/cos(c/2 + (d*x)/2))*9i)/4)/(d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4))

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