3.686 \(\int \frac{\sec (c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=135 \[ \frac{2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{C \tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(b^2*d) + (2*a*(A*b^2 - a^2*C + 2*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt
[a + b]])/((a - b)^(3/2)*b^2*(a + b)^(3/2)*d) - ((A*b^2 + a^2*C)*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c +
d*x]))

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Rubi [A]  time = 0.272399, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4081, 3998, 3770, 3831, 2659, 208} \[ \frac{2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{C \tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*ArcTanh[Sin[c + d*x]])/(b^2*d) + (2*a*(A*b^2 - a^2*C + 2*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt
[a + b]])/((a - b)^(3/2)*b^2*(a + b)^(3/2)*d) - ((A*b^2 + a^2*C)*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c +
d*x]))

Rule 4081

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-a b (A+C)-\left (a^2-b^2\right ) C \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{C \int \sec (c+d x) \, dx}{b^2}-\frac{\left (a \left (a^2-b^2\right ) C-a b^2 (A+C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (a \left (a^2-b^2\right ) C-a b^2 (A+C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 a \left (C-\frac{b^2 (A+C)}{a^2-b^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^3 d}\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac{2 a \left (A b^2-a^2 C+2 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 2.3422, size = 331, normalized size = 2.45 \[ \frac{2 (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{b \left (a^2 C+A b^2\right ) (b \sin (c)-a \sin (d x))}{a (a-b) (a+b) \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right )}+\frac{2 a (\sin (c)+i \cos (c)) \left (C \left (a^2-2 b^2\right )-A b^2\right ) (a \cos (c+d x)+b) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^{3/2} \sqrt{(\cos (c)-i \sin (c))^2}}-C (a \cos (c+d x)+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+C (a \cos (c+d x)+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{b^2 d (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*(-(C*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/
2]]) + C*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*a*(-(A*b^2) + (a^2 - 2*b^2)*C)*Arc
Tan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^
2])]*(b + a*Cos[c + d*x])*(I*Cos[c] + Sin[c]))/((a^2 - b^2)^(3/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (b*(A*b^2 + a
^2*C)*(b*Sin[c] - a*Sin[d*x]))/(a*(a - b)*(a + b)*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]))))/(b^2*d*(A + 2
*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)

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Maple [B]  time = 0.089, size = 350, normalized size = 2.6 \begin{align*} 2\,{\frac{b\tan \left ( 1/2\,dx+c/2 \right ) A}{d \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ){a}^{2}C}{db \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+2\,{\frac{Aa}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{{a}^{3}C}{d{b}^{2} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+4\,{\frac{aC}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+{\frac{C}{d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*b/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)*A+2/d/b/(a^2-b^2)*tan(1
/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)*a^2*C+2/d*a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*
arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-2/d*a^3/b^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a
-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C+4/d*a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+
1/2*c)/((a+b)*(a-b))^(1/2))*C+1/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)*C-1/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*C

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.4835, size = 1513, normalized size = 11.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((C*a^3*b - (A + 2*C)*a*b^3 + (C*a^4 - (A + 2*C)*a^2*b^2)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*
x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^
2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^
4)*cos(d*x + c))*log(sin(d*x + c) + 1) - (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(
d*x + c))*log(-sin(d*x + c) + 1) - 2*(C*a^4*b + (A - C)*a^2*b^3 - A*b^5)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 +
a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b^7)*d), -1/2*(2*(C*a^3*b - (A + 2*C)*a*b^3 + (C*a^4 - (A + 2*
C)*a^2*b^2)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x
+ c))) - (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1)
+ (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(C
*a^4*b + (A - C)*a^2*b^3 - A*b^5)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a
^2*b^5 + b^7)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27165, size = 312, normalized size = 2.31 \begin{align*} \frac{\frac{2 \,{\left (C a^{3} - A a b^{2} - 2 \, C a b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac{C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac{2 \,{\left (C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(C*a^3 - A*a*b^2 - 2*C*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/
2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^2*b^2 - b^4)*sqrt(-a^2 + b^2)) + C*log(abs(tan(1/2*d*x +
1/2*c) + 1))/b^2 - C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2 + 2*(C*a^2*tan(1/2*d*x + 1/2*c) + A*b^2*tan(1/2*d
*x + 1/2*c))/((a^2*b - b^3)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)))/d