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

Optimal. Leaf size=148 \[ \frac{(3 A-3 B+4 C) \tan ^3(c+d x)}{3 a d}+\frac{(3 A-3 B+4 C) \tan (c+d x)}{a d}-\frac{(2 A-3 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(2 A-3 B+3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.198302, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4084, 3787, 3768, 3770, 3767} \[ \frac{(3 A-3 B+4 C) \tan ^3(c+d x)}{3 a d}+\frac{(3 A-3 B+4 C) \tan (c+d x)}{a d}-\frac{(2 A-3 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(2 A-3 B+3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \]

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]),x]

[Out]

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

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 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 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 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]

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

Mathematica [B]  time = 6.31357, size = 898, normalized size = 6.07 \[ \frac{2 (2 A-3 B+3 C) \cos (c+d x) \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 ^2\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)}-\frac{2 (2 A-3 B+3 C) \cos (c+d x) \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 ^2\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)}+\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 (-6 A \sin \left (\frac{d x}{2}\right )+6 B \sin \left (\frac{d x}{2}\right )+6 C \sin \left (\frac{d x}{2}\right )+30 A \sin \left (\frac{3 d x}{2}\right )-27 B \sin \left (\frac{3 d x}{2}\right )+39 C \sin \left (\frac{3 d x}{2}\right )-12 A \sin \left (c-\frac{d x}{2}\right )+12 B \sin \left (c-\frac{d x}{2}\right )-24 C \sin \left (c-\frac{d x}{2}\right )-6 A \sin \left (c+\frac{d x}{2}\right )+6 B \sin \left (c+\frac{d x}{2}\right )-6 C \sin \left (c+\frac{d x}{2}\right )-24 A \sin \left (2 c+\frac{d x}{2}\right )+24 B \sin \left (2 c+\frac{d x}{2}\right )-24 C \sin \left (2 c+\frac{d x}{2}\right )+12 A \sin \left (c+\frac{3 d x}{2}\right )-9 B \sin \left (c+\frac{3 d x}{2}\right )+21 C \sin \left (c+\frac{3 d x}{2}\right )+12 A \sin \left (2 c+\frac{3 d x}{2}\right )-9 B \sin \left (2 c+\frac{3 d x}{2}\right )+9 C \sin \left (2 c+\frac{3 d x}{2}\right )-6 A \sin \left (3 c+\frac{3 d x}{2}\right )+9 B \sin \left (3 c+\frac{3 d x}{2}\right )-9 C \sin \left (3 c+\frac{3 d x}{2}\right )+6 A \sin \left (c+\frac{5 d x}{2}\right )-3 B \sin \left (c+\frac{5 d x}{2}\right )+7 C \sin \left (c+\frac{5 d x}{2}\right )+3 B \sin \left (2 c+\frac{5 d x}{2}\right )+C \sin \left (2 c+\frac{5 d x}{2}\right )+3 B \sin \left (3 c+\frac{5 d x}{2}\right )-3 C \sin \left (3 c+\frac{5 d x}{2}\right )-6 A \sin \left (4 c+\frac{5 d x}{2}\right )+9 B \sin \left (4 c+\frac{5 d x}{2}\right )-9 C \sin \left (4 c+\frac{5 d x}{2}\right )+12 A \sin \left (2 c+\frac{7 d x}{2}\right )-12 B \sin \left (2 c+\frac{7 d x}{2}\right )+16 C \sin \left (2 c+\frac{7 d x}{2}\right )+6 A \sin \left (3 c+\frac{7 d x}{2}\right )-6 B \sin \left (3 c+\frac{7 d x}{2}\right )+10 C \sin \left (3 c+\frac{7 d x}{2}\right )+6 A \sin \left (4 c+\frac{7 d x}{2}\right )-6 B \sin \left (4 c+\frac{7 d x}{2}\right )+6 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)} \]

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]),x]

[Out]

