3.102 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=197 \[ \frac {a^3 (30 A+23 C) \tan ^3(c+d x)}{120 d}+\frac {a^3 (30 A+23 C) \tan (c+d x)}{10 d}+\frac {a^3 (30 A+23 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {3 a^3 (30 A+23 C) \tan (c+d x) \sec (c+d x)}{80 d}+\frac {(30 A+7 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{120 d}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{10 a d} \]

[Out]

1/16*a^3*(30*A+23*C)*arctanh(sin(d*x+c))/d+1/10*a^3*(30*A+23*C)*tan(d*x+c)/d+3/80*a^3*(30*A+23*C)*sec(d*x+c)*t
an(d*x+c)/d+1/120*(30*A+7*C)*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+1/6*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^3*tan(d*x+c)/
d+1/10*C*(a+a*sec(d*x+c))^4*tan(d*x+c)/a/d+1/120*a^3*(30*A+23*C)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.42, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4089, 4010, 4001, 3791, 3770, 3767, 8, 3768} \[ \frac {a^3 (30 A+23 C) \tan ^3(c+d x)}{120 d}+\frac {a^3 (30 A+23 C) \tan (c+d x)}{10 d}+\frac {a^3 (30 A+23 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {3 a^3 (30 A+23 C) \tan (c+d x) \sec (c+d x)}{80 d}+\frac {(30 A+7 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{120 d}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{10 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(30*A + 23*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^3*(30*A + 23*C)*Tan[c + d*x])/(10*d) + (3*a^3*(30*A + 23
*C)*Sec[c + d*x]*Tan[c + d*x])/(80*d) + ((30*A + 7*C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(120*d) + (C*Sec[c
+ d*x]^2*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(6*d) + (C*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(10*a*d) + (a^3*
(30*A + 23*C)*Tan[c + d*x]^3)/(120*d)

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 3768

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 3770

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

Rule 3791

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 4001

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 4010

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)), I
nt[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] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4089

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

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac {\int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (2 a (3 A+C)+3 a C \sec (c+d x)) \, dx}{6 a}\\ &=\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac {\int \sec (c+d x) (a+a \sec (c+d x))^3 \left (12 a^2 C+a^2 (30 A+7 C) \sec (c+d x)\right ) \, dx}{30 a^2}\\ &=\frac {(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac {1}{40} (30 A+23 C) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac {(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac {1}{40} (30 A+23 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 {(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac {1}{40} \left (a^3 (30 A+23 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{40} \left (a^3 (30 A+23 C)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{40} \left (3 a^3 (30 A+23 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{40} \left (3 a^3 (30 A+23 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {a^3 (30 A+23 C) \tanh ^{-1}(\sin (c+d x))}{40 d}+\frac {3 a^3 (30 A+23 C) \sec (c+d x) \tan (c+d x)}{80 d}+\frac {(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac {1}{80} \left (3 a^3 (30 A+23 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (30 A+23 C)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{40 d}-\frac {\left (3 a^3 (30 A+23 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{40 d}\\ &=\frac {a^3 (30 A+23 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 (30 A+23 C) \tan (c+d x)}{10 d}+\frac {3 a^3 (30 A+23 C) \sec (c+d x) \tan (c+d x)}{80 d}+\frac {(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac {a^3 (30 A+23 C) \tan ^3(c+d x)}{120 d}\\ \end {align*}

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Mathematica [A]  time = 3.24, size = 387, normalized size = 1.96 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (480 (30 A+23 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-160 (45 A+34 C) \sin (c)+1140 A \sin (2 c+d x)+8160 A \sin (c+2 d x)-2640 A \sin (3 c+2 d x)+1590 A \sin (2 c+3 d x)+1590 A \sin (4 c+3 d x)+4080 A \sin (3 c+4 d x)-240 A \sin (5 c+4 d x)+450 A \sin (4 c+5 d x)+450 A \sin (6 c+5 d x)+720 A \sin (5 c+6 d x)+30 (38 A+75 C) \sin (d x)+2250 C \sin (2 c+d x)+7680 C \sin (c+2 d x)-480 C \sin (3 c+2 d x)+1955 C \sin (2 c+3 d x)+1955 C \sin (4 c+3 d x)+3264 C \sin (3 c+4 d x)+345 C \sin (4 c+5 d x)+345 C \sin (6 c+5 d x)+544 C \sin (5 c+6 d x))\right )}{30720 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30720*(a^3*(1 + Cos[c + d*x])^3*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^6*Sec[c + d*x]^6*(480*(30*A + 23*C)
*Cos[c + d*x]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]
*(-160*(45*A + 34*C)*Sin[c] + 30*(38*A + 75*C)*Sin[d*x] + 1140*A*Sin[2*c + d*x] + 2250*C*Sin[2*c + d*x] + 8160
*A*Sin[c + 2*d*x] + 7680*C*Sin[c + 2*d*x] - 2640*A*Sin[3*c + 2*d*x] - 480*C*Sin[3*c + 2*d*x] + 1590*A*Sin[2*c
+ 3*d*x] + 1955*C*Sin[2*c + 3*d*x] + 1590*A*Sin[4*c + 3*d*x] + 1955*C*Sin[4*c + 3*d*x] + 4080*A*Sin[3*c + 4*d*
x] + 3264*C*Sin[3*c + 4*d*x] - 240*A*Sin[5*c + 4*d*x] + 450*A*Sin[4*c + 5*d*x] + 345*C*Sin[4*c + 5*d*x] + 450*
A*Sin[6*c + 5*d*x] + 345*C*Sin[6*c + 5*d*x] + 720*A*Sin[5*c + 6*d*x] + 544*C*Sin[5*c + 6*d*x])))/(d*(A + 2*C +
A*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.46, size = 181, normalized size = 0.92 \[ \frac {15 \, {\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (45 \, A + 34 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} + 15 \, {\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 16 \, {\left (15 \, A + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 10 \, {\left (6 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 144 \, C a^{3} \cos \left (d x + c\right ) + 40 \, C a^{3}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*(30*A + 23*C)*a^3*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(30*A + 23*C)*a^3*cos(d*x + c)^6*log(-si
n(d*x + c) + 1) + 2*(16*(45*A + 34*C)*a^3*cos(d*x + c)^5 + 15*(30*A + 23*C)*a^3*cos(d*x + c)^4 + 16*(15*A + 17
*C)*a^3*cos(d*x + c)^3 + 10*(6*A + 23*C)*a^3*cos(d*x + c)^2 + 144*C*a^3*cos(d*x + c) + 40*C*a^3)*sin(d*x + c))
/(d*cos(d*x + c)^6)

