3.190 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=79 \[ \frac {2 a \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f \sqrt {c-d} (c+d)^{3/2}}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))} \]

[Out]

2*a*arctanh((c-d)^(1/2)*tan(1/2*e+1/2*f*x)/(c+d)^(1/2))/(c+d)^(3/2)/f/(c-d)^(1/2)+a*tan(f*x+e)/(c+d)/f/(c+d*se
c(f*x+e))

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Rubi [A]  time = 0.14, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4003, 12, 3831, 2659, 208} \[ \frac {2 a \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f \sqrt {c-d} (c+d)^{3/2}}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c + d*Sec[e + f*x])^2,x]

[Out]

(2*a*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(Sqrt[c - d]*(c + d)^(3/2)*f) + (a*Tan[e + f*x])/((c
+ d)*f*(c + d*Sec[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4003

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a*B
)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] &
& LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx &=\frac {a \tan (e+f x)}{(c+d) f (c+d \sec (e+f x))}-\frac {\int \frac {a (c-d) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{-c^2+d^2}\\ &=\frac {a \tan (e+f x)}{(c+d) f (c+d \sec (e+f x))}+\frac {a \int \frac {\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{c+d}\\ &=\frac {a \tan (e+f x)}{(c+d) f (c+d \sec (e+f x))}+\frac {a \int \frac {1}{1+\frac {c \cos (e+f x)}{d}} \, dx}{d (c+d)}\\ &=\frac {a \tan (e+f x)}{(c+d) f (c+d \sec (e+f x))}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c}{d}+\left (1-\frac {c}{d}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{d (c+d) f}\\ &=\frac {2 a \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{\sqrt {c-d} (c+d)^{3/2} f}+\frac {a \tan (e+f x)}{(c+d) f (c+d \sec (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 75, normalized size = 0.95 \[ \frac {a \left (\frac {\sin (e+f x)}{c \cos (e+f x)+d}-\frac {2 \tanh ^{-1}\left (\frac {(d-c) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}\right )}{f (c+d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c + d*Sec[e + f*x])^2,x]

[Out]

(a*((-2*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] + Sin[e + f*x]/(d + c*Cos[e + f*
x])))/((c + d)*f)

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fricas [B]  time = 0.47, size = 357, normalized size = 4.52 \[ \left [\frac {{\left (a c \cos \left (f x + e\right ) + a d\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \, {\left (a c^{2} - a d^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f \cos \left (f x + e\right ) + {\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f\right )}}, \frac {{\left (a c \cos \left (f x + e\right ) + a d\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) + {\left (a c^{2} - a d^{2}\right )} \sin \left (f x + e\right )}{{\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f \cos \left (f x + e\right ) + {\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/2*((a*c*cos(f*x + e) + a*d)*sqrt(c^2 - d^2)*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 + 2*sqrt
(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x + e) + d^2))
+ 2*(a*c^2 - a*d^2)*sin(f*x + e))/((c^4 + c^3*d - c^2*d^2 - c*d^3)*f*cos(f*x + e) + (c^3*d + c^2*d^2 - c*d^3 -
d^4)*f), ((a*c*cos(f*x + e) + a*d)*sqrt(-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2
)*sin(f*x + e))) + (a*c^2 - a*d^2)*sin(f*x + e))/((c^4 + c^3*d - c^2*d^2 - c*d^3)*f*cos(f*x + e) + (c^3*d + c^
2*d^2 - c*d^3 - d^4)*f)]

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giac [B]  time = 0.34, size = 143, normalized size = 1.81 \[ -\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} a}{\sqrt {-c^{2} + d^{2}} {\left (c + d\right )}} + \frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )} {\left (c + d\right )}}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))^2,x, algorithm="giac")

[Out]

-2*((pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c*tan(1/2*f*x + 1/2*e) - d*tan(1/2*f*x + 1/2*e)
)/sqrt(-c^2 + d^2)))*a/(sqrt(-c^2 + d^2)*(c + d)) + a*tan(1/2*f*x + 1/2*e)/((c*tan(1/2*f*x + 1/2*e)^2 - d*tan(
1/2*f*x + 1/2*e)^2 - c - d)*(c + d)))/f

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maple [A]  time = 0.72, size = 105, normalized size = 1.33 \[ \frac {4 a \left (-\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{2 \left (c +d \right ) \left (\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c -\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d -c -d \right )}+\frac {\arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{2 \left (c +d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))^2,x)

[Out]

4/f*a*(-1/2*tan(1/2*e+1/2*f*x)/(c+d)/(tan(1/2*e+1/2*f*x)^2*c-tan(1/2*e+1/2*f*x)^2*d-c-d)+1/2/(c+d)/((c+d)*(c-d
))^(1/2)*arctanh(tan(1/2*e+1/2*f*x)*(c-d)/((c+d)*(c-d))^(1/2)))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*d^2>0)', see `assume?`
for more details)Is 4*c^2-4*d^2 positive or negative?

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mupad [B]  time = 1.91, size = 85, normalized size = 1.08 \[ \frac {2\,a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (c+d\right )\,\left (\left (d-c\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+c+d\right )}+\frac {2\,a\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c-d}}{\sqrt {c+d}}\right )}{f\,{\left (c+d\right )}^{3/2}\,\sqrt {c-d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))/(cos(e + f*x)*(c + d/cos(e + f*x))^2),x)

[Out]

(2*a*tan(e/2 + (f*x)/2))/(f*(c + d)*(c + d - tan(e/2 + (f*x)/2)^2*(c - d))) + (2*a*atanh((tan(e/2 + (f*x)/2)*(
c - d)^(1/2))/(c + d)^(1/2)))/(f*(c + d)^(3/2)*(c - d)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))**2,x)

[Out]

a*(Integral(sec(e + f*x)/(c**2 + 2*c*d*sec(e + f*x) + d**2*sec(e + f*x)**2), x) + Integral(sec(e + f*x)**2/(c*
*2 + 2*c*d*sec(e + f*x) + d**2*sec(e + f*x)**2), x))

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