(2*(2*A - 3*B + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*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
*(2*A - 3*B + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*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])) + (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)*(-6*A*Sin[(d*x)/2] + 6*
B*Sin[(d*x)/2] + 6*C*Sin[(d*x)/2] + 30*A*Sin[(3*d*x)/2] - 27*B*Sin[(3*d*x)/2] + 39*C*Sin[(3*d*x)/2] - 12*A*Sin
[c - (d*x)/2] + 12*B*Sin[c - (d*x)/2] - 24*C*Sin[c - (d*x)/2] - 6*A*Sin[c + (d*x)/2] + 6*B*Sin[c + (d*x)/2] -
6*C*Sin[c + (d*x)/2] - 24*A*Sin[2*c + (d*x)/2] + 24*B*Sin[2*c + (d*x)/2] - 24*C*Sin[2*c + (d*x)/2] + 12*A*Sin[
c + (3*d*x)/2] - 9*B*Sin[c + (3*d*x)/2] + 21*C*Sin[c + (3*d*x)/2] + 12*A*Sin[2*c + (3*d*x)/2] - 9*B*Sin[2*c +
(3*d*x)/2] + 9*C*Sin[2*c + (3*d*x)/2] - 6*A*Sin[3*c + (3*d*x)/2] + 9*B*Sin[3*c + (3*d*x)/2] - 9*C*Sin[3*c + (3
*d*x)/2] + 6*A*Sin[c + (5*d*x)/2] - 3*B*Sin[c + (5*d*x)/2] + 7*C*Sin[c + (5*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2]
+ C*Sin[2*c + (5*d*x)/2] + 3*B*Sin[3*c + (5*d*x)/2] - 3*C*Sin[3*c + (5*d*x)/2] - 6*A*Sin[4*c + (5*d*x)/2] + 9
*B*Sin[4*c + (5*d*x)/2] - 9*C*Sin[4*c + (5*d*x)/2] + 12*A*Sin[2*c + (7*d*x)/2] - 12*B*Sin[2*c + (7*d*x)/2] + 1
6*C*Sin[2*c + (7*d*x)/2] + 6*A*Sin[3*c + (7*d*x)/2] - 6*B*Sin[3*c + (7*d*x)/2] + 10*C*Sin[3*c + (7*d*x)/2] + 6
*A*Sin[4*c + (7*d*x)/2] - 6*B*Sin[4*c + (7*d*x)/2] + 6*C*Sin[4*c + (7*d*x)/2]))/(24*d*(A + 2*C + 2*B*Cos[c + d
*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x]))

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Maple [B]  time = 0.073, size = 442, normalized size = 3. \begin{align*}{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{3\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{B}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{C}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3\,C}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3\,B}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{A}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{5\,C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,B}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{A}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{C}{3\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}+{\frac{B}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{C}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3\,C}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{3\,B}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{A}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{5\,C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3\,B}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{A}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \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)),x)

[Out]

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

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Maxima [B]  time = 0.964716, size = 655, normalized size = 4.43 \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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(C*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*B*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c
) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*si
n(d*x + c)/(a*(cos(d*x + c) + 1))) - 6*A*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*
x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 0.530972, size = 489, normalized size = 3.3 \begin{align*} -\frac{3 \,{\left ({\left (2 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (2 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (2 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (2 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (3 \, A - 3 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (6 \, A - 3 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, B - C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\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)),x, algorithm="fricas")

[Out]

-1/12*(3*((2*A - 3*B + 3*C)*cos(d*x + c)^4 + (2*A - 3*B + 3*C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((2*A
- 3*B + 3*C)*cos(d*x + c)^4 + (2*A - 3*B + 3*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(3*A - 3*B + 4*
C)*cos(d*x + c)^3 + (6*A - 3*B + 7*C)*cos(d*x + c)^2 + (3*B - C)*cos(d*x + c) + 2*C)*sin(d*x + c))/(a*d*cos(d*
x + c)^4 + a*d*cos(d*x + c)^3)

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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{\left (c + d x \right )} + 1}\, dx + \int \frac{B \sec ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{5}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \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)),x)

[Out]

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

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Giac [A]  time = 1.25946, size = 328, normalized size = 2.22 \begin{align*} -\frac{\frac{3 \,{\left (2 \, A - 3 \, B + 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{3 \,{\left (2 \, A - 3 \, B + 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{6 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} + \frac{2 \,{\left (6 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, 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 )}^{3} a}}{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)),x, algorithm="giac")

[Out]

-1/6*(3*(2*A - 3*B + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*(2*A - 3*B + 3*C)*log(abs(tan(1/2*d*x + 1/2
*c) - 1))/a - 6*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a + 2*(6*A*tan(1/2*
d*x + 1/2*c)^5 - 9*B*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1/2*c)^5 - 12*A*tan(1/2*d*x + 1/2*c)^3 + 12*B
*tan(1/2*d*x + 1/2*c)^3 - 16*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) - 3*B*tan(1/2*d*x + 1/2*c) +
9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d