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

Optimal. Leaf size=169 \[ -\frac{2 (2 A-5 B+8 C) \tan (c+d x)}{3 a^2 d}+\frac{(2 A-4 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{(2 A-4 B+7 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

((2*A - 4*B + 7*C)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) - (2*(2*A - 5*B + 8*C)*Tan[c + d*x])/(3*a^2*d) + ((2*A - 4
*B + 7*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) - ((2*A - 5*B + 8*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 +
Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.337234, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{2 (2 A-5 B+8 C) \tan (c+d x)}{3 a^2 d}+\frac{(2 A-4 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{(2 A-4 B+7 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A - 4*B + 7*C)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) - (2*(2*A - 5*B + 8*C)*Tan[c + d*x])/(3*a^2*d) + ((2*A - 4
*B + 7*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*d) - ((2*A - 5*B + 8*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 +
Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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

Mathematica [B]  time = 6.32553, size = 901, normalized size = 5.33 \[ -\frac{4 (2 A-4 B+7 C) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{4 (2 A-4 B+7 C) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{\sec \left (\frac{c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (20 A \sin \left (\frac{d x}{2}\right )-14 B \sin \left (\frac{d x}{2}\right )+14 C \sin \left (\frac{d x}{2}\right )-22 A \sin \left (\frac{3 d x}{2}\right )+64 B \sin \left (\frac{3 d x}{2}\right )-97 C \sin \left (\frac{3 d x}{2}\right )+36 A \sin \left (c-\frac{d x}{2}\right )-84 B \sin \left (c-\frac{d x}{2}\right )+126 C \sin \left (c-\frac{d x}{2}\right )-36 A \sin \left (c+\frac{d x}{2}\right )+42 B \sin \left (c+\frac{d x}{2}\right )-42 C \sin \left (c+\frac{d x}{2}\right )+20 A \sin \left (2 c+\frac{d x}{2}\right )-56 B \sin \left (2 c+\frac{d x}{2}\right )+98 C \sin \left (2 c+\frac{d x}{2}\right )+18 A \sin \left (c+\frac{3 d x}{2}\right )-6 B \sin \left (c+\frac{3 d x}{2}\right )+3 C \sin \left (c+\frac{3 d x}{2}\right )-22 A \sin \left (2 c+\frac{3 d x}{2}\right )+34 B \sin \left (2 c+\frac{3 d x}{2}\right )-37 C \sin \left (2 c+\frac{3 d x}{2}\right )+18 A \sin \left (3 c+\frac{3 d x}{2}\right )-36 B \sin \left (3 c+\frac{3 d x}{2}\right )+63 C \sin \left (3 c+\frac{3 d x}{2}\right )-18 A \sin \left (c+\frac{5 d x}{2}\right )+48 B \sin \left (c+\frac{5 d x}{2}\right )-75 C \sin \left (c+\frac{5 d x}{2}\right )+6 A \sin \left (2 c+\frac{5 d x}{2}\right )+6 B \sin \left (2 c+\frac{5 d x}{2}\right )-15 C \sin \left (2 c+\frac{5 d x}{2}\right )-18 A \sin \left (3 c+\frac{5 d x}{2}\right )+30 B \sin \left (3 c+\frac{5 d x}{2}\right )-39 C \sin \left (3 c+\frac{5 d x}{2}\right )+6 A \sin \left (4 c+\frac{5 d x}{2}\right )-12 B \sin \left (4 c+\frac{5 d x}{2}\right )+21 C \sin \left (4 c+\frac{5 d x}{2}\right )-8 A \sin \left (2 c+\frac{7 d x}{2}\right )+20 B \sin \left (2 c+\frac{7 d x}{2}\right )-32 C \sin \left (2 c+\frac{7 d x}{2}\right )+6 B \sin \left (3 c+\frac{7 d x}{2}\right )-12 C \sin \left (3 c+\frac{7 d x}{2}\right )-8 A \sin \left (4 c+\frac{7 d x}{2}\right )+14 B \sin \left (4 c+\frac{7 d x}{2}\right )-20 C \sin \left (4 c+\frac{7 d x}{2}\right )\right ) \cos \left (\frac{c}{2}+\frac{d x}{2}\right )}{24 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(2*A - 4*B + 7*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(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])^2) + (4*(2*A - 4*
B + 7*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(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])^2) + (Cos[c/2 + (d*x)/2]*Sec[c
/2]*Sec[c]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(20*A*Sin[(d*x)/2] - 14*B*Sin[(d*x)/2] + 14*
C*Sin[(d*x)/2] - 22*A*Sin[(3*d*x)/2] + 64*B*Sin[(3*d*x)/2] - 97*C*Sin[(3*d*x)/2] + 36*A*Sin[c - (d*x)/2] - 84*
B*Sin[c - (d*x)/2] + 126*C*Sin[c - (d*x)/2] - 36*A*Sin[c + (d*x)/2] + 42*B*Sin[c + (d*x)/2] - 42*C*Sin[c + (d*
x)/2] + 20*A*Sin[2*c + (d*x)/2] - 56*B*Sin[2*c + (d*x)/2] + 98*C*Sin[2*c + (d*x)/2] + 18*A*Sin[c + (3*d*x)/2]
- 6*B*Sin[c + (3*d*x)/2] + 3*C*Sin[c + (3*d*x)/2] - 22*A*Sin[2*c + (3*d*x)/2] + 34*B*Sin[2*c + (3*d*x)/2] - 37
*C*Sin[2*c + (3*d*x)/2] + 18*A*Sin[3*c + (3*d*x)/2] - 36*B*Sin[3*c + (3*d*x)/2] + 63*C*Sin[3*c + (3*d*x)/2] -
18*A*Sin[c + (5*d*x)/2] + 48*B*Sin[c + (5*d*x)/2] - 75*C*Sin[c + (5*d*x)/2] + 6*A*Sin[2*c + (5*d*x)/2] + 6*B*S
in[2*c + (5*d*x)/2] - 15*C*Sin[2*c + (5*d*x)/2] - 18*A*Sin[3*c + (5*d*x)/2] + 30*B*Sin[3*c + (5*d*x)/2] - 39*C
*Sin[3*c + (5*d*x)/2] + 6*A*Sin[4*c + (5*d*x)/2] - 12*B*Sin[4*c + (5*d*x)/2] + 21*C*Sin[4*c + (5*d*x)/2] - 8*A
*Sin[2*c + (7*d*x)/2] + 20*B*Sin[2*c + (7*d*x)/2] - 32*C*Sin[2*c + (7*d*x)/2] + 6*B*Sin[3*c + (7*d*x)/2] - 12*
C*Sin[3*c + (7*d*x)/2] - 8*A*Sin[4*c + (7*d*x)/2] + 14*B*Sin[4*c + (7*d*x)/2] - 20*C*Sin[4*c + (7*d*x)/2]))/(2
4*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2)

