3.349 \(\int \frac {\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=156 \[ -\frac {(7 B-27 C) \tan (c+d x)}{15 a^3 d}+\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(B-C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

(B-3*C)*arctanh(sin(d*x+c))/a^3/d-1/15*(7*B-27*C)*tan(d*x+c)/a^3/d+1/5*(B-C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*se
c(d*x+c))^3+1/15*(4*B-9*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-(B-3*C)*tan(d*x+c)/d/(a^3+a^3*sec(d*
x+c))

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Rubi [A]  time = 0.50, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4072, 4019, 4008, 3787, 3770, 3767, 8} \[ -\frac {(7 B-27 C) \tan (c+d x)}{15 a^3 d}+\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(B-C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

((B - 3*C)*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((7*B - 27*C)*Tan[c + d*x])/(15*a^3*d) + ((B - C)*Sec[c + d*x]^3*T
an[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((4*B - 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d
*x])^2) - ((B - 3*C)*Tan[c + d*x])/(d*(a^3 + a^3*Sec[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(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)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4019

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

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=\int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^3(c+d x) (3 a (B-C)-a (B-6 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a^2 (4 B-9 C)-a^2 (7 B-27 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \sec (c+d x) \left (-15 a^3 (B-3 C)+a^3 (7 B-27 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(7 B-27 C) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac {(B-3 C) \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(7 B-27 C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 B-27 C) \tan (c+d x)}{15 a^3 d}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.85, size = 294, normalized size = 1.88 \[ \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left ((22 B-57 C) \tan ^3\left (\frac {1}{2} (c+d x)\right )+(87 C-22 B) \tan \left (\frac {1}{2} (c+d x)\right )+96 (B-C) \sin ^{10}\left (\frac {1}{2} (c+d x)\right ) \csc ^7(c+d x)-4 (7 B-12 C) \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)-15 (B-3 C) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\frac {1}{4} \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) ((7 B-12 C) \cos (c+d x)-4 B+9 C)+15 (B-3 C) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{15 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(-15*(B - 3*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]]) - 4*(7*B - 12*C)*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 96*(B - C)*Csc[c + d*x]^7*Sin[(c
+ d*x)/2]^10 + (-22*B + 87*C)*Tan[(c + d*x)/2] - ((-4*B + 9*C + (7*B - 12*C)*Cos[c + d*x])*Sec[(c + d*x)/2]^4*
Tan[(c + d*x)/2])/4 + 15*(B - 3*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c +
d*x)/2]])*Tan[(c + d*x)/2]^2 + (22*B - 57*C)*Tan[(c + d*x)/2]^3))/(15*a^3*d)

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fricas [A]  time = 0.47, size = 256, normalized size = 1.64 \[ \frac {15 \, {\left ({\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 3 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 3 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (11 \, B - 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (17 \, B - 57 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, B - 117 \, C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/30*(15*((B - 3*C)*cos(d*x + c)^4 + 3*(B - 3*C)*cos(d*x + c)^3 + 3*(B - 3*C)*cos(d*x + c)^2 + (B - 3*C)*cos(d
*x + c))*log(sin(d*x + c) + 1) - 15*((B - 3*C)*cos(d*x + c)^4 + 3*(B - 3*C)*cos(d*x + c)^3 + 3*(B - 3*C)*cos(d
*x + c)^2 + (B - 3*C)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(11*B - 36*C)*cos(d*x + c)^3 + 3*(17*B - 57*
C)*cos(d*x + c)^2 + (32*B - 117*C)*cos(d*x + c) - 15*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x
+ c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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giac [A]  time = 1.21, size = 186, normalized size = 1.19 \[ \frac {\frac {60 \, {\left (B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(B - 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(B - 3*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^
3 - 120*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^1
2*tan(1/2*d*x + 1/2*c)^5 + 20*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*B*a^12*ta
n(1/2*d*x + 1/2*c) - 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [A]  time = 0.66, size = 245, normalized size = 1.57 \[ -\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{20 d \,a^{3}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{2 d \,a^{3}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{3}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{3}}-\frac {C}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{3}}-\frac {C}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*C-1/3/d/a^3*B*tan(1/2*d*x+1/2*c)^3+1/2/d/a^
3*tan(1/2*d*x+1/2*c)^3*C-7/4/d/a^3*B*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)-1/d/a^3*ln(tan(1/2*d*x
+1/2*c)-1)*B+3/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*C-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*C+1/d/a^3*ln(tan(1/2*d*x+1/2*c)
+1)*B-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*C-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*C

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maxima [A]  time = 0.39, size = 286, normalized size = 1.83 \[ \frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(3*C*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 -
60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - B*((105*s
in(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^
5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

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mupad [B]  time = 2.86, size = 168, normalized size = 1.08 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-3\,C\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{4\,a^3}-\frac {3\,C}{2\,a^3}+\frac {2\,B-4\,C}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{6\,a^3}+\frac {2\,B-4\,C}{12\,a^3}\right )}{d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^3),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*(B - 3*C))/(a^3*d) - (tan(c/2 + (d*x)/2)*((3*(B - C))/(4*a^3) - (3*C)/(2*a^3) + (
2*B - 4*C)/(2*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(B - C))/(20*a^3*d) - (tan(c/2 + (d*x)/2)^3*((B - C)/(6*a^3) +
(2*B - 4*C)/(12*a^3)))/d - (2*C*tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x)/2)^2 - a^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {B \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(B*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c
+ d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

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