3.191 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=131 \[ \frac {a (2 c-d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f (c-d)^{3/2} (c+d)^{5/2}}+\frac {a (c-2 d) \tan (e+f x)}{2 f (c-d) (c+d)^2 (c+d \sec (e+f x))}+\frac {a \tan (e+f x)}{2 f (c+d) (c+d \sec (e+f x))^2} \]

[Out]

a*(2*c-d)*arctanh((c-d)^(1/2)*tan(1/2*e+1/2*f*x)/(c+d)^(1/2))/(c-d)^(3/2)/(c+d)^(5/2)/f+1/2*a*tan(f*x+e)/(c+d)
/f/(c+d*sec(f*x+e))^2+1/2*a*(c-2*d)*tan(f*x+e)/(c-d)/(c+d)^2/f/(c+d*sec(f*x+e))

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Rubi [A]  time = 0.27, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {a (2 c-d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f (c-d)^{3/2} (c+d)^{5/2}}+\frac {a (c-2 d) \tan (e+f x)}{2 f (c-d) (c+d)^2 (c+d \sec (e+f x))}+\frac {a \tan (e+f x)}{2 f (c+d) (c+d \sec (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c + d*Sec[e + f*x])^3,x]

[Out]

(a*(2*c - d)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/((c - d)^(3/2)*(c + d)^(5/2)*f) + (a*Tan[e +
f*x])/(2*(c + d)*f*(c + d*Sec[e + f*x])^2) + (a*(c - 2*d)*Tan[e + f*x])/(2*(c - d)*(c + 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+a \sec (e+f x))}{(c+d \sec (e+f x))^3} \, dx &=\frac {a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}-\frac {\int \frac {\sec (e+f x) (-2 a (c-d)-a (c-d) \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx}{2 \left (c^2-d^2\right )}\\ &=\frac {a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac {a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}+\frac {\int \frac {a (c-d) (2 c-d) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2}\\ &=\frac {a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac {a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}+\frac {(a (2 c-d)) \int \frac {\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{2 (c-d) (c+d)^2}\\ &=\frac {a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac {a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}+\frac {(a (2 c-d)) \int \frac {1}{1+\frac {c \cos (e+f x)}{d}} \, dx}{2 (c-d) d (c+d)^2}\\ &=\frac {a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac {a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}+\frac {(a (2 c-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 )}{(c-d) d (c+d)^2 f}\\ &=\frac {a (2 c-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)^{5/2} f}+\frac {a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac {a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 1.24, size = 167, normalized size = 1.27 \[ \frac {a (\cos (e+f x)+1) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {c^2-d^2} \sin (e+f x) \left (\left (2 c^2-2 c d-d^2\right ) \cos (e+f x)+d (c-2 d)\right )-2 (2 c-d) (c \cos (e+f x)+d)^2 \tanh ^{-1}\left (\frac {(d-c) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )\right )}{4 f (c-d) (c+d)^2 \sqrt {c^2-d^2} (c \cos (e+f x)+d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c + d*Sec[e + f*x])^3,x]

[Out]

(a*(1 + Cos[e + f*x])*Sec[(e + f*x)/2]^2*(-2*(2*c - d)*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(d
+ c*Cos[e + f*x])^2 + Sqrt[c^2 - d^2]*((c - 2*d)*d + (2*c^2 - 2*c*d - d^2)*Cos[e + f*x])*Sin[e + f*x]))/(4*(c
- d)*(c + d)^2*Sqrt[c^2 - d^2]*f*(d + c*Cos[e + f*x])^2)

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fricas [B]  time = 0.52, size = 736, normalized size = 5.62 \[ \left [\frac {{\left (2 \, a c d^{2} - a d^{3} + {\left (2 \, a c^{3} - a c^{2} d\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a c^{2} d - a c d^{2}\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 (a c^{3} d - 2 \, a c^{2} d^{2} - a c d^{3} + 2 \, a d^{4} + {\left (2 \, a c^{4} - 2 \, a c^{3} d - 3 \, a c^{2} d^{2} + 2 \, a c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (c^{7} + c^{6} d - 2 \, c^{5} d^{2} - 2 \, c^{4} d^{3} + c^{3} d^{4} + c^{2} d^{5}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (c^{6} d + c^{5} d^{2} - 2 \, c^{4} d^{3} - 2 \, c^{3} d^{4} + c^{2} d^{5} + c d^{6}\right )} f \cos \left (f x + e\right ) + {\left (c^{5} d^{2} + c^{4} d^{3} - 2 \, c^{3} d^{4} - 2 \, c^{2} d^{5} + c d^{6} + d^{7}\right )} f\right )}}, \frac {{\left (2 \, a c d^{2} - a d^{3} + {\left (2 \, a c^{3} - a c^{2} d\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a c^{2} d - a c d^{2}\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 (a c^{3} d - 2 \, a c^{2} d^{2} - a c d^{3} + 2 \, a d^{4} + {\left (2 \, a c^{4} - 2 \, a c^{3} d - 3 \, a c^{2} d^{2} + 2 \, a c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c^{7} + c^{6} d - 2 \, c^{5} d^{2} - 2 \, c^{4} d^{3} + c^{3} d^{4} + c^{2} d^{5}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (c^{6} d + c^{5} d^{2} - 2 \, c^{4} d^{3} - 2 \, c^{3} d^{4} + c^{2} d^{5} + c d^{6}\right )} f \cos \left (f x + e\right ) + {\left (c^{5} d^{2} + c^{4} d^{3} - 2 \, c^{3} d^{4} - 2 \, c^{2} d^{5} + c d^{6} + d^{7}\right )} f\right )}}\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))^3,x, algorithm="fricas")

