3.4.32 \(\int \frac {\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [332]

Optimal. Leaf size=131 \[ \frac {3 (B-C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(3 B-4 C) \tan (c+d x)}{a d}+\frac {3 (B-C) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 B-4 C) \tan ^3(c+d x)}{3 a d} \]

[Out]

3/2*(B-C)*arctanh(sin(d*x+c))/a/d-(3*B-4*C)*tan(d*x+c)/a/d+3/2*(B-C)*sec(d*x+c)*tan(d*x+c)/a/d+(B-C)*sec(d*x+c
)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))-1/3*(3*B-4*C)*tan(d*x+c)^3/a/d

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Rubi [A]
time = 0.17, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4157, 4104, 3872, 3853, 3855, 3852} \begin {gather*} -\frac {(3 B-4 C) \tan ^3(c+d x)}{3 a d}-\frac {(3 B-4 C) \tan (c+d x)}{a d}+\frac {3 (B-C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {(B-C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac {3 (B-C) \tan (c+d x) \sec (c+d x)}{2 a d} \end {gather*}

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]),x]

[Out]

(3*(B - C)*ArcTanh[Sin[c + d*x]])/(2*a*d) - ((3*B - 4*C)*Tan[c + d*x])/(a*d) + (3*(B - C)*Sec[c + d*x]*Tan[c +
d*x])/(2*a*d) + ((B - C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) - ((3*B - 4*C)*Tan[c + d*x]^3)
/(3*a*d)

Rule 3852

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rule 3872

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 4104

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 4157

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)} \, dx &=\int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{a+a \sec (c+d x)} \, dx\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {\int \sec ^3(c+d x) (3 a (B-C)-a (3 B-4 C) \sec (c+d x)) \, dx}{a^2}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 B-4 C) \int \sec ^4(c+d x) \, dx}{a}+\frac {(3 (B-C)) \int \sec ^3(c+d x) \, dx}{a}\\ &=\frac {3 (B-C) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(3 (B-C)) \int \sec (c+d x) \, dx}{2 a}+\frac {(3 B-4 C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a d}\\ &=\frac {3 (B-C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(3 B-4 C) \tan (c+d x)}{a d}+\frac {3 (B-C) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 B-4 C) \tan ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(550\) vs. \(2(131)=262\).
time = 1.15, size = 550, normalized size = 4.20 \begin {gather*} -\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (9 B \cos \left (\frac {5}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 C \cos \left (\frac {5}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 B \cos \left (\frac {7}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 C \cos \left (\frac {7}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+27 (B-C) \cos \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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+27 (B-C) \cos \left (\frac {3}{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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-9 B \cos \left (\frac {5}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 C \cos \left (\frac {5}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 B \cos \left (\frac {7}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 C \cos \left (\frac {7}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-12 B \sin \left (\frac {1}{2} (c+d x)\right )+18 B \sin \left (\frac {3}{2} (c+d x)\right )-30 C \sin \left (\frac {3}{2} (c+d x)\right )-6 B \sin \left (\frac {5}{2} (c+d x)\right )+2 C \sin \left (\frac {5}{2} (c+d x)\right )+12 B \sin \left (\frac {7}{2} (c+d x)\right )-16 C \sin \left (\frac {7}{2} (c+d x)\right )\right )}{24 a d (1+\cos (c+d x))} \end {gather*}

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]),x]

[Out]

-1/24*(Cos[(c + d*x)/2]*Sec[c + d*x]^3*(9*B*Cos[(5*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 9*
C*Cos[(5*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*B*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2
] - Sin[(c + d*x)/2]] - 9*C*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 27*(B - C)*Cos[(c
+ d*x)/2]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 27*(B - C)*C
os[(3*(c + d*x))/2]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 9*
B*Cos[(5*(c + d*x))/2]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*C*Cos[(5*(c + d*x))/2]*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] - 9*B*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*C*Cos[(7*(c + d*
x))/2]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 12*B*Sin[(c + d*x)/2] + 18*B*Sin[(3*(c + d*x))/2] - 30*C*Sin
[(3*(c + d*x))/2] - 6*B*Sin[(5*(c + d*x))/2] + 2*C*Sin[(5*(c + d*x))/2] + 12*B*Sin[(7*(c + d*x))/2] - 16*C*Sin
[(7*(c + d*x))/2]))/(a*d*(1 + Cos[c + d*x]))

