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

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

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4010, 4001, 3791, 3770, 3767, 8, 3768} \[ \frac {a^3 (15 A+13 B) \tan ^3(c+d x)}{60 d}+\frac {a^3 (15 A+13 B) \tan (c+d x)}{5 d}+\frac {a^3 (15 A+13 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 (15 A+13 B) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {(5 A-B) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.46, size = 294, normalized size = 1.80 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (15 A+13 B) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (5 A+3 B) \sin (2 c+d x)+80 (30 A+29 B) \sin (d x)+570 A \sin (c+2 d x)+570 A \sin (3 c+2 d x)+1680 A \sin (2 c+3 d x)-120 A \sin (4 c+3 d x)+225 A \sin (3 c+4 d x)+225 A \sin (5 c+4 d x)+360 A \sin (4 c+5 d x)+750 B \sin (c+2 d x)+750 B \sin (3 c+2 d x)+1520 B \sin (2 c+3 d x)+195 B \sin (3 c+4 d x)+195 B \sin (5 c+4 d x)+304 B \sin (4 c+5 d x))\right )}{15360 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15360*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^5*(240*(15*A + 13*B)*Cos[c + d*x]^5*(Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(80*(30*A + 29*B)*Sin[
d*x] - 240*(5*A + 3*B)*Sin[2*c + d*x] + 570*A*Sin[c + 2*d*x] + 750*B*Sin[c + 2*d*x] + 570*A*Sin[3*c + 2*d*x] +
750*B*Sin[3*c + 2*d*x] + 1680*A*Sin[2*c + 3*d*x] + 1520*B*Sin[2*c + 3*d*x] - 120*A*Sin[4*c + 3*d*x] + 225*A*S
in[3*c + 4*d*x] + 195*B*Sin[3*c + 4*d*x] + 225*A*Sin[5*c + 4*d*x] + 195*B*Sin[5*c + 4*d*x] + 360*A*Sin[4*c + 5
*d*x] + 304*B*Sin[4*c + 5*d*x])))/d

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fricas [A]  time = 0.47, size = 165, normalized size = 1.01 \[ \frac {15 \, {\left (15 \, A + 13 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (15 \, A + 13 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (45 \, A + 38 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (15 \, A + 13 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, A + 19 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 24 \, B a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

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+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.77, size = 246, normalized size = 1.51 \[ \frac {15 \, {\left (15 \, A a^{3} + 13 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (15 \, A a^{3} + 13 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (225 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 195 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1050 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 910 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1920 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1664 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1830 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1330 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 735 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 765 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

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+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(15*(15*A*a^3 + 13*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(15*A*a^3 + 13*B*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(225*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 195*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 1050*A*a^3*tan(1
/2*d*x + 1/2*c)^7 - 910*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 1920*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1664*B*a^3*tan(1/2*
d*x + 1/2*c)^5 - 1830*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 1330*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 735*A*a^3*tan(1/2*d*x
+ 1/2*c) + 765*B*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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maple [A]  time = 1.70, size = 234, normalized size = 1.44 \[ \frac {3 A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {13 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {13 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {15 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {38 a^{3} B \tan \left (d x +c \right )}{15 d}+\frac {19 a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d} \]

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+B*sec(d*x+c)),x)

[Out]

3/d*A*a^3*tan(d*x+c)+13/8/d*a^3*B*sec(d*x+c)*tan(d*x+c)+13/8/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+15/8/d*A*a^3*se
c(d*x+c)*tan(d*x+c)+15/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+38/15/d*a^3*B*tan(d*x+c)+19/15/d*a^3*B*tan(d*x+c)*s
ec(d*x+c)^2+1/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+3/4/d*a^3*B*tan(d*x+c)*sec(d*x+c)^3+1/4/d*A*a^3*tan(d*x+c)*sec(d
*x+c)^3+1/5/d*a^3*B*tan(d*x+c)*sec(d*x+c)^4

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maxima [B]  time = 0.37, size = 337, normalized size = 2.07 \[ \frac {240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 15 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 45 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{3} \tan \left (d x + c\right )}{240 \, d} \]

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+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.60, size = 224, normalized size = 1.37 \[ \frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (15\,A+13\,B\right )}{4\,d}-\frac {\left (\frac {15\,A\,a^3}{4}+\frac {13\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {35\,A\,a^3}{2}-\frac {91\,B\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (32\,A\,a^3+\frac {416\,B\,a^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {61\,A\,a^3}{2}-\frac {133\,B\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,A\,a^3}{4}+\frac {51\,B\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*atanh(tan(c/2 + (d*x)/2))*(15*A + 13*B))/(4*d) - (tan(c/2 + (d*x)/2)*((49*A*a^3)/4 + (51*B*a^3)/4) + tan(
c/2 + (d*x)/2)^9*((15*A*a^3)/4 + (13*B*a^3)/4) - tan(c/2 + (d*x)/2)^7*((35*A*a^3)/2 + (91*B*a^3)/6) - tan(c/2
+ (d*x)/2)^3*((61*A*a^3)/2 + (133*B*a^3)/6) + tan(c/2 + (d*x)/2)^5*(32*A*a^3 + (416*B*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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

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+B*sec(d*x+c)),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(B*sec(c + d*x)**3, x) + Integral(3*B*sec(c + d*x)**4, x) + Integral(3
*B*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**6, x))

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