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Maple [B]  time = 0.074, size = 373, normalized size = 2.2 \begin{align*} -{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{B}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{5\,C}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{A}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) B}{d{a}^{2}}}+{\frac{7\,C}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{B}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{5\,C}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{A}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) B}{d{a}^{2}}}-{\frac{7\,C}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{C}{2\,d{a}^{2}} \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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)

[Out]

-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A+1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*B-1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3-3/2/d/a^2*
A*tan(1/2*d*x+1/2*c)+5/2/d/a^2*B*tan(1/2*d*x+1/2*c)-7/2/d/a^2*C*tan(1/2*d*x+1/2*c)-1/d/a^2/(tan(1/2*d*x+1/2*c)
+1)*B+5/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)*C+1/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*A-2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*B
+7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^2*C/(tan(1/2*d*x+1/2*c)+1)^2-1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*B+5/
2/d/a^2/(tan(1/2*d*x+1/2*c)-1)*C-1/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*A+2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B-7/2/d/a
^2*ln(tan(1/2*d*x+1/2*c)-1)*C+1/2/d/a^2*C/(tan(1/2*d*x+1/2*c)-1)^2

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Maxima [B]  time = 0.972351, size = 582, normalized size = 3.44 \begin{align*} -\frac{C{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{21 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{21 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B{\left (\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + A{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(C*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c
)/(cos(d*x + c) + 1) - 1)/a^2) - B*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)
/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*
sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))) + A*((9*sin(d*x + c)/(cos(d
*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*l
og(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2))/d

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Fricas [A]  time = 0.525568, size = 630, normalized size = 3.73 \begin{align*} \frac{3 \,{\left ({\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (2 \, A - 5 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (10 \, A - 28 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (B - C\right )} \cos \left (d x + c\right ) - 3 \, C\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} 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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(3*((2*A - 4*B + 7*C)*cos(d*x + c)^4 + 2*(2*A - 4*B + 7*C)*cos(d*x + c)^3 + (2*A - 4*B + 7*C)*cos(d*x + c
)^2)*log(sin(d*x + c) + 1) - 3*((2*A - 4*B + 7*C)*cos(d*x + c)^4 + 2*(2*A - 4*B + 7*C)*cos(d*x + c)^3 + (2*A -
4*B + 7*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(2*A - 5*B + 8*C)*cos(d*x + c)^3 + (10*A - 28*B + 43
*C)*cos(d*x + c)^2 - 6*(B - C)*cos(d*x + c) - 3*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^
3 + a^2*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 ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{B \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26907, size = 317, normalized size = 1.88 \begin{align*} \frac{\frac{3 \,{\left (2 \, A - 4 \, B + 7 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \,{\left (2 \, A - 4 \, B + 7 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac{6 \,{\left (2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, 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 ) + 3 \, 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^{2}} - \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*A - 4*B + 7*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(2*A - 4*B + 7*C)*log(abs(tan(1/2*d*x + 1/
2*c) - 1))/a^2 - 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + 3*C*t
an(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2) - (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x
+ 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/2*c) - 15*B*a^4*tan(1/2*d*x + 1/2*c) + 21*
C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d