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

Optimal. Leaf size=210 \[ \frac {a^3 (19 A+17 B) \tan ^3(c+d x)}{15 d}+\frac {a^3 (19 A+17 B) \tan (c+d x)}{5 d}+\frac {a^3 (26 A+23 B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac {(3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {a^3 (26 A+23 B) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

[Out]

1/16*a^3*(26*A+23*B)*arctanh(sin(d*x+c))/d+1/5*a^3*(19*A+17*B)*tan(d*x+c)/d+1/16*a^3*(26*A+23*B)*sec(d*x+c)*ta
n(d*x+c)/d+1/40*a^3*(22*A+21*B)*sec(d*x+c)^3*tan(d*x+c)/d+1/6*a*B*sec(d*x+c)^3*(a+a*sec(d*x+c))^2*tan(d*x+c)/d
+1/15*(3*A+4*B)*sec(d*x+c)^3*(a^3+a^3*sec(d*x+c))*tan(d*x+c)/d+1/15*a^3*(19*A+17*B)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.40, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4018, 3997, 3787, 3768, 3770, 3767} \[ \frac {a^3 (19 A+17 B) \tan ^3(c+d x)}{15 d}+\frac {a^3 (19 A+17 B) \tan (c+d x)}{5 d}+\frac {a^3 (26 A+23 B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac {(3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {a^3 (26 A+23 B) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(26*A + 23*B)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^3*(19*A + 17*B)*Tan[c + d*x])/(5*d) + (a^3*(26*A + 23*B)
*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^3*(22*A + 21*B)*Sec[c + d*x]^3*Tan[c + d*x])/(40*d) + (a*B*Sec[c + d*x
]^3*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(6*d) + ((3*A + 4*B)*Sec[c + d*x]^3*(a^3 + a^3*Sec[c + d*x])*Tan[c +
d*x])/(15*d) + (a^3*(19*A + 17*B)*Tan[c + d*x]^3)/(15*d)

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 3787

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 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 4018

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

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {a B \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{6} \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (3 a (2 A+B)+2 a (3 A+4 B) \sec (c+d x)) \, dx\\ &=\frac {a B \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 A+4 B) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {1}{30} \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (3 a^2 (16 A+13 B)+3 a^2 (22 A+21 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^3 (22 A+21 B) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a B \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 A+4 B) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {1}{120} \int \sec ^3(c+d x) \left (15 a^3 (26 A+23 B)+24 a^3 (19 A+17 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^3 (22 A+21 B) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a B \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 A+4 B) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {1}{5} \left (a^3 (19 A+17 B)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{8} \left (a^3 (26 A+23 B)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {a^3 (26 A+23 B) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^3 (22 A+21 B) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a B \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 A+4 B) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {1}{16} \left (a^3 (26 A+23 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (19 A+17 B)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {a^3 (26 A+23 B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 (19 A+17 B) \tan (c+d x)}{5 d}+\frac {a^3 (26 A+23 B) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^3 (22 A+21 B) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a B \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 A+4 B) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {a^3 (19 A+17 B) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 1.89, size = 346, normalized size = 1.65 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (480 (26 A+23 B) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-320 (19 A+17 B) \sin (c)+750 (2 A+3 B) \sin (d x)+1500 A \sin (2 c+d x)+7680 A \sin (c+2 d x)-1440 A \sin (3 c+2 d x)+1890 A \sin (2 c+3 d x)+1890 A \sin (4 c+3 d x)+3648 A \sin (3 c+4 d x)+390 A \sin (4 c+5 d x)+390 A \sin (6 c+5 d x)+608 A \sin (5 c+6 d x)+2250 B \sin (2 c+d x)+7680 B \sin (c+2 d x)-480 B \sin (3 c+2 d x)+1955 B \sin (2 c+3 d x)+1955 B \sin (4 c+3 d x)+3264 B \sin (3 c+4 d x)+345 B \sin (4 c+5 d x)+345 B \sin (6 c+5 d x)+544 B \sin (5 c+6 d x))\right )}{61440 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/61440*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^6*(480*(26*A + 23*B)*Cos[c + d*x]^6*(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]*(-320*(19*A + 17*B)*Si
n[c] + 750*(2*A + 3*B)*Sin[d*x] + 1500*A*Sin[2*c + d*x] + 2250*B*Sin[2*c + d*x] + 7680*A*Sin[c + 2*d*x] + 7680
*B*Sin[c + 2*d*x] - 1440*A*Sin[3*c + 2*d*x] - 480*B*Sin[3*c + 2*d*x] + 1890*A*Sin[2*c + 3*d*x] + 1955*B*Sin[2*
c + 3*d*x] + 1890*A*Sin[4*c + 3*d*x] + 1955*B*Sin[4*c + 3*d*x] + 3648*A*Sin[3*c + 4*d*x] + 3264*B*Sin[3*c + 4*
d*x] + 390*A*Sin[4*c + 5*d*x] + 345*B*Sin[4*c + 5*d*x] + 390*A*Sin[6*c + 5*d*x] + 345*B*Sin[6*c + 5*d*x] + 608
*A*Sin[5*c + 6*d*x] + 544*B*Sin[5*c + 6*d*x])))/d

