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

Optimal. Leaf size=61 \[ \frac {(a C+b B) \tan (c+d x)}{d}+\frac {(2 a B+b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4048, 3770, 3767, 8} \[ \frac {(a C+b B) \tan (c+d x)}{d}+\frac {(2 a B+b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 75, normalized size = 1.23 \[ \frac {a B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \tan (c+d x)}{d}+\frac {b B \tan (c+d x)}{d}+\frac {b C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 96, normalized size = 1.57 \[ \frac {{\left (2 \, B a + C b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, B a + C b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b + 2 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.21, size = 153, normalized size = 2.51 \[ \frac {{\left (2 \, B a + C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, B a + C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*B*a + C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*B*a + C*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(
2*C*a*tan(1/2*d*x + 1/2*c)^3 + 2*B*b*tan(1/2*d*x + 1/2*c)^3 - C*b*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x +
1/2*c) - 2*B*b*tan(1/2*d*x + 1/2*c) - C*b*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 = 0.93, size = 86, normalized size = 1.41 \[ \frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a C \tan \left (d x +c \right )}{d}+\frac {b B \tan \left (d x +c \right )}{d}+\frac {b C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 88, normalized size = 1.44 \[ -\frac {C b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, B a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, C a \tan \left (d x + c\right ) - 4 \, B b \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(C*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*B*a*log(se
c(d*x + c) + tan(d*x + c)) - 4*C*a*tan(d*x + c) - 4*B*b*tan(d*x + c))/d

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mupad [B]  time = 4.77, size = 104, normalized size = 1.70 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B\,b+2\,C\,a+C\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B\,b+2\,C\,a-C\,b\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,B\,a+C\,b\right )}{d} \]

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)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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), x)

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