3.8.66 \(\int \sec ^2(c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [766]

Optimal. Leaf size=114 \[ \frac {(4 a B+3 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(b B+a C) \tan (c+d x)}{d}+\frac {(4 a B+3 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {(b B+a C) \tan ^3(c+d x)}{3 d} \]

[Out]

1/8*(4*B*a+3*C*b)*arctanh(sin(d*x+c))/d+(B*b+C*a)*tan(d*x+c)/d+1/8*(4*B*a+3*C*b)*sec(d*x+c)*tan(d*x+c)/d+1/4*b
*C*sec(d*x+c)^3*tan(d*x+c)/d+1/3*(B*b+C*a)*tan(d*x+c)^3/d

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Rubi [A]
time = 0.14, antiderivative size = 114, 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, 4082, 3872, 3853, 3855, 3852} \begin {gather*} \frac {(a C+b B) \tan ^3(c+d x)}{3 d}+\frac {(a C+b B) \tan (c+d x)}{d}+\frac {(4 a B+3 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 a B+3 b C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b C \tan (c+d x) \sec ^3(c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*B + 3*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((b*B + a*C)*Tan[c + d*x])/d + ((4*a*B + 3*b*C)*Sec[c + d*x]*T
an[c + d*x])/(8*d) + (b*C*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((b*B + a*C)*Tan[c + d*x]^3)/(3*d)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4082

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Dist[1/(n + 1), Int[(d
*Csc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*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 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 \sec ^2(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^3(c+d x) (a+b \sec (c+d x)) (B+C \sec (c+d x)) \, dx\\ &=\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \sec ^3(c+d x) (4 a B+3 b C+4 (b B+a C) \sec (c+d x)) \, dx\\ &=\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}+(b B+a C) \int \sec ^4(c+d x) \, dx+\frac {1}{4} (4 a B+3 b C) \int \sec ^3(c+d x) \, dx\\ &=\frac {(4 a B+3 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} (4 a B+3 b C) \int \sec (c+d x) \, dx-\frac {(b B+a C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {(4 a B+3 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(b B+a C) \tan (c+d x)}{d}+\frac {(4 a B+3 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {(b B+a C) \tan ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.68, size = 85, normalized size = 0.75 \begin {gather*} \frac {3 (4 a B+3 b C) \tanh ^{-1}(\sin (c+d x))+\sec (c+d x) \left (12 a B+9 b C+8 (b B+a C) (2+\cos (2 (c+d x))) \sec (c+d x)+6 b C \sec ^2(c+d x)\right ) \tan (c+d x)}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(4*a*B + 3*b*C)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(12*a*B + 9*b*C + 8*(b*B + a*C)*(2 + Cos[2*(c + d*x)])
*Sec[c + d*x] + 6*b*C*Sec[c + d*x]^2)*Tan[c + d*x])/(24*d)

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Maple [A]
time = 0.09, size = 131, normalized size = 1.15

method	result	size
derivativedivides	\(\frac {-b B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(131\)
default	\(\frac {-b B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(131\)
norman	\(\frac {\frac {\left (4 B a -8 b B -8 a C +5 C b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 B a +8 b B +8 a C +5 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 B a -40 b B -40 a C -9 C b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {\left (12 B a +40 b B +40 a C -9 C b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (4 B a +3 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 B a +3 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(201\)
risch	\(-\frac {i \left (12 B a \,{\mathrm e}^{7 i \left (d x +c \right )}+9 C b \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B a \,{\mathrm e}^{5 i \left (d x +c \right )}+33 C b \,{\mathrm e}^{5 i \left (d x +c \right )}-48 B b \,{\mathrm e}^{4 i \left (d x +c \right )}-48 C a \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B a \,{\mathrm e}^{3 i \left (d x +c \right )}-33 C b \,{\mathrm e}^{3 i \left (d x +c \right )}-64 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-64 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B a \,{\mathrm e}^{i \left (d x +c \right )}-9 C b \,{\mathrm e}^{i \left (d x +c \right )}-16 b B -16 a C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a}{2 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a}{2 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(266\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-b*B*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C*b*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*
x+c)+tan(d*x+c)))+B*a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-a*C*(-2/3-1/3*sec(d*x+c)^2)*ta
n(d*x+c))

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Maxima [A]
time = 0.28, size = 163, normalized size = 1.43 \begin {gather*} \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b - 3 \, C b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a {\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 )}}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*b - 3*C*b*(2*(3*sin(d*
x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x +
c) - 1)) - 12*B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]
time = 2.49, size = 136, normalized size = 1.19 \begin {gather*} \frac {3 \, {\left (4 \, B a + 3 \, C b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, B a + 3 \, C b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, B a + 3 \, C b\right )} \cos \left (d x + c\right )^{2} + 6 \, C b + 8 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(4*B*a + 3*C*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*B*a + 3*C*b)*cos(d*x + c)^4*log(-sin(d*x +
c) + 1) + 2*(16*(C*a + B*b)*cos(d*x + c)^3 + 3*(4*B*a + 3*C*b)*cos(d*x + c)^2 + 6*C*b + 8*(C*a + B*b)*cos(d*x
+ c))*sin(d*x + c))/(d*cos(d*x + c)^4)

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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 ) \sec ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (106) = 212\).
time = 0.49, size = 304, normalized size = 2.67 \begin {gather*} \frac {3 \, {\left (4 \, B a + 3 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, B a + 3 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C b \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 )}^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*B*a + 3*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*B*a + 3*C*b)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) + 2*(12*B*a*tan(1/2*d*x + 1/2*c)^7 - 24*C*a*tan(1/2*d*x + 1/2*c)^7 - 24*B*b*tan(1/2*d*x + 1/2*c)^7 + 15*C*
b*tan(1/2*d*x + 1/2*c)^7 - 12*B*a*tan(1/2*d*x + 1/2*c)^5 + 40*C*a*tan(1/2*d*x + 1/2*c)^5 + 40*B*b*tan(1/2*d*x
+ 1/2*c)^5 + 9*C*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 40*C*a*tan(1/2*d*x + 1/2*c)^3 - 40
*B*b*tan(1/2*d*x + 1/2*c)^3 + 9*C*b*tan(1/2*d*x + 1/2*c)^3 + 12*B*a*tan(1/2*d*x + 1/2*c) + 24*C*a*tan(1/2*d*x
+ 1/2*c) + 24*B*b*tan(1/2*d*x + 1/2*c) + 15*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 7.70, size = 194, normalized size = 1.70 \begin {gather*} \frac {\left (B\,a-2\,B\,b-2\,C\,a+\frac {5\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {10\,B\,b}{3}-B\,a+\frac {10\,C\,a}{3}+\frac {3\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,C\,b}{4}-\frac {10\,B\,b}{3}-\frac {10\,C\,a}{3}-B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B\,a+2\,B\,b+2\,C\,a+\frac {5\,C\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B\,a+\frac {3\,C\,b}{4}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(B*a + 2*B*b + 2*C*a + (5*C*b)/4) + tan(c/2 + (d*x)/2)^7*(B*a - 2*B*b - 2*C*a + (5*C*b)/4)
- tan(c/2 + (d*x)/2)^3*(B*a + (10*B*b)/3 + (10*C*a)/3 - (3*C*b)/4) + tan(c/2 + (d*x)/2)^5*((10*B*b)/3 - B*a +
(10*C*a)/3 + (3*C*b)/4))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c
/2 + (d*x)/2)^8 + 1)) + (atanh(tan(c/2 + (d*x)/2))*(B*a + (3*C*b)/4))/d

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