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

Optimal. Leaf size=58 \[ a A x+\frac {a C \tan (c+d x)}{d}+\frac {b (2 A+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

a*A*x+1/2*b*(2*A+C)*arctanh(sin(d*x+c))/d+a*C*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 = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4049, 3770, 3767, 8} \[ a A x+\frac {a C \tan (c+d x)}{d}+\frac {b (2 A+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*x + (b*(2*A + C)*ArcTanh[Sin[c + d*x]])/(2*d) + (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 4049

Int[((A_.) + 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 + b*(2*A + C)*Csc[e + f*x] + 2*a*C*Csc[e + f*x
]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x)) \left (A+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 A+b (2 A+C) \sec (c+d x)+2 a C \sec ^2(c+d x)\right ) \, dx\\ &=a A x+\frac {b C \sec (c+d x) \tan (c+d x)}{2 d}+(a C) \int \sec ^2(c+d x) \, dx+\frac {1}{2} (b (2 A+C)) \int \sec (c+d x) \, dx\\ &=a A x+\frac {b (2 A+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b C \sec (c+d x) \tan (c+d x)}{2 d}-\frac {(a C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=a A x+\frac {b (2 A+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {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 = 0.02, size = 67, normalized size = 1.16 \[ a A x+\frac {a C \tan (c+d x)}{d}+\frac {A b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*x + (A*b*ArcTanh[Sin[c + d*x]])/d + (b*C*ArcTanh[Sin[c + d*x]])/(2*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 = 0.45, size = 101, normalized size = 1.74 \[ \frac {4 \, A a d x \cos \left (d x + c\right )^{2} + {\left (2 \, A + C\right )} b \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + C\right )} b \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a \cos \left (d x + c\right ) + C b\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.22, size = 134, normalized size = 2.31 \[ \frac {2 \, {\left (d x + c\right )} A a + {\left (2 \, A b + C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A b + C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*x + c)*A*a + (2*A*b + C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A*b + 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 - C*b*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*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 = 0.94, size = 85, normalized size = 1.47 \[ a A x +\frac {A a c}{d}+\frac {a C \tan \left (d x +c \right )}{d}+\frac {A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 88, normalized size = 1.52 \[ \frac {4 \, {\left (d x + c\right )} A a - C b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, C a \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(d*x + c)*A*a - 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*A*b*log(sec(d*x + c) + tan(d*x + c)) + 4*C*a*tan(d*x + c))/d

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mupad [B]  time = 3.63, size = 135, normalized size = 2.33 \[ \frac {2\,A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,b\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {A\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (A*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
2i)/d - (C*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/d + (C*a*sin(c + d*x))/(d*cos(c + d*x)) + (C
*b*sin(c + d*x))/(2*d*cos(c + d*x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x)), x)

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