[Out]

[1/4*((2*a*c*d^2 - a*d^3 + (2*a*c^3 - a*c^2*d)*cos(f*x + e)^2 + 2*(2*a*c^2*d - a*c*d^2)*cos(f*x + e))*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^3*d - 2*a*c^2*d^2 - a*c*d^3 +
2*a*d^4 + (2*a*c^4 - 2*a*c^3*d - 3*a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e))*sin(f*x + e))/((c^7 + c^6*d -
2*c^5*d^2 - 2*c^4*d^3 + c^3*d^4 + c^2*d^5)*f*cos(f*x + e)^2 + 2*(c^6*d + c^5*d^2 - 2*c^4*d^3 - 2*c^3*d^4 + c^2
*d^5 + c*d^6)*f*cos(f*x + e) + (c^5*d^2 + c^4*d^3 - 2*c^3*d^4 - 2*c^2*d^5 + c*d^6 + d^7)*f), 1/2*((2*a*c*d^2 -
a*d^3 + (2*a*c^3 - a*c^2*d)*cos(f*x + e)^2 + 2*(2*a*c^2*d - a*c*d^2)*cos(f*x + e))*sqrt(-c^2 + d^2)*arctan(-s
qrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) + (a*c^3*d - 2*a*c^2*d^2 - a*c*d^3 + 2*a*d^4
+ (2*a*c^4 - 2*a*c^3*d - 3*a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e))*sin(f*x + e))/((c^7 + c^6*d - 2*c^5*d^
2 - 2*c^4*d^3 + c^3*d^4 + c^2*d^5)*f*cos(f*x + e)^2 + 2*(c^6*d + c^5*d^2 - 2*c^4*d^3 - 2*c^3*d^4 + c^2*d^5 + c
*d^6)*f*cos(f*x + e) + (c^5*d^2 + c^4*d^3 - 2*c^3*d^4 - 2*c^2*d^5 + c*d^6 + d^7)*f)]

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giac [B]  time = 0.42, size = 274, normalized size = 2.09 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, c + 2 \, d\right ) + \arctan \left (-\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} {\left (2 \, a c - a d\right )}}{{\left (c^{3} + c^{2} d - c d^{2} - d^{3}\right )} \sqrt {-c^{2} + d^{2}}} - \frac {2 \, a c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c^{3} + c^{2} d - c d^{2} - d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2}}}{f} \]

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))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.74, size = 178, normalized size = 1.36 \[ \frac {4 a \left (\frac {-\frac {\left (2 c -d \right ) \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{4 \left (c^{2}+2 c d +d^{2}\right )}+\frac {\left (2 c -3 d \right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{4 \left (c +d \right ) \left (c -d \right )}}{\left (\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c -\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d -c -d \right )^{2}}+\frac {\left (2 c -d \right ) \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{4 \left (c^{3}+c^{2} d -c \,d^{2}-d^{3}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f} \]

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))^3,x)

[Out]

4/f*a*((-1/4*(2*c-d)/(c^2+2*c*d+d^2)*tan(1/2*e+1/2*f*x)^3+1/4*(2*c-3*d)/(c+d)/(c-d)*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+1/4*(2*c-d)/(c^3+c^2*d-c*d^2-d^3)/((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))^3,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 = 3.78, size = 171, normalized size = 1.31 \[ \frac {a\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c-d}}{\sqrt {c+d}}\right )\,\left (2\,c-d\right )}{f\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{3/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,a\,c-a\,d\right )}{{\left (c+d\right )}^2}-\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c-3\,d\right )}{\left (c+d\right )\,\left (c-d\right )}}{f\,\left (2\,c\,d-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2-2\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (c^2-2\,c\,d+d^2\right )+c^2+d^2\right )} \]

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))^3),x)

[Out]

(a*atanh((tan(e/2 + (f*x)/2)*(c - d)^(1/2))/(c + d)^(1/2))*(2*c - d))/(f*(c + d)^(5/2)*(c - d)^(3/2)) - ((tan(
e/2 + (f*x)/2)^3*(2*a*c - a*d))/(c + d)^2 - (a*tan(e/2 + (f*x)/2)*(2*c - 3*d))/((c + d)*(c - d)))/(f*(2*c*d -
tan(e/2 + (f*x)/2)^2*(2*c^2 - 2*d^2) + tan(e/2 + (f*x)/2)^4*(c^2 - 2*c*d + d^2) + c^2 + d^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \frac {\sec {\left (e + f x \right )}}{c^{3} + 3 c^{2} d \sec {\left (e + f x \right )} + 3 c d^{2} \sec ^{2}{\left (e + f x \right )} + d^{3} \sec ^{3}{\left (e + f x \right )}}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{c^{3} + 3 c^{2} d \sec {\left (e + f x \right )} + 3 c d^{2} \sec ^{2}{\left (e + f x \right )} + d^{3} \sec ^{3}{\left (e + f x \right )}}\, dx\right ) \]

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))**3,x)

[Out]

a*(Integral(sec(e + f*x)/(c**3 + 3*c**2*d*sec(e + f*x) + 3*c*d**2*sec(e + f*x)**2 + d**3*sec(e + f*x)**3), x)
+ Integral(sec(e + f*x)**2/(c**3 + 3*c**2*d*sec(e + f*x) + 3*c*d**2*sec(e + f*x)**2 + d**3*sec(e + f*x)**3), x
))

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