3.466 \(\int \frac{\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=216 \[ -\frac{2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac{(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{(11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(2 A-6 B+13 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

((2*A - 6*B + 13*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (2*(11*A - 36*B + 76*C)*Tan[c + d*x])/(15*a^3*d) + ((2*
A - 6*B + 13*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*S
ec[c + d*x])^3) - ((A - 6*B + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((11*A - 36
*B + 76*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.516282, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4084, 4019, 3787, 3767, 8, 3768, 3770} \[ -\frac{2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac{(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{(11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(2 A-6 B+13 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

((2*A - 6*B + 13*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (2*(11*A - 36*B + 76*C)*Tan[c + d*x])/(15*a^3*d) + ((2*
A - 6*B + 13*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*S
ec[c + d*x])^3) - ((A - 6*B + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((11*A - 36
*B + 76*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 4084

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

Rule 4019

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

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 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 8

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

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]

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^4(c+d x) (a (A+4 B-4 C)+a (2 A-2 B+7 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec ^3(c+d x) \left (-3 a^2 (A-6 B+11 C)+a^2 (8 A-18 B+43 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \sec ^2(c+d x) \left (-2 a^3 (11 A-36 B+76 C)+15 a^3 (2 A-6 B+13 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(2 A-6 B+13 C) \int \sec ^3(c+d x) \, dx}{a^3}-\frac{(2 (11 A-36 B+76 C)) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=\frac{(2 A-6 B+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(2 A-6 B+13 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac{(2 (11 A-36 B+76 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac{(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac{(2 A-6 B+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 6.45349, size = 1081, normalized size = 5. \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

(-8*(2*A - 6*B + 13*C)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]*(A + B*S
ec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3)
+ (8*(2*A - 6*B + 13*C)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]*(A + B*
Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3)
+ (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(490*A*Sin[(d*x)
/2] - 870*B*Sin[(d*x)/2] + 1235*C*Sin[(d*x)/2] - 530*A*Sin[(3*d*x)/2] + 1830*B*Sin[(3*d*x)/2] - 3805*C*Sin[(3*
d*x)/2] + 654*A*Sin[c - (d*x)/2] - 2094*B*Sin[c - (d*x)/2] + 4329*C*Sin[c - (d*x)/2] - 654*A*Sin[c + (d*x)/2]
+ 1314*B*Sin[c + (d*x)/2] - 1989*C*Sin[c + (d*x)/2] + 490*A*Sin[2*c + (d*x)/2] - 1650*B*Sin[2*c + (d*x)/2] + 3
575*C*Sin[2*c + (d*x)/2] + 350*A*Sin[c + (3*d*x)/2] - 450*B*Sin[c + (3*d*x)/2] + 475*C*Sin[c + (3*d*x)/2] - 53
0*A*Sin[2*c + (3*d*x)/2] + 1230*B*Sin[2*c + (3*d*x)/2] - 2005*C*Sin[2*c + (3*d*x)/2] + 350*A*Sin[3*c + (3*d*x)
/2] - 1050*B*Sin[3*c + (3*d*x)/2] + 2275*C*Sin[3*c + (3*d*x)/2] - 378*A*Sin[c + (5*d*x)/2] + 1278*B*Sin[c + (5
*d*x)/2] - 2673*C*Sin[c + (5*d*x)/2] + 150*A*Sin[2*c + (5*d*x)/2] - 90*B*Sin[2*c + (5*d*x)/2] - 105*C*Sin[2*c
+ (5*d*x)/2] - 378*A*Sin[3*c + (5*d*x)/2] + 918*B*Sin[3*c + (5*d*x)/2] - 1593*C*Sin[3*c + (5*d*x)/2] + 150*A*S
in[4*c + (5*d*x)/2] - 450*B*Sin[4*c + (5*d*x)/2] + 975*C*Sin[4*c + (5*d*x)/2] - 190*A*Sin[2*c + (7*d*x)/2] + 6
30*B*Sin[2*c + (7*d*x)/2] - 1325*C*Sin[2*c + (7*d*x)/2] + 30*A*Sin[3*c + (7*d*x)/2] + 60*B*Sin[3*c + (7*d*x)/2
] - 255*C*Sin[3*c + (7*d*x)/2] - 190*A*Sin[4*c + (7*d*x)/2] + 480*B*Sin[4*c + (7*d*x)/2] - 875*C*Sin[4*c + (7*
d*x)/2] + 30*A*Sin[5*c + (7*d*x)/2] - 90*B*Sin[5*c + (7*d*x)/2] + 195*C*Sin[5*c + (7*d*x)/2] - 44*A*Sin[3*c +
(9*d*x)/2] + 144*B*Sin[3*c + (9*d*x)/2] - 304*C*Sin[3*c + (9*d*x)/2] + 30*B*Sin[4*c + (9*d*x)/2] - 90*C*Sin[4*
c + (9*d*x)/2] - 44*A*Sin[5*c + (9*d*x)/2] + 114*B*Sin[5*c + (9*d*x)/2] - 214*C*Sin[5*c + (9*d*x)/2]))/(240*d*
(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3)

