3.933 \(\int \frac{a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=231 \[ -\frac{b \left (-5 a^2 b^3 B+4 a^3 b^2 C+6 a^4 b B-8 a^5 C-2 a b^4 C+2 b^5 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{b^2 \left (5 a^2 b B-8 a^3 C+2 a b^2 C-2 b^3 B\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{x (b B-a C)}{a^3} \]

[Out]

((b*B - a*C)*x)/a^3 - (b*(6*a^4*b*B - 5*a^2*b^3*B + 2*b^5*B - 8*a^5*C + 4*a^3*b^2*C - 2*a*b^4*C)*ArcTanh[(Sqrt
[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^(5/2)*(a + b)^(5/2)*d) + (b^2*(b*B - 2*a*C)*Tan[c + d*x])
/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (b^2*(5*a^2*b*B - 2*b^3*B - 8*a^3*C + 2*a*b^2*C)*Tan[c + d*x])/(
2*a^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 1.10007, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {24, 3923, 4060, 3919, 3831, 2659, 208} \[ -\frac{b \left (-5 a^2 b^3 B+4 a^3 b^2 C+6 a^4 b B-8 a^5 C-2 a b^4 C+2 b^5 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{b^2 \left (5 a^2 b B-8 a^3 C+2 a b^2 C-2 b^3 B\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{x (b B-a C)}{a^3} \]

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])^4,x]

[Out]

((b*B - a*C)*x)/a^3 - (b*(6*a^4*b*B - 5*a^2*b^3*B + 2*b^5*B - 8*a^5*C + 4*a^3*b^2*C - 2*a*b^4*C)*ArcTanh[(Sqrt
[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^(5/2)*(a + b)^(5/2)*d) + (b^2*(b*B - 2*a*C)*Tan[c + d*x])
/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (b^2*(5*a^2*b*B - 2*b^3*B - 8*a^3*C + 2*a*b^2*C)*Tan[c + d*x])/(
2*a^2*(a^2 - b^2)^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 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]

Rule 4060

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

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 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{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))^4} \, dx &=\frac{\int \frac{b^2 (b B-a C)+b^3 C \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx}{b^2}\\ &=\frac{b^2 (b B-2 a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{-2 b^2 \left (a^2-b^2\right ) (b B-a C)+2 a b^3 (b B-2 a C) \sec (c+d x)-b^4 (b B-2 a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx}{2 a b^2 \left (a^2-b^2\right )}\\ &=\frac{b^2 (b B-2 a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{b^2 \left (5 a^2 b B-2 b^3 B-8 a^3 C+2 a b^2 C\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{2 b^2 \left (a^2-b^2\right )^2 (b B-a C)-a b^3 \left (4 a^2 b B-b^3 B-6 a^3 C\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{(b B-a C) x}{a^3}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{b^2 \left (5 a^2 b B-2 b^3 B-8 a^3 C+2 a b^2 C\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (b \left (6 a^4 b B-5 a^2 b^3 B+2 b^5 B-8 a^5 C+4 a^3 b^2 C-2 a b^4 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{(b B-a C) x}{a^3}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{b^2 \left (5 a^2 b B-2 b^3 B-8 a^3 C+2 a b^2 C\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (6 a^4 b B-5 a^2 b^3 B+2 b^5 B-8 a^5 C+4 a^3 b^2 C-2 a b^4 C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{(b B-a C) x}{a^3}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{b^2 \left (5 a^2 b B-2 b^3 B-8 a^3 C+2 a b^2 C\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (6 a^4 b B-5 a^2 b^3 B+2 b^5 B-8 a^5 C+4 a^3 b^2 C-2 a b^4 C\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^3 \left (a^2-b^2\right )^2 d}\\ &=\frac{(b B-a C) x}{a^3}-\frac{b \left (6 a^4 b B-5 a^2 b^3 B+2 b^5 B-8 a^5 C+4 a^3 b^2 C-2 a b^4 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{b^2 \left (5 a^2 b B-2 b^3 B-8 a^3 C+2 a b^2 C\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.85421, size = 302, normalized size = 1.31 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) (-a C+b B+b C \sec (c+d x)) \left (-\frac{a b^2 \left (-6 a^2 b B+10 a^3 C-4 a b^2 C+3 b^3 B\right ) \sin (c+d x) (a \cos (c+d x)+b)}{(a-b)^2 (a+b)^2}+\frac{2 b \left (-5 a^2 b^3 B+4 a^3 b^2 C+6 a^4 b B-8 a^5 C-2 a b^4 C+2 b^5 B\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a b^3 (2 a C-b B) \sin (c+d x)}{(a-b) (a+b)}+2 (c+d x) (b B-a C) (a \cos (c+d x)+b)^2\right )}{2 a^3 d (a+b \sec (c+d x))^3 ((b B-a C) \cos (c+d x)+b C)} \]

