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[(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]))

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Rubi [A]  time = 0.270406, antiderivative size = 131, normalized size of antiderivative = 1., 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 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]

Rule 12

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

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 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 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]

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*}

Mathematica [A]  time = 1.17543, 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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Maple [A]  time = 0.102, size = 178, normalized size = 1.4 \begin{align*} 4\,{\frac{a}{f} \left ({\frac{1}{ \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) ^{2}} \left ( -1/4\,{\frac{ \left ( 2\,c-d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}}{{c}^{2}+2\,cd+{d}^{2}}}+1/4\,{\frac{ \left ( 2\,c-3\,d \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( c+d \right ) \left ( c-d \right ) }} \right ) }+1/4\,{\frac{2\,c-d}{ \left ({c}^{3}+{c}^{2}d-{d}^{2}c-{d}^{3} \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) } \right ) } \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 0.579293, size = 1604, normalized size = 12.24 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 ) \end{align*}

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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Giac [B]  time = 1.39031, size = 370, normalized size = 2.82 \begin{align*} \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} \end{align*}

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