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Maple [B]  time = 0.082, size = 433, normalized size = 2. \begin{align*} -{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{2\,C}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{31\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{7\,C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{A}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) B}{d{a}^{3}}}+{\frac{13\,C}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{7\,C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{A}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) B}{d{a}^{3}}}-{\frac{13\,C}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*A+1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*B-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-1/3/d/a
^3*tan(1/2*d*x+1/2*c)^3*A+1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*B-2/3/d/a^3*C*tan(1/2*d*x+1/2*c)^3-7/4/d/a^3*A*tan(1/
2*d*x+1/2*c)+17/4/d/a^3*B*tan(1/2*d*x+1/2*c)-31/4/d/a^3*C*tan(1/2*d*x+1/2*c)+7/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)*
C-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*B+1/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*A-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*B+13/2/
d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^3*C/(tan(1/2*d*x+1/2*c)+1)^2+7/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)*C-1/d/a
^3/(tan(1/2*d*x+1/2*c)-1)*B-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*A+3/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*B-13/2/d/a^3*l
n(tan(1/2*d*x+1/2*c)-1)*C+1/2/d/a^3*C/(tan(1/2*d*x+1/2*c)-1)^2

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Maxima [B]  time = 0.988907, size = 666, normalized size = 3.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(C*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*B*(40*sin(d*x + c)/((a^3 -
a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) +
1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d
*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

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Fricas [A]  time = 0.532569, size = 849, normalized size = 3.93 \begin{align*} \frac{15 \,{\left ({\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (11 \, A - 36 \, B + 76 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (34 \, A - 114 \, B + 239 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (64 \, A - 234 \, B + 479 \, C\right )} \cos \left (d x + c\right )^{2} - 15 \,{\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sin \left (d x + c\right )}{60 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*((2*A - 6*B + 13*C)*cos(d*x + c)^5 + 3*(2*A - 6*B + 13*C)*cos(d*x + c)^4 + 3*(2*A - 6*B + 13*C)*cos(d
*x + c)^3 + (2*A - 6*B + 13*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((2*A - 6*B + 13*C)*cos(d*x + c)^5 +
3*(2*A - 6*B + 13*C)*cos(d*x + c)^4 + 3*(2*A - 6*B + 13*C)*cos(d*x + c)^3 + (2*A - 6*B + 13*C)*cos(d*x + c)^2
)*log(-sin(d*x + c) + 1) - 2*(4*(11*A - 36*B + 76*C)*cos(d*x + c)^4 + 3*(34*A - 114*B + 239*C)*cos(d*x + c)^3
+ (64*A - 234*B + 479*C)*cos(d*x + c)^2 - 15*(2*B - 3*C)*cos(d*x + c) - 15*C)*sin(d*x + c))/(a^3*d*cos(d*x + c
)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c
+ d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**6/(sec(c +
d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

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Giac [A]  time = 1.25943, size = 389, normalized size = 1.8 \begin{align*} \frac{\frac{30 \,{\left (2 \, A - 6 \, B + 13 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{30 \,{\left (2 \, A - 6 \, B + 13 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac{60 \,{\left (2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, C \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 )}^{2} a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 465 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(2*A - 6*B + 13*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(2*A - 6*B + 13*C)*log(abs(tan(1/2*d*x
+ 1/2*c) - 1))/a^3 - 60*(2*B*tan(1/2*d*x + 1/2*c)^3 - 7*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) +
5*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*
tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^12*tan(1/
2*d*x + 1/2*c)^3 + 40*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x
+ 1/2*c) + 465*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d