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])^4,x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(b*B - a*C + b*C*Sec[c + d*x])*(2*(b*B - a*C)*(c + d*x)*(b + a*Cos[c + d*
x])^2 + (2*b*(6*a^4*b*B - 5*a^2*b^3*B + 2*b^5*B - 8*a^5*C + 4*a^3*b^2*C - 2*a*b^4*C)*ArcTanh[((-a + b)*Tan[(c
+ d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])^2)/(a^2 - b^2)^(5/2) + (a*b^3*(-(b*B) + 2*a*C)*Sin[c + d*x])/
((a - b)*(a + b)) - (a*b^2*(-6*a^2*b*B + 3*b^3*B + 10*a^3*C - 4*a*b^2*C)*(b + a*Cos[c + d*x])*Sin[c + d*x])/((
a - b)^2*(a + b)^2)))/(2*a^3*d*(b*C + (b*B - a*C)*Cos[c + d*x])*(a + b*Sec[c + d*x])^3)

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Maple [B]  time = 0.114, size = 1308, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

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))^4,x)

[Out]

2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B*b-2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C-6/d*b^3/(tan(1/2*d*x+1/2*c)^2*a-ta
n(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B-1/d/a*b^4/(tan(1/2*d*x+1/2*c)^2*a-tan
(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B+2/d/a^2*b^5/(tan(1/2*d*x+1/2*c)^2*a-ta
n(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B+10/d*a*b^2/(tan(1/2*d*x+1/2*c)^2*a-ta
n(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+2/d*b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1
/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C-2/d/a*b^4/(tan(1/2*d*x+1/2*c)^2*a-tan(1/
2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+6/d*b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d
*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-1/d/a*b^4/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2
*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-2/d/a^2*b^5/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2
/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-10/d*a*b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-
b)^2*tan(1/2*d*x+1/2*c)*C+2/d*b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*
d*x+1/2*c)*C+2/d/a*b^4/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*
C-6/d*a*b^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+5/
d/a*b^4/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-2/d/a^
3*b^6/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+8/d*a^2*
b/(a^4-2*a^2*b^2+b^4)/((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*b^3/(a^4
-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C+2/d/a^2*b^5/(a^4-2
*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*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((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))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.80603, size = 3089, normalized size = 13.37 \begin{align*} \text{result too large to display} \end{align*}

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))^4,x, algorithm="fricas")

[Out]