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fricas [A]  time = 0.45, size = 185, normalized size = 0.88 \[ \frac {15 \, {\left (26 \, A + 23 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (26 \, A + 23 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, {\left (19 \, A + 17 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} + 15 \, {\left (26 \, A + 23 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 16 \, {\left (19 \, A + 17 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 10 \, {\left (18 \, A + 23 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 48 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 40 \, B a^{3}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*(26*A + 23*B)*a^3*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(26*A + 23*B)*a^3*cos(d*x + c)^6*log(-si
n(d*x + c) + 1) + 2*(32*(19*A + 17*B)*a^3*cos(d*x + c)^5 + 15*(26*A + 23*B)*a^3*cos(d*x + c)^4 + 16*(19*A + 17
*B)*a^3*cos(d*x + c)^3 + 10*(18*A + 23*B)*a^3*cos(d*x + c)^2 + 48*(A + 3*B)*a^3*cos(d*x + c) + 40*B*a^3)*sin(d
*x + c))/(d*cos(d*x + c)^6)

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giac [A]  time = 1.45, size = 280, normalized size = 1.33 \[ \frac {15 \, {\left (26 \, A a^{3} + 23 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (26 \, A a^{3} + 23 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (390 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 345 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 2210 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1955 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 5148 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4554 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5988 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5814 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4190 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3165 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1530 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(26*A*a^3 + 23*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(26*A*a^3 + 23*B*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(390*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 345*B*a^3*tan(1/2*d*x + 1/2*c)^11 - 2210*A*a^3*tan
(1/2*d*x + 1/2*c)^9 - 1955*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 5148*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 4554*B*a^3*tan(1
/2*d*x + 1/2*c)^7 - 5988*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 5814*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 4190*A*a^3*tan(1/2
*d*x + 1/2*c)^3 + 3165*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 1530*A*a^3*tan(1/2*d*x + 1/2*c) - 1575*B*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 = 1.63, size = 281, normalized size = 1.34 \[ \frac {13 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {13 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {34 a^{3} B \tan \left (d x +c \right )}{15 d}+\frac {17 a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {38 A \,a^{3} \tan \left (d x +c \right )}{15 d}+\frac {19 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {23 a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {23 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {23 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {3 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{3} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/8/d*A*a^3*sec(d*x+c)*tan(d*x+c)+13/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+34/15/d*a^3*B*tan(d*x+c)+17/15/d*a^3
*B*tan(d*x+c)*sec(d*x+c)^2+38/15/d*A*a^3*tan(d*x+c)+19/15/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+23/24/d*a^3*B*tan(d*
x+c)*sec(d*x+c)^3+23/16/d*a^3*B*sec(d*x+c)*tan(d*x+c)+23/16/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+3/4/d*A*a^3*tan(
d*x+c)*sec(d*x+c)^3+3/5/d*a^3*B*tan(d*x+c)*sec(d*x+c)^4+1/5/d*A*a^3*tan(d*x+c)*sec(d*x+c)^4+1/6/d*a^3*B*tan(d*
x+c)*sec(d*x+c)^5

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maxima [B]  time = 0.36, size = 405, normalized size = 1.93 \[ \frac {32 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 96 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{3} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 5 \, B a^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 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))*B*a^3 + 160*(tan(d*x + c)^3 + 3*tan(d*x
+ c))*B*a^3 - 5*B*a^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*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)) - 90*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*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(s
in(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 120*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)))/d

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mupad [B]  time = 4.63, size = 262, normalized size = 1.25 \[ \frac {\left (-\frac {13\,A\,a^3}{4}-\frac {23\,B\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {221\,A\,a^3}{12}+\frac {391\,B\,a^3}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {429\,A\,a^3}{10}-\frac {759\,B\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {499\,A\,a^3}{10}+\frac {969\,B\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {419\,A\,a^3}{12}-\frac {211\,B\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+\frac {105\,B\,a^3}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (26\,A+23\,B\right )}{8\,d} \]

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)^3,x)

[Out]

(tan(c/2 + (d*x)/2)*((51*A*a^3)/4 + (105*B*a^3)/8) - tan(c/2 + (d*x)/2)^11*((13*A*a^3)/4 + (23*B*a^3)/8) - tan
(c/2 + (d*x)/2)^3*((419*A*a^3)/12 + (211*B*a^3)/8) + tan(c/2 + (d*x)/2)^9*((221*A*a^3)/12 + (391*B*a^3)/24) -
tan(c/2 + (d*x)/2)^7*((429*A*a^3)/10 + (759*B*a^3)/20) + tan(c/2 + (d*x)/2)^5*((499*A*a^3)/10 + (969*B*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))*(26*A + 23*B))/(8*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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