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giac [A]  time = 0.38, size = 280, normalized size = 1.42 \[ \frac {15 \, {\left (30 \, A a^{3} + 23 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (30 \, A a^{3} + 23 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (450 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 345 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 2550 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1955 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 5940 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4554 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 7500 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5814 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5130 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1470 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(30*A*a^3 + 23*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(30*A*a^3 + 23*C*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(450*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 345*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 2550*A*a^3*tan
(1/2*d*x + 1/2*c)^9 - 1955*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 5940*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 4554*C*a^3*tan(1
/2*d*x + 1/2*c)^7 - 7500*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 5814*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 5130*A*a^3*tan(1/2
*d*x + 1/2*c)^3 + 3165*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1470*A*a^3*tan(1/2*d*x + 1/2*c) - 1575*C*a^3*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d

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maple [A]  time = 1.94, size = 257, normalized size = 1.30 \[ \frac {3 A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {34 a^{3} C \tan \left (d x +c \right )}{15 d}+\frac {17 C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {15 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {23 C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {23 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {23 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/d*A*a^3*tan(d*x+c)+34/15*a^3*C*tan(d*x+c)/d+17/15/d*C*a^3*tan(d*x+c)*sec(d*x+c)^2+15/8/d*A*a^3*sec(d*x+c)*ta
n(d*x+c)+15/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+23/24/d*C*a^3*tan(d*x+c)*sec(d*x+c)^3+23/16/d*C*a^3*sec(d*x+c)
*tan(d*x+c)+23/16/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+3/5/d*C*a^3*tan(d*x+c)*s
ec(d*x+c)^4+1/4/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+1/6/d*C*a^3*tan(d*x+c)*sec(d*x+c)^5

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maxima [B]  time = 0.36, size = 382, normalized size = 1.94 \[ \frac {480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 96 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{3} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 5 \, C a^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 480 \, A a^{3} \tan \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(480*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 96*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*C*a^3 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 5*C*a^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 3
3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log
(sin(d*x + c) - 1)) - 30*A*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)) - 90*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)) - 360*A*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)) + 480*A*a^3*tan(d*x + c))/d

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mupad [B]  time = 5.28, size = 262, normalized size = 1.33 \[ \frac {\left (-\frac {15\,A\,a^3}{4}-\frac {23\,C\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {85\,A\,a^3}{4}+\frac {391\,C\,a^3}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {99\,A\,a^3}{2}-\frac {759\,C\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {125\,A\,a^3}{2}+\frac {969\,C\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {171\,A\,a^3}{4}-\frac {211\,C\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,A\,a^3}{4}+\frac {105\,C\,a^3}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (30\,A+23\,C\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((49*A*a^3)/4 + (105*C*a^3)/8) - tan(c/2 + (d*x)/2)^11*((15*A*a^3)/4 + (23*C*a^3)/8) - tan
(c/2 + (d*x)/2)^3*((171*A*a^3)/4 + (211*C*a^3)/8) + tan(c/2 + (d*x)/2)^9*((85*A*a^3)/4 + (391*C*a^3)/24) - tan
(c/2 + (d*x)/2)^7*((99*A*a^3)/2 + (759*C*a^3)/20) + tan(c/2 + (d*x)/2)^5*((125*A*a^3)/2 + (969*C*a^3)/20))/(d*
(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(
c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (a^3*atanh(tan(c/2 + (d*x)/2))*(30*A + 23*C))/(8*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{5}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{7}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(A*sec(c + d*x)**2, x) + Integral(3*A*sec(c + d*x)**3, x) + Integral(3*A*sec(c + d*x)**4, x) + I
ntegral(A*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**4, x) + Integral(3*C*sec(c + d*x)**5, x) + Integral(3
*C*sec(c + d*x)**6, x) + Integral(C*sec(c + d*x)**7, x))

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