[-1/4*(4*(C*a^9 - B*a^8*b - 3*C*a^7*b^2 + 3*B*a^6*b^3 + 3*C*a^5*b^4 - 3*B*a^4*b^5 - C*a^3*b^6 + B*a^2*b^7)*d*x
*cos(d*x + c)^2 + 8*(C*a^8*b - B*a^7*b^2 - 3*C*a^6*b^3 + 3*B*a^5*b^4 + 3*C*a^4*b^5 - 3*B*a^3*b^6 - C*a^2*b^7 +
B*a*b^8)*d*x*cos(d*x + c) + 4*(C*a^7*b^2 - B*a^6*b^3 - 3*C*a^5*b^4 + 3*B*a^4*b^5 + 3*C*a^3*b^6 - 3*B*a^2*b^7
- C*a*b^8 + B*b^9)*d*x + (8*C*a^5*b^3 - 6*B*a^4*b^4 - 4*C*a^3*b^5 + 5*B*a^2*b^6 + 2*C*a*b^7 - 2*B*b^8 + (8*C*a
^7*b - 6*B*a^6*b^2 - 4*C*a^5*b^3 + 5*B*a^4*b^4 + 2*C*a^3*b^5 - 2*B*a^2*b^6)*cos(d*x + c)^2 + 2*(8*C*a^6*b^2 -
6*B*a^5*b^3 - 4*C*a^4*b^4 + 5*B*a^3*b^5 + 2*C*a^2*b^6 - 2*B*a*b^7)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*co
s(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*(8*C*a^6*b^3 - 5*B*a^5*b^4 - 10*C*a^4*b^5 + 7*B*a^3*b^6
+ 2*C*a^2*b^7 - 2*B*a*b^8 + (10*C*a^7*b^2 - 6*B*a^6*b^3 - 14*C*a^5*b^4 + 9*B*a^4*b^5 + 4*C*a^3*b^6 - 3*B*a^2*b
^7)*cos(d*x + c))*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8
*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c) + (a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*d), -1/2*(2*(C*a^9 -
B*a^8*b - 3*C*a^7*b^2 + 3*B*a^6*b^3 + 3*C*a^5*b^4 - 3*B*a^4*b^5 - C*a^3*b^6 + B*a^2*b^7)*d*x*cos(d*x + c)^2 +
4*(C*a^8*b - B*a^7*b^2 - 3*C*a^6*b^3 + 3*B*a^5*b^4 + 3*C*a^4*b^5 - 3*B*a^3*b^6 - C*a^2*b^7 + B*a*b^8)*d*x*cos(
d*x + c) + 2*(C*a^7*b^2 - B*a^6*b^3 - 3*C*a^5*b^4 + 3*B*a^4*b^5 + 3*C*a^3*b^6 - 3*B*a^2*b^7 - C*a*b^8 + B*b^9)
*d*x - (8*C*a^5*b^3 - 6*B*a^4*b^4 - 4*C*a^3*b^5 + 5*B*a^2*b^6 + 2*C*a*b^7 - 2*B*b^8 + (8*C*a^7*b - 6*B*a^6*b^2
- 4*C*a^5*b^3 + 5*B*a^4*b^4 + 2*C*a^3*b^5 - 2*B*a^2*b^6)*cos(d*x + c)^2 + 2*(8*C*a^6*b^2 - 6*B*a^5*b^3 - 4*C*
a^4*b^4 + 5*B*a^3*b^5 + 2*C*a^2*b^6 - 2*B*a*b^7)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*co
s(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (8*C*a^6*b^3 - 5*B*a^5*b^4 - 10*C*a^4*b^5 + 7*B*a^3*b^6 + 2*C*a^
2*b^7 - 2*B*a*b^8 + (10*C*a^7*b^2 - 6*B*a^6*b^3 - 14*C*a^5*b^4 + 9*B*a^4*b^5 + 4*C*a^3*b^6 - 3*B*a^2*b^7)*cos(
d*x + c))*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3
*a^6*b^5 - a^4*b^7)*d*cos(d*x + c) + (a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{B b}{a^{3} + 3 a^{2} b \sec{\left (c + d x \right )} + 3 a b^{2} \sec ^{2}{\left (c + d x \right )} + b^{3} \sec ^{3}{\left (c + d x \right )}}\, dx - \int \frac{C a}{a^{3} + 3 a^{2} b \sec{\left (c + d x \right )} + 3 a b^{2} \sec ^{2}{\left (c + d x \right )} + b^{3} \sec ^{3}{\left (c + d x \right )}}\, dx - \int - \frac{C b \sec{\left (c + d x \right )}}{a^{3} + 3 a^{2} b \sec{\left (c + d x \right )} + 3 a b^{2} \sec ^{2}{\left (c + d x \right )} + b^{3} \sec ^{3}{\left (c + d x \right )}}\, dx \end{align*}

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))**4,x)

[Out]

-Integral(-B*b/(a**3 + 3*a**2*b*sec(c + d*x) + 3*a*b**2*sec(c + d*x)**2 + b**3*sec(c + d*x)**3), x) - Integral
(C*a/(a**3 + 3*a**2*b*sec(c + d*x) + 3*a*b**2*sec(c + d*x)**2 + b**3*sec(c + d*x)**3), x) - Integral(-C*b*sec(
c + d*x)/(a**3 + 3*a**2*b*sec(c + d*x) + 3*a*b**2*sec(c + d*x)**2 + b**3*sec(c + d*x)**3), x)

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Giac [B]  time = 1.42383, size = 695, normalized size = 3.01 \begin{align*} \text{result too large to display} \end{align*}

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))^4,x, algorithm="giac")

[Out]

((8*C*a^5*b - 6*B*a^4*b^2 - 4*C*a^3*b^3 + 5*B*a^2*b^4 + 2*C*a*b^5 - 2*B*b^6)*(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^7 - 2*a^5*
b^2 + a^3*b^4)*sqrt(-a^2 + b^2)) - (C*a - B*b)*(d*x + c)/a^3 + (10*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 6*B*a^3*
b^3*tan(1/2*d*x + 1/2*c)^3 - 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 5*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^2
*b^4*tan(1/2*d*x + 1/2*c)^3 + 3*B*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*C*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 2*B*b^6*ta
n(1/2*d*x + 1/2*c)^3 - 10*C*a^4*b^2*tan(1/2*d*x + 1/2*c) + 6*B*a^3*b^3*tan(1/2*d*x + 1/2*c) - 8*C*a^3*b^3*tan(
1/2*d*x + 1/2*c) + 5*B*a^2*b^4*tan(1/2*d*x + 1/2*c) + 4*C*a^2*b^4*tan(1/2*d*x + 1/2*c) - 3*B*a*b^5*tan(1/2*d*x
+ 1/2*c) + 2*C*a*b^5*tan(1/2*d*x + 1/2*c) - 2*B*b^6*tan(1/2*d*x + 1/2*c))/((a^6 - 2*a^4*b^2 + a^2*b^4)*(a*tan
(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2))/d