3.4.23 \(\int \sec (c+d x) (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [323]

Optimal. Leaf size=163 \[ \frac {a^3 (15 B+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (15 B+13 C) \tan (c+d x)}{5 d}+\frac {3 a^3 (15 B+13 C) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {a^3 (15 B+13 C) \tan ^3(c+d x)}{60 d} \]

[Out]

1/8*a^3*(15*B+13*C)*arctanh(sin(d*x+c))/d+1/5*a^3*(15*B+13*C)*tan(d*x+c)/d+3/40*a^3*(15*B+13*C)*sec(d*x+c)*tan
(d*x+c)/d+1/20*(5*B-C)*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+1/5*C*(a+a*sec(d*x+c))^4*tan(d*x+c)/a/d+1/60*a^3*(15*B+
13*C)*tan(d*x+c)^3/d

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Rubi [A]
time = 0.22, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {4157, 4095, 4086, 3876, 3855, 3852, 8, 3853} \begin {gather*} \frac {a^3 (15 B+13 C) \tan ^3(c+d x)}{60 d}+\frac {a^3 (15 B+13 C) \tan (c+d x)}{5 d}+\frac {a^3 (15 B+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 (15 B+13 C) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {(5 B-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(15*B + 13*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(15*B + 13*C)*Tan[c + d*x])/(5*d) + (3*a^3*(15*B + 13*C
)*Sec[c + d*x]*Tan[c + d*x])/(40*d) + ((5*B - C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(20*d) + (C*(a + a*Sec[c
+ d*x])^4*Tan[c + d*x])/(5*a*d) + (a^3*(15*B + 13*C)*Tan[c + d*x]^3)/(60*d)

Rule 8

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

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 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 4086

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

Rule 4095

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

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 \sec (c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {\int \sec (c+d x) (a+a \sec (c+d x))^3 (4 a C+a (5 B-C) \sec (c+d x)) \, dx}{5 a}\\ &=\frac {(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} (15 B+13 C) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac {(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} (15 B+13 C) \int \left (a^3 \sec (c+d x)+3 a^3 \sec ^2(c+d x)+3 a^3 \sec ^3(c+d x)+a^3 \sec ^4(c+d x)\right ) \, dx\\ &=\frac {(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} \left (a^3 (15 B+13 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{20} \left (a^3 (15 B+13 C)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 B+13 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 B+13 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {a^3 (15 B+13 C) \tanh ^{-1}(\sin (c+d x))}{20 d}+\frac {3 a^3 (15 B+13 C) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{40} \left (3 a^3 (15 B+13 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (15 B+13 C)\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{20 d}-\frac {\left (3 a^3 (15 B+13 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{20 d}\\ &=\frac {a^3 (15 B+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (15 B+13 C) \tan (c+d x)}{5 d}+\frac {3 a^3 (15 B+13 C) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {a^3 (15 B+13 C) \tan ^3(c+d x)}{60 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(391\) vs. \(2(163)=326\).
time = 0.85, size = 391, normalized size = 2.40 \begin {gather*} -\frac {a^3 \sec ^5(c+d x) \left (225 B \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+195 C \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+150 (15 B+13 C) \cos (c+d x) \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 )+75 (15 B+13 C) \cos (3 (c+d x)) \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 )-225 B \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-195 C \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-1200 B \sin (c+d x)-1600 C \sin (c+d x)-1140 B \sin (2 (c+d x))-1500 C \sin (2 (c+d x))-1560 B \sin (3 (c+d x))-1520 C \sin (3 (c+d x))-450 B \sin (4 (c+d x))-390 C \sin (4 (c+d x))-360 B \sin (5 (c+d x))-304 C \sin (5 (c+d x))\right )}{1920 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1920*(a^3*Sec[c + d*x]^5*(225*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 195*C*Cos[5*(c
+ d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 150*(15*B + 13*C)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 75*(15*B + 13*C)*Cos[3*(c + d*x)]*(Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 225*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]] - 195*C*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 1200*B*Sin[c + d
*x] - 1600*C*Sin[c + d*x] - 1140*B*Sin[2*(c + d*x)] - 1500*C*Sin[2*(c + d*x)] - 1560*B*Sin[3*(c + d*x)] - 1520
*C*Sin[3*(c + d*x)] - 450*B*Sin[4*(c + d*x)] - 390*C*Sin[4*(c + d*x)] - 360*B*Sin[5*(c + d*x)] - 304*C*Sin[5*(
c + d*x)]))/d

