3.932 \(\int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\)
Optimal. Leaf size=140 \[ \frac {b^2 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {x (b B-a C)}{a^2}-\frac {2 b \left (-3 a^3 C+2 a^2 b B+a b^2 C-b^3 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}} \]
[Out]
(B*b-C*a)*x/a^2-2*b*(2*B*a^2*b-B*b^3-3*C*a^3+C*a*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^2/
(a-b)^(3/2)/(a+b)^(3/2)/d+b^2*(B*b-2*C*a)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))
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Rubi [A] time = 0.39, antiderivative size = 140, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used =
{24, 3923, 3919, 3831, 2659, 208} \[ -\frac {2 b \left (2 a^2 b B-3 a^3 C+a b^2 C-b^3 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {x (b B-a C)}{a^2} \]
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])^3,x]
[Out]
((b*B - a*C)*x)/a^2 - (2*b*(2*a^2*b*B - b^3*B - 3*a^3*C + a*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt
[a + b]])/(a^2*(a - b)^(3/2)*(a + b)^(3/2)*d) + (b^2*(b*B - 2*a*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c
+ d*x]))
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]
Rule 3923
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(b*(
b*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 -
b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]
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))^3} \, dx &=\frac {\int \frac {b^2 (b B-a C)+b^3 C \sec (c+d x)}{(a+b \sec (c+d x))^2} \, dx}{b^2}\\ &=\frac {b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {-b^2 \left (a^2-b^2\right ) (b B-a C)+a b^3 (b B-2 a C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a b^2 \left (a^2-b^2\right )}\\ &=\frac {(b B-a C) x}{a^2}+\frac {b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (b \left (2 a^2 b B-b^3 B-3 a^3 C+a b^2 C\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {(b B-a C) x}{a^2}+\frac {b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (2 a^2 b B-b^3 B-3 a^3 C+a b^2 C\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {(b B-a C) x}{a^2}+\frac {b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (2 \left (2 a^2 b B-b^3 B-3 a^3 C+a b^2 C\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 )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac {(b B-a C) x}{a^2}-\frac {2 b \left (2 a^2 b B-b^3 B-3 a^3 C+a 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^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.86, size = 211, normalized size = 1.51 \[ \frac {\sec (c+d x) (a \cos (c+d x)+b) (-a C+b B+b C \sec (c+d x)) \left (-\frac {2 b \left (3 a^3 C-2 a^2 b B-a b^2 C+b^3 B\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {a b^2 (b B-2 a C) \sin (c+d x)}{(a-b) (a+b)}+(c+d x) (b B-a C) (a \cos (c+d x)+b)\right )}{a^2 d (a+b \sec (c+d x))^2 ((b B-a C) \cos (c+d x)+b C)} \]
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])^3,x]
[Out]
((b + a*Cos[c + d*x])*Sec[c + d*x]*(b*B - a*C + b*C*Sec[c + d*x])*((b*B - a*C)*(c + d*x)*(b + a*Cos[c + d*x])
- (2*b*(-2*a^2*b*B + b^3*B + 3*a^3*C - a*b^2*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Co
s[c + d*x]))/(a^2 - b^2)^(3/2) + (a*b^2*(b*B - 2*a*C)*Sin[c + d*x])/((a - b)*(a + b))))/(a^2*d*(b*C + (b*B - a
*C)*Cos[c + d*x])*(a + b*Sec[c + d*x])^2)
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fricas [B] time = 0.61, size = 714, normalized size = 5.10 \[ \left [-\frac {2 \, {\left (C a^{6} - B a^{5} b - 2 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + C a^{2} b^{4} - B a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (C a^{5} b - B a^{4} b^{2} - 2 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + C a b^{5} - B b^{6}\right )} d x - {\left (3 \, C a^{3} b^{2} - 2 \, B a^{2} b^{3} - C a b^{4} + B b^{5} + {\left (3 \, C a^{4} b - 2 \, B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\right )} \cos \left (d x + c\right )\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 (2 \, C a^{4} b^{2} - B a^{3} b^{3} - 2 \, C a^{2} b^{4} + B a b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, -\frac {{\left (C a^{6} - B a^{5} b - 2 