3.113 \(\int (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=177 \[ \frac {a^4 (10 A+7 C) \tan (c+d x)}{2 d}+\frac {a^4 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(8 A+7 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+a^4 A x+\frac {(5 A+7 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{15 d}+\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^3}{5 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

[Out]

a^4*A*x+1/2*a^4*(12*A+7*C)*arctanh(sin(d*x+c))/d+1/2*a^4*(10*A+7*C)*tan(d*x+c)/d+1/5*a*C*(a+a*sec(d*x+c))^3*ta
n(d*x+c)/d+1/5*C*(a+a*sec(d*x+c))^4*tan(d*x+c)/d+1/15*(5*A+7*C)*(a^2+a^2*sec(d*x+c))^2*tan(d*x+c)/d+1/6*(8*A+7
*C)*(a^4+a^4*sec(d*x+c))*tan(d*x+c)/d

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Rubi [A]  time = 0.29, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4055, 3917, 3914, 3767, 8, 3770} \[ \frac {a^4 (10 A+7 C) \tan (c+d x)}{2 d}+\frac {a^4 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(5 A+7 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{15 d}+\frac {(8 A+7 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+a^4 A x+\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^3}{5 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

a^4*A*x + (a^4*(12*A + 7*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a^4*(10*A + 7*C)*Tan[c + d*x])/(2*d) + (a*C*(a + a
*Sec[c + d*x])^3*Tan[c + d*x])/(5*d) + (C*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(5*d) + ((5*A + 7*C)*(a^2 + a^2
*Sec[c + d*x])^2*Tan[c + d*x])/(15*d) + ((8*A + 7*C)*(a^4 + a^4*Sec[c + d*x])*Tan[c + d*x])/(6*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 3770

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

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4055

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

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {\int (a+a \sec (c+d x))^4 (5 a A+4 a C \sec (c+d x)) \, dx}{5 a}\\ &=\frac {a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {\int (a+a \sec (c+d x))^3 \left (20 a^2 A+4 a^2 (5 A+7 C) \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac {a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac {\int (a+a \sec (c+d x))^2 \left (60 a^3 A+20 a^3 (8 A+7 C) \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac {a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac {\int (a+a \sec (c+d x)) \left (120 a^4 A+60 a^4 (10 A+7 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=a^4 A x+\frac {a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac {1}{2} \left (a^4 (10 A+7 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (a^4 (12 A+7 C)\right ) \int \sec (c+d x) \, dx\\ &=a^4 A x+\frac {a^4 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{6 d}-\frac {\left (a^4 (10 A+7 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^4 A x+\frac {a^4 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^4 (10 A+7 C) \tan (c+d x)}{2 d}+\frac {a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [B]  time = 3.16, size = 418, normalized size = 2.36 \[ \frac {a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (\sec (c) (-780 A \sin (2 c+d x)+120 A \sin (c+2 d x)+120 A \sin (3 c+2 d x)+820 A \sin (2 c+3 d x)-180 A \sin (4 c+3 d x)+60 A \sin (3 c+4 d x)+60 A \sin (5 c+4 d x)+200 A \sin (4 c+5 d x)+150 A d x \cos (2 c+d x)+75 A d x \cos (2 c+3 d x)+75 A d x \cos (4 c+3 d x)+15 A d x \cos (4 c+5 d x)+15 A d x \cos (6 c+5 d x)+1220 A \sin (d x)+150 A d x \cos (d x)-480 C \sin (2 c+d x)+330 C \sin (c+2 d x)+330 C \sin (3 c+2 d x)+800 C \sin (2 c+3 d x)-30 C \sin (4 c+3 d x)+105 C \sin (3 c+4 d x)+105 C \sin (5 c+4 d x)+166 C \sin (4 c+5 d x)+1180 C \sin (d x))-240 (12 A+7 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3840 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^8*Sec[c + d*x]^5*(-240*(12*A + 7*C)*Cos[c +
d*x]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*(150*A*d
*x*Cos[d*x] + 150*A*d*x*Cos[2*c + d*x] + 75*A*d*x*Cos[2*c + 3*d*x] + 75*A*d*x*Cos[4*c + 3*d*x] + 15*A*d*x*Cos[
4*c + 5*d*x] + 15*A*d*x*Cos[6*c + 5*d*x] + 1220*A*Sin[d*x] + 1180*C*Sin[d*x] - 780*A*Sin[2*c + d*x] - 480*C*Si
n[2*c + d*x] + 120*A*Sin[c + 2*d*x] + 330*C*Sin[c + 2*d*x] + 120*A*Sin[3*c + 2*d*x] + 330*C*Sin[3*c + 2*d*x] +
820*A*Sin[2*c + 3*d*x] + 800*C*Sin[2*c + 3*d*x] - 180*A*Sin[4*c + 3*d*x] - 30*C*Sin[4*c + 3*d*x] + 60*A*Sin[3
*c + 4*d*x] + 105*C*Sin[3*c + 4*d*x] + 60*A*Sin[5*c + 4*d*x] + 105*C*Sin[5*c + 4*d*x] + 200*A*Sin[4*c + 5*d*x]
+ 166*C*Sin[4*c + 5*d*x])))/(3840*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.45, size = 177, normalized size = 1.00 \[ \frac {60 \, A a^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left (12 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (12 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (100 \, A + 83 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, A + 34 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, C a^{4} \cos \left (d x + c\right ) + 6 \, C a^{4}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/60*(60*A*a^4*d*x*cos(d*x + c)^5 + 15*(12*A + 7*C)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(12*A + 7*C)
*a^4*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(2*(100*A + 83*C)*a^4*cos(d*x + c)^4 + 15*(4*A + 7*C)*a^4*cos(d
*x + c)^3 + 2*(5*A + 34*C)*a^4*cos(d*x + c)^2 + 30*C*a^4*cos(d*x + c) + 6*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)
^5)