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Maple [A]
time = 1.39, size = 271, normalized size = 1.66

method	result	size
norman	\(\frac {-\frac {32 a^{3} \left (15 B +13 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {7 a^{3} \left (15 B +13 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {a^{3} \left (15 B +13 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a^{3} \left (49 B +51 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (183 B +133 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {a^{3} \left (15 B +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{3} \left (15 B +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(201\)
derivativedivides	\(\frac {a^{3} B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{3} C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-3 a^{3} B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a^{3} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} B \tan \left (d x +c \right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(271\)
default	\(\frac {a^{3} B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{3} C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-3 a^{3} B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a^{3} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} B \tan \left (d x +c \right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(271\)
risch	\(-\frac {i a^{3} \left (225 B \,{\mathrm e}^{9 i \left (d x +c \right )}+195 C \,{\mathrm e}^{9 i \left (d x +c \right )}-120 B \,{\mathrm e}^{8 i \left (d x +c \right )}+570 B \,{\mathrm e}^{7 i \left (d x +c \right )}+750 C \,{\mathrm e}^{7 i \left (d x +c \right )}-1200 B \,{\mathrm e}^{6 i \left (d x +c \right )}-720 C \,{\mathrm e}^{6 i \left (d x +c \right )}-2400 B \,{\mathrm e}^{4 i \left (d x +c \right )}-2320 C \,{\mathrm e}^{4 i \left (d x +c \right )}-570 B \,{\mathrm e}^{3 i \left (d x +c \right )}-750 C \,{\mathrm e}^{3 i \left (d x +c \right )}-1680 B \,{\mathrm e}^{2 i \left (d x +c \right )}-1520 C \,{\mathrm e}^{2 i \left (d x +c \right )}-225 B \,{\mathrm e}^{i \left (d x +c \right )}-195 C \,{\mathrm e}^{i \left (d x +c \right )}-360 B -304 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}-\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}+\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(299\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*B*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-a^3*C*(-8/15-1/5*sec
(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)-3*a^3*B*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+3*a^3*C*(-(-1/4*sec(d*x+c)^
3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+3*a^3*B*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+
c)+tan(d*x+c)))-3*a^3*C*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a^3*B*tan(d*x+c)+a^3*C*(1/2*sec(d*x+c)*tan(d*x+c)+1
/2*ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (151) = 302\).
time = 0.27, size = 337, normalized size = 2.07 \begin {gather*} \frac {240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 15 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 45 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B a^{3} \tan \left (d x + c\right )}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*C*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 15*B*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(
d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 45*C*a^3*(2*(3*sin(d
*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x
+ c) - 1)) - 180*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) -
60*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*B*a^3*ta
n(d*x + c))/d

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Fricas [A]
time = 3.12, size = 165, normalized size = 1.01 \begin {gather*} \frac {15 \, {\left (15 \, B + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (15 \, B + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (45 \, B + 38 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (15 \, B + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, B + 19 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 24 \, C a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(15*B + 13*C)*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(15*B + 13*C)*a^3*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 2*(8*(45*B + 38*C)*a^3*cos(d*x + c)^4 + 15*(15*B + 13*C)*a^3*cos(d*x + c)^3 + 8*(15*B + 19*C
)*a^3*cos(d*x + c)^2 + 30*(B + 3*C)*a^3*cos(d*x + c) + 24*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int B \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(B*sec(c + d*x)**2, x) + Integral(3*B*sec(c + d*x)**3, x) + Integral(3*B*sec(c + d*x)**4, x) + I
ntegral(B*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**3, x) + Integral(3*C*sec(c + d*x)**4, x) + Integral(3
*C*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))

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Giac [A]
time = 0.52, size = 246, normalized size = 1.51 \begin {gather*} \frac {15 \, {\left (15 \, B a^{3} + 13 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (15 \, B a^{3} + 13 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (225 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 195 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1050 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 910 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1920 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1664 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1830 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1330 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 735 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 765 \, C a^{3} \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 )}^{5}}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(15*B*a^3 + 13*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(15*B*a^3 + 13*C*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(225*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 195*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 1050*B*a^3*tan(1
/2*d*x + 1/2*c)^7 - 910*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 1920*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 1664*C*a^3*tan(1/2*
d*x + 1/2*c)^5 - 1830*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 1330*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 735*B*a^3*tan(1/2*d*x
+ 1/2*c) + 765*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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Mupad [B]
time = 5.52, size = 224, normalized size = 1.37 \begin {gather*} \frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (15\,B+13\,C\right )}{4\,d}-\frac {\left (\frac {15\,B\,a^3}{4}+\frac {13\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {35\,B\,a^3}{2}-\frac {91\,C\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (32\,B\,a^3+\frac {416\,C\,a^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {61\,B\,a^3}{2}-\frac {133\,C\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,B\,a^3}{4}+\frac {51\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \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)*(a + a/cos(c + d*x))^3)/cos(c + d*x),x)

[Out]

(a^3*atanh(tan(c/2 + (d*x)/2))*(15*B + 13*C))/(4*d) - (tan(c/2 + (d*x)/2)*((49*B*a^3)/4 + (51*C*a^3)/4) + tan(
c/2 + (d*x)/2)^9*((15*B*a^3)/4 + (13*C*a^3)/4) - tan(c/2 + (d*x)/2)^7*((35*B*a^3)/2 + (91*C*a^3)/6) - tan(c/2
+ (d*x)/2)^3*((61*B*a^3)/2 + (133*C*a^3)/6) + tan(c/2 + (d*x)/2)^5*(32*B*a^3 + (416*C*a^3)/15))/(d*(5*tan(c/2
+ (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)
^10 - 1))

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