3.931 \(\int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=75 \[ \frac {x (b B-a C)}{a}-\frac {2 b (b B-2 a C) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}} \]

[Out]

(B*b-C*a)*x/a-2*b*(B*b-2*C*a)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/d/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {24, 3919, 3831, 2659, 208} \[ \frac {x (b B-a C)}{a}-\frac {2 b (b B-2 a C) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - a*C)*x)/a - (2*b*(b*B - 2*a*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqr
t[a + b]*d)

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

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 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 76, normalized size = 1.01 \[ \frac {\frac {2 b (b B-2 a C) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+(c+d x) (b B-a C)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - a*C)*(c + d*x) + (2*b*(b*B - 2*a*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b
^2])/(a*d)

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fricas [A]  time = 0.54, size = 285, normalized size = 3.80 \[ \left [-\frac {2 \, {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} d x + {\left (2 \, C a b - B b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}, -\frac {{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} d x - {\left (2 \, C a b - B b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right )}{{\left (a^{3} - a b^{2}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*d*x + (2*C*a*b - B*b^2)*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)))/((a^3 - a*b^2)*d), -((C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*d*x - (2*C*a*b
- B*b^2)*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))))/((a^3 -
a*b^2)*d)]

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giac [B]  time = 0.25, size = 309, normalized size = 4.12 \[ \frac {\frac {{\left (\sqrt {-a^{2} + b^{2}} C {\left (a + b\right )} {\left | a \right |} {\left | -a + b \right |} - \sqrt {-a^{2} + b^{2}} B b {\left | a \right |} {\left | -a + b \right |} + \sqrt {-a^{2} + b^{2}} {\left (a b - 2 \, b^{2}\right )} B {\left | -a + b \right |} - {\left (a^{2} - 3 \, a b\right )} \sqrt {-a^{2} + b^{2}} C {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b + \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} a^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} {\left | a \right |}} - \frac {{\left (C a^{2} - B a b - 3 \, C a b + 2 \, B b^{2} + C a {\left | a \right |} - B b {\left | a \right |} + C b {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b - \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{a^{2} - b {\left | a \right |}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.73, size = 133, normalized size = 1.77 \[ -\frac {2 b^{2} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B b}{d a}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*b^2/a/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B+4/d*b/((a-b)*(a+b))^(1/
2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*C+2/d/a*arctan(tan(1/2*d*x+1/2*c))*B*b-2/d*arctan(tan
(1/2*d*x+1/2*c))*C

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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((a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 7.94, size = 1169, normalized size = 15.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b)/(a + b/cos(c + d*x))^2,x)

[Out]

(2*C*b^2*atan((C^2*a^4*sin(c/2 + (d*x)/2) + B^2*a^2*b^2*sin(c/2 + (d*x)/2) + 3*C^2*a^2*b^2*sin(c/2 + (d*x)/2)
- 2*B*C*a*b^3*sin(c/2 + (d*x)/2) - 2*B*C*a^3*b*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(C^2*a^3 + B^2*a*b^2
+ 3*C^2*a*b^2 - 2*B*C*b^3 - 2*B*C*a^2*b))))/(d*(a^2 - b^2)) - (2*C*a^2*atan((C^2*a^4*sin(c/2 + (d*x)/2) + B^2*
a^2*b^2*sin(c/2 + (d*x)/2) + 3*C^2*a^2*b^2*sin(c/2 + (d*x)/2) - 2*B*C*a*b^3*sin(c/2 + (d*x)/2) - 2*B*C*a^3*b*s
in(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(C^2*a^3 + B^2*a*b^2 + 3*C^2*a*b^2 - 2*B*C*b^3 - 2*B*C*a^2*b))))/(d*(
a^2 - b^2)) - (2*C*b^3*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)
)/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) + (2*B*a*b*atan((C^2*a^4*sin(c/2 + (d*x)/2) + B^2*a^2*b^2*sin(c/2
+ (d*x)/2) + 3*C^2*a^2*b^2*sin(c/2 + (d*x)/2) - 2*B*C*a*b^3*sin(c/2 + (d*x)/2) - 2*B*C*a^3*b*sin(c/2 + (d*x)/
2))/(a*cos(c/2 + (d*x)/2)*(C^2*a^3 + B^2*a*b^2 + 3*C^2*a*b^2 - 2*B*C*b^3 - 2*B*C*a^2*b))))/(d*(a^2 - b^2)) - (
B*a*b^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*
x)/2)))/(d*(a^2 - b^2)^(3/2)) + (2*C*a^2*b*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/
2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) + (B*b^4*log((a*sin(c/2 + (d*x)/2) - b*sin(c/
2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(a*d*(a^2 - b^2)^(3/2)) - (2*B*b^3*a
tan((C^2*a^4*sin(c/2 + (d*x)/2) + B^2*a^2*b^2*sin(c/2 + (d*x)/2) + 3*C^2*a^2*b^2*sin(c/2 + (d*x)/2) - 2*B*C*a*
b^3*sin(c/2 + (d*x)/2) - 2*B*C*a^3*b*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(C^2*a^3 + B^2*a*b^2 + 3*C^2*a*
b^2 - 2*B*C*b^3 - 2*B*C*a^2*b))))/(a*d*(a^2 - b^2)) - (2*C*b*log((a*cos(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2)
- sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2))*((a + b)*(a - b))^(1/2))/(d*(a^2 - b^2)) + (B*b^2*
log((a*cos(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2))*(
(a + b)*(a - b))^(1/2))/(a*d*(a^2 - b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-B*b/(a + b*sec(c + d*x)), x) - Integral(C*a/(a + b*sec(c + d*x)), x) - Integral(-C*b*sec(c + d*x)/(
a + b*sec(c + d*x)), x)

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