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giac [A]  time = 0.57, size = 257, normalized size = 1.45 \[ \frac {30 \, {\left (d x + c\right )} A a^{4} + 15 \, {\left (12 \, A a^{4} + 7 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (12 \, A a^{4} + 7 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (150 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 680 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 490 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1180 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 896 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 920 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 790 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 270 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/30*(30*(d*x + c)*A*a^4 + 15*(12*A*a^4 + 7*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(12*A*a^4 + 7*C*a^4
)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(150*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 105*C*a^4*tan(1/2*d*x + 1/2*c)^9
- 680*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 490*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 1180*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 89
6*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 920*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 790*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 270*A*a
^4*tan(1/2*d*x + 1/2*c) + 375*C*a^4*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 = 1.64, size = 226, normalized size = 1.28 \[ A \,a^{4} x +\frac {A \,a^{4} c}{d}+\frac {83 a^{4} C \tan \left (d x +c \right )}{15 d}+\frac {6 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {7 a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {7 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {20 A \,a^{4} \tan \left (d x +c \right )}{3 d}+\frac {34 a^{4} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {2 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

A*a^4*x+1/d*A*a^4*c+83/15/d*a^4*C*tan(d*x+c)+6/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+7/2/d*a^4*C*sec(d*x+c)*tan(d*
x+c)+7/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+20/3/d*A*a^4*tan(d*x+c)+34/15/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+2/d*A
*a^4*sec(d*x+c)*tan(d*x+c)+1/d*a^4*C*tan(d*x+c)*sec(d*x+c)^3+1/3/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/5/d*a^4*C*t
an(d*x+c)*sec(d*x+c)^4

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maxima [A]  time = 0.35, size = 308, normalized size = 1.74 \[ \frac {20 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 60 \, {\left (d x + c\right )} A a^{4} + 4 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{4} + 120 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 15 \, C a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 360 \, A a^{4} \tan \left (d x + c\right ) + 60 \, C a^{4} \tan \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/60*(20*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 60*(d*x + c)*A*a^4 + 4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^
3 + 15*tan(d*x + c))*C*a^4 + 120*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 - 15*C*a^4*(2*(3*sin(d*x + c)^3 - 5*s
in(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)) - 60
*A*a^4*(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^4*(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^4*log(sec(d*x + c) +
tan(d*x + c)) + 360*A*a^4*tan(d*x + c) + 60*C*a^4*tan(d*x + c))/d

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mupad [B]  time = 2.81, size = 277, normalized size = 1.56 \[ \frac {2\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {83\,C\,a^4\,\sin \left (c+d\,x\right )}{15\,d\,\cos \left (c+d\,x\right )}+\frac {7\,C\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {34\,C\,a^4\,\sin \left (c+d\,x\right )}{15\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^4}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{5\,d\,{\cos \left (c+d\,x\right )}^5} \]

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))^4,x)

[Out]

(2*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
)))/d + (7*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (20*A*a^4*sin(c + d*x))/(3*d*cos(c + d*x))
+ (2*A*a^4*sin(c + d*x))/(d*cos(c + d*x)^2) + (A*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (83*C*a^4*sin(c + d*
x))/(15*d*cos(c + d*x)) + (7*C*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (34*C*a^4*sin(c + d*x))/(15*d*cos(c +
d*x)^3) + (C*a^4*sin(c + d*x))/(d*cos(c + d*x)^4) + (C*a^4*sin(c + d*x))/(5*d*cos(c + d*x)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

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

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