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + C a^{2} b^{4} - B a b^{5}\right )} d x \cos \left (d x + c\right ) + {\left (C a^{5} b - B a^{4} b^{2} - 2 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + C a b^{5} - B b^{6}\right )} d x - {\left (3 \, C a^{3} b^{2} - 2 \, B a^{2} b^{3} - C a b^{4} + B b^{5} + {\left (3 \, C a^{4} b - 2 \, B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\right )} \cos \left (d x + c\right )\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 (2 \, C a^{4} b^{2} - B a^{3} b^{3} - 2 \, C a^{2} b^{4} + B a b^{5}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\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))^3,x, algorithm="fricas")
[Out]
[-1/2*(2*(C*a^6 - B*a^5*b - 2*C*a^4*b^2 + 2*B*a^3*b^3 + C*a^2*b^4 - B*a*b^5)*d*x*cos(d*x + c) + 2*(C*a^5*b - B
*a^4*b^2 - 2*C*a^3*b^3 + 2*B*a^2*b^4 + C*a*b^5 - B*b^6)*d*x - (3*C*a^3*b^2 - 2*B*a^2*b^3 - C*a*b^4 + B*b^5 + (
3*C*a^4*b - 2*B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*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)) + 2*(2*C*a^4*b^2 - B*a^3*b^3 - 2*C*a^2*b^4 + B*a*b^5)*sin(d*x + c))/((a^7 - 2*a^
5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), -((C*a^6 - B*a^5*b - 2*C*a^4*b^2 + 2*B*a^3
*b^3 + C*a^2*b^4 - B*a*b^5)*d*x*cos(d*x + c) + (C*a^5*b - B*a^4*b^2 - 2*C*a^3*b^3 + 2*B*a^2*b^4 + C*a*b^5 - B*
b^6)*d*x - (3*C*a^3*b^2 - 2*B*a^2*b^3 - C*a*b^4 + B*b^5 + (3*C*a^4*b - 2*B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*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))) + (2*C*a^
4*b^2 - B*a^3*b^3 - 2*C*a^2*b^4 + B*a*b^5)*sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b
- 2*a^4*b^3 + a^2*b^5)*d)]
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giac [A] time = 0.40, size = 223, normalized size = 1.59 \[ \frac {\frac {2 \, {\left (3 \, C a^{3} b - 2 \, B a^{2} b^{2} - C a b^{3} + B b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {{\left (C a - B b\right )} {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (2 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]
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))^3,x, algorithm="giac")
[Out]
(2*(3*C*a^3*b - 2*B*a^2*b^2 - C*a*b^3 + B*b^4)*(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^4 - a^2*b^2)*sqrt(-a^2 + b^2)) - (C*a -
B*b)*(d*x + c)/a^2 + 2*(2*C*a*b^2*tan(1/2*d*x + 1/2*c) - B*b^3*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(a*tan(1/2
*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)))/d
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maple [B] time = 0.71, size = 415, normalized size = 2.96 \[ -\frac {2 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {4 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {4 b^{2} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 b^{4} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {6 b a \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d a \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d \,a^{2}}-\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a} \]
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))^3,x)
[Out]
-2/d*b^3/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)*B+4/d*b^2/(a^2-b^2
)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)*C-4/d*b^2/(a-b)/(a+b)/((a-b)*(a+b))^(
1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B+2/d*b^4/a^2/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arcta
nh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B+6/d*b*a/(a-b)/(a+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*b^3/a/(a-b)/(a+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^2*B*arctan(tan(1/2*d*x+1/2*c))*b-2/d/a*C*arctan(tan(1/2*d*x+1/2*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))^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*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 = 12.86, size = 5260, normalized size = 37.57 \[ \text {result too large to display} \]
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))^3,x)
[Out]
(2*atan(((((32*tan(c/2 + (d*x)/2)*(2*B^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*C^2*a^7*b - 5*B^2*a^2*b^6 + 4*B^2*a^3
