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

Optimal. Leaf size=99 \[ \frac {(b c-a d) \tan (e+f x)}{f \left (c^2-d^2\right ) (c+d \sec (e+f x))}+\frac {2 (a c-b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f (c-d)^{3/2} (c+d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 99, 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} \[ \frac {(b c-a d) \tan (e+f x)}{f \left (c^2-d^2\right ) (c+d \sec (e+f x))}+\frac {2 (a c-b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f (c-d)^{3/2} (c+d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*c - b*d)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/((c - d)^(3/2)*(c + d)^(3/2)*f) + ((b*c -
a*d)*Tan[e + f*x])/((c^2 - d^2)*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+b \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx &=\frac {(b c-a d) \tan (e+f x)}{\left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac {\int \frac {(-a c+b d) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{-c^2+d^2}\\ &=\frac {(b c-a d) \tan (e+f x)}{\left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac {(a c-b d) \int \frac {\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{c^2-d^2}\\ &=\frac {(b c-a d) \tan (e+f x)}{\left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac {(a c-b d) \int \frac {1}{1+\frac {c \cos (e+f x)}{d}} \, dx}{d \left (c^2-d^2\right )}\\ &=\frac {(b c-a d) \tan (e+f x)}{\left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac {(2 (a c-b d)) \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 \left (c^2-d^2\right ) f}\\ &=\frac {2 (a c-b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{(c-d)^{3/2} (c+d)^{3/2} f}+\frac {(b c-a d) \tan (e+f x)}{\left (c^2-d^2\right ) f (c+d \sec (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 97, normalized size = 0.98 \[ \frac {\frac {(b c-a d) \sin (e+f x)}{(c-d) (c+d) (c \cos (e+f x)+d)}-\frac {2 (a c-b d) \tanh ^{-1}\left (\frac {(d-c) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 389, normalized size = 3.93 \[ \left [\frac {{\left (a c d - b d^{2} + {\left (a c^{2} - b c d\right )} \cos \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \, {\left (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f \cos \left (f x + e\right ) + {\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f\right )}}, \frac {{\left (a c d - b d^{2} + {\left (a c^{2} - b c d\right )} \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) + {\left (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \sin \left (f x + e\right )}{{\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f \cos \left (f x + e\right ) + {\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a*c*d - b*d^2 + (a*c^2 - b*c*d)*cos(f*x + e))*sqrt(c^2 - d^2)*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*c
os(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*(b*c^3 - a*c^2*d - b*c*d^2 + a*d^3)*sin(f*x + e))/((c^5 - 2*c^3*d^2 + c*d^4)*f*cos(f
*x + e) + (c^4*d - 2*c^2*d^3 + d^5)*f), ((a*c*d - b*d^2 + (a*c^2 - b*c*d)*cos(f*x + e))*sqrt(-c^2 + d^2)*arcta
n(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) + (b*c^3 - a*c^2*d - b*c*d^2 + a*d^3)*sin
(f*x + e))/((c^5 - 2*c^3*d^2 + c*d^4)*f*cos(f*x + e) + (c^4*d - 2*c^2*d^3 + d^5)*f)]

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giac [A]  time = 0.66, size = 179, normalized size = 1.81 \[ -\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} {\left (a c - b d\right )}}{{\left (c^{2} - d^{2}\right )} \sqrt {-c^{2} + d^{2}}} + \frac {b c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )} {\left (c^{2} - d^{2}\right )}}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+b*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*c - b*d)/((c^2 - d^2)*sqrt(-c^2 + d^2)) + (b*c*tan(1/2*f*x + 1/2*e) - a*d*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^2 - d^2)))/f

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maple [A]  time = 0.66, size = 132, normalized size = 1.33 \[ \frac {\frac {2 \left (d a -c b \right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{\left (c^{2}-d^{2}\right ) \left (\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c -\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d -c -d \right )}+\frac {2 \left (c a -b d \right ) \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c +d \right ) \left (c -d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(2*(a*d-b*c)/(c^2-d^2)*tan(1/2*e+1/2*f*x)/(tan(1/2*e+1/2*f*x)^2*c-tan(1/2*e+1/2*f*x)^2*d-c-d)+2*(a*c-b*d)/
(c+d)/(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+b*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 = 2.12, size = 106, normalized size = 1.07 \[ \frac {2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c-d}}{\sqrt {c+d}}\right )\,\left (a\,c-b\,d\right )}{f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{3/2}}-\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,d-b\,c\right )}{f\,\left (c+d\right )\,\left (c-d\right )\,\left (\left (d-c\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+c+d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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