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Maple [A]
time = 0.56, size = 190, normalized size = 1.45

method	result	size
derivativedivides	\(\frac {-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {B -2 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-\frac {3 C}{2}+\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\frac {5 C}{2}-\frac {3 B}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {2 C -B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (\frac {3 C}{2}-\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 C}{2}-\frac {3 B}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{d a}\)	\(190\)
default	\(\frac {-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {B -2 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-\frac {3 C}{2}+\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\frac {5 C}{2}-\frac {3 B}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {2 C -B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (\frac {3 C}{2}-\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 C}{2}-\frac {3 B}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{d a}\)	\(190\)
norman	\(\frac {\frac {\left (-9 C +7 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 \left (B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\left (B -C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (27 B -37 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (-49 C +39 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {3 \left (B -C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {3 \left (B -C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(192\)
risch	\(-\frac {i \left (9 B \,{\mathrm e}^{6 i \left (d x +c \right )}-9 C \,{\mathrm e}^{6 i \left (d x +c \right )}+9 B \,{\mathrm e}^{5 i \left (d x +c \right )}-9 C \,{\mathrm e}^{5 i \left (d x +c \right )}+24 B \,{\mathrm e}^{4 i \left (d x +c \right )}-24 C \,{\mathrm e}^{4 i \left (d x +c \right )}+12 B \,{\mathrm e}^{3 i \left (d x +c \right )}-24 C \,{\mathrm e}^{3 i \left (d x +c \right )}+27 B \,{\mathrm e}^{2 i \left (d x +c \right )}-39 C \,{\mathrm e}^{2 i \left (d x +c \right )}+3 B \,{\mathrm e}^{i \left (d x +c \right )}-7 C \,{\mathrm e}^{i \left (d x +c \right )}+12 B -16 C \right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a d}\)	\(276\)

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)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 368 vs. \(2 (125) = 250\).
time = 0.29, size = 368, normalized size = 2.81 \begin {gather*} \frac {C {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \end {gather*}

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)),x, algorithm="maxima")

[Out]

1/6*(C*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*B*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c
) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*si
n(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]
time = 2.31, size = 170, normalized size = 1.30 \begin {gather*} \frac {9 \, {\left ({\left (B - C\right )} \cos \left (d x + c\right )^{4} + {\left (B - C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left ({\left (B - C\right )} \cos \left (d x + c\right )^{4} + {\left (B - C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (3 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (3 \, B - C\right )} \cos \left (d x + c\right ) - 2 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}

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)),x, algorithm="fricas")

[Out]

1/12*(9*((B - C)*cos(d*x + c)^4 + (B - C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 9*((B - C)*cos(d*x + c)^4 +
(B - C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(3*B - 4*C)*cos(d*x + c)^3 + (3*B - 7*C)*cos(d*x + c)^2
- (3*B - C)*cos(d*x + c) - 2*C)*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)

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

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)),x)

[Out]

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

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Giac [A]
time = 0.47, size = 182, normalized size = 1.39 \begin {gather*} \frac {\frac {9 \, {\left (B - C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {9 \, {\left (B - C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, {\left (B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \end {gather*}

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)),x, algorithm="giac")

[Out]

1/6*(9*(B - C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 9*(B - C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 6*(B*ta
n(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a + 2*(9*B*tan(1/2*d*x + 1/2*c)^5 - 15*C*tan(1/2*d*x + 1/2*c)^5 -
12*B*tan(1/2*d*x + 1/2*c)^3 + 16*C*tan(1/2*d*x + 1/2*c)^3 + 3*B*tan(1/2*d*x + 1/2*c) - 9*C*tan(1/2*d*x + 1/2*
c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d

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Mupad [B]
time = 3.26, size = 152, normalized size = 1.16 \begin {gather*} \frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-C\right )}{a\,d}-\frac {\left (3\,B-5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {16\,C}{3}-4\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B-3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \end {gather*}

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))),x)

[Out]

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

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