*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 7*C^2*a^4*b^4 + 4*C^2*a^5
*b^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C*a^3*b^5 - 8*B*C*a^4*b^4 - 10*B*C*a^5
*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) + (((32*(C*a^10 + B*a^4*b^6 - 3*B*a^6*b^4 + B*a^7*b^3
+ 2*B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4 + 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2)
- (tan(c/2 + (d*x)/2)*(B*b - C*a)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/
(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*(B*b - C*a)*1i)/a^2)*(B*b - C*a))/a^2 + (((32*tan(c/2 + (d*x)/2)*(2*B
^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*C^2*a^7*b - 5*B^2*a^2*b^6 + 4*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 +
B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 7*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2
*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C*a^3*b^5 - 8*B*C*a^4*b^4 - 10*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 -
a^2*b^3 - a^3*b^2) - (((32*(C*a^10 + B*a^4*b^6 - 3*B*a^6*b^4 + B*a^7*b^3 + 2*B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4
+ 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (tan(c/2 + (d*x)/2)*(B*b - C*a)*(2*
a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)
))*(B*b - C*a)*1i)/a^2)*(B*b - C*a))/a^2)/((64*(B^3*b^8 - B^3*a*b^7 - 3*C^3*a^7*b - 3*B^3*a^2*b^6 + 2*B^3*a^3*
b^5 + 2*B^3*a^4*b^4 - C^3*a^3*b^5 + 2*C^3*a^4*b^4 + 4*C^3*a^5*b^3 - 6*C^3*a^6*b^2 - 3*B^2*C*a*b^7 + 3*B*C^2*a^
2*b^6 - 5*B*C^2*a^3*b^5 - 11*B*C^2*a^4*b^4 + 13*B*C^2*a^5*b^3 + 8*B*C^2*a^6*b^2 + 4*B^2*C*a^2*b^6 + 10*B^2*C*a
^3*b^5 - 9*B^2*C*a^4*b^4 - 7*B^2*C*a^5*b^3))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (((32*tan(c/2 + (d*x)/2)*(2*B
^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*C^2*a^7*b - 5*B^2*a^2*b^6 + 4*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 +
B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 7*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2
*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C*a^3*b^5 - 8*B*C*a^4*b^4 - 10*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 -
a^2*b^3 - a^3*b^2) + (((32*(C*a^10 + B*a^4*b^6 - 3*B*a^6*b^4 + B*a^7*b^3 + 2*B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4
+ 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (tan(c/2 + (d*x)/2)*(B*b - C*a)*(2*
a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)
))*(B*b - C*a)*1i)/a^2)*(B*b - C*a)*1i)/a^2 - (((32*tan(c/2 + (d*x)/2)*(2*B^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*
C^2*a^7*b - 5*B^2*a^2*b^6 + 4*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^
2*a^3*b^5 - 7*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C
*a^3*b^5 - 8*B*C*a^4*b^4 - 10*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (((32*(C*a^10
+ B*a^4*b^6 - 3*B*a^6*b^4 + B*a^7*b^3 + 2*B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4 + 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*
b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (tan(c/2 + (d*x)/2)*(B*b - C*a)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a
^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*(B*b - C*a)*1i)/a^2)*(B*b - C*a)
*1i)/a^2))*(B*b - C*a))/(a^2*d) - (2*tan(c/2 + (d*x)/2)*(B*b^3 - 2*C*a*b^2))/(d*(a + b)*(a*b - a^2)*(a + b - t
an(c/2 + (d*x)/2)^2*(a - b))) + (b*atan(((b*((32*tan(c/2 + (d*x)/2)*(2*B^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*C^2
*a^7*b - 5*B^2*a^2*b^6 + 4*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a
^3*b^5 - 7*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C*a^
3*b^5 - 8*B*C*a^4*b^4 - 10*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) + (b*((32*(C*a^10 +
B*a^4*b^6 - 3*B*a^6*b^4 + B*a^7*b^3 + 2*B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4 + 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*b
))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*b*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 -
2*B*a^2*b - C*a*b^2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a
^2*b^3 - a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B
*a^2*b - C*a*b^2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B
*a^2*b - C*a*b^2)*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2) + (b*((32*tan(c/2 + (d*x)/2)*(2*B^2*b^8 + C^2*a^
8 - 2*B^2*a*b^7 - 2*C^2*a^7*b - 5*B^2*a^2*b^6 + 4*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + B^2*a^6*b^2 +
2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 7*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2*B*C*a^7*b + 4*
B*C*a^2*b^6 + 12*B*C*a^3*b^5 - 8*B*C*a^4*b^4 - 10*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b
^2) - (b*((32*(C*a^10 + B*a^4*b^6 - 3*B*a^6*b^4 + B*a^7*b^3 + 2*B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4 + 4*C*a^7*b^
3 - B*a^9*b - 3*C*a^9*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*b*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(
1/2)*(B*b^3 + 3*C*a^3 - 2*B*a^2*b - C*a*b^2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*
b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)
*(B*b^3 + 3*C*a^3 - 2*B*a^2*b - C*a*b^2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)
*(B*b^3 + 3*C*a^3 - 2*B*a^2*b - C*a*b^2)*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))/((64*(B^3*b^8 - B^3*a*b^
7 - 3*C^3*a^7*b - 3*B^3*a^2*b^6 + 2*B^3*a^3*b^5 + 2*B^3*a^4*b^4 - C^3*a^3*b^5 + 2*C^3*a^4*b^4 + 4*C^3*a^5*b^3
- 6*C^3*a^6*b^2 - 3*B^2*C*a*b^7 + 3*B*C^2*a^2*b^6 - 5*B*C^2*a^3*b^5 - 11*B*C^2*a^4*b^4 + 13*B*C^2*a^5*b^3 + 8*
B*C^2*a^6*b^2 + 4*B^2*C*a^2*b^6 + 10*B^2*C*a^3*b^5 - 9*B^2*C*a^4*b^4 - 7*B^2*C*a^5*b^3))/(a^5*b + a^6 - a^3*b^
3 - a^4*b^2) + (b*((32*tan(c/2 + (d*x)/2)*(2*B^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*C^2*a^7*b - 5*B^2*a^2*b^6 + 4
*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 7*C^2*a^4*b^4 + 4
*C^2*a^5*b^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C*a^3*b^5 - 8*B*C*a^4*b^4 - 10
*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) + (b*((32*(C*a^10 + B*a^4*b^6 - 3*B*a^6*b^4 +
B*a^7*b^3 + 2*B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4 + 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*b))/(a^5*b + a^6 - a^3*b^3
- a^4*b^2) - (32*b*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B*a^2*b - C*a*b^2)*(2*a
^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a
^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B*a^2*b - C*a*b^2))/(a^8 -
a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B*a^2*b - C*a*b^2))/(a^8 -
a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2) - (b*((32*tan(c/2 + (d*x)/2)*(2*B^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*C^2*a^7*b
- 5*B^2*a^2*b^6 + 4*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5
- 7*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C*a^3*b^5
- 8*B*C*a^4*b^4 - 10*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (b*((32*(C*a^10 + B*a^4
*b^6 - 3*B*a^6*b^4 + B*a^7*b^3 + 2*B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4 + 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*b))/(a^
5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*b*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B*a
^2*b - C*a*b^2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3
- a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B*a^2*b
- C*a*b^2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B*a^2*b
- C*a*b^2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + 3*C*a^3 - 2*B*a^2*
b - C*a*b^2)*2i)/(d*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {B b}{a^{2} + 2 a b \sec {\left (c + d x \right )} + b^{2} \sec ^{2}{\left (c + d x \right )}}\right )\, dx - \int \frac {C a}{a^{2} + 2 a b \sec {\left (c + d x \right )} + b^{2} \sec ^{2}{\left (c + d x \right )}}\, dx - \int \left (- \frac {C b \sec {\left (c + d x \right )}}{a^{2} + 2 a b \sec {\left (c + d x \right )} + b^{2} \sec ^{2}{\left (c + d x \right )}}\right )\, dx \]
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))**3,x)
[Out]
-Integral(-B*b/(a**2 + 2*a*b*sec(c + d*x) + b**2*sec(c + d*x)**2), x) - Integral(C*a/(a**2 + 2*a*b*sec(c + d*x
) + b**2*sec(c + d*x)**2), x) - Integral(-C*b*sec(c + d*x)/(a**2 + 2*a*b*sec(c + d*x) + b**2*sec(c + d*x)**2),
x)
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