3.5.59 \(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [459]

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

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 169, normalized size of antiderivative = 1.00, 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 = {4169, 4104, 3872, 3852, 8, 3853, 3855} \begin {gather*} -\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} \end {gather*}

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 8

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

Rule 3852

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 3853

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 3855

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

Rule 3872

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 4104

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 4169

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*C
sc[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)]

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)) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(901\) vs. \(2(169)=338\).
time = 6.37, size = 901, normalized size = 5.33 \begin {gather*} -\frac {4 (2 A-4 B+7 C) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {4 (2 A-4 B+7 C) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\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 )}{24 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2} \end {gather*}

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 [A]
time = 0.60, size = 209, normalized size = 1.24

method	result	size
derivativedivides	\(\frac {-\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-7 C +4 B -2 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (2 A -4 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{2 d \,a^{2}}\)	\(209\)
default	\(\frac {-\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-7 C +4 B -2 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (2 A -4 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{2 d \,a^{2}}\)	\(209\)
norman	\(\frac {\frac {\left (5 A -11 B +18 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (A -B +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (3 A -9 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (5 A -11 B +17 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (25 A -61 B +100 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (35 A -95 B +149 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} a}-\frac {\left (2 A -4 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}+\frac {\left (2 A -4 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\)	\(249\)
risch	\(-\frac {i \left (6 A \,{\mathrm e}^{6 i \left (d x +c \right )}-12 B \,{\mathrm e}^{6 i \left (d x +c \right )}+21 C \,{\mathrm e}^{6 i \left (d x +c \right )}+18 A \,{\mathrm e}^{5 i \left (d x +c \right )}-36 B \,{\mathrm e}^{5 i \left (d x +c \right )}+63 C \,{\mathrm e}^{5 i \left (d x +c \right )}+20 A \,{\mathrm e}^{4 i \left (d x +c \right )}-56 B \,{\mathrm e}^{4 i \left (d x +c \right )}+98 C \,{\mathrm e}^{4 i \left (d x +c \right )}+36 A \,{\mathrm e}^{3 i \left (d x +c \right )}-84 B \,{\mathrm e}^{3 i \left (d x +c \right )}+126 C \,{\mathrm e}^{3 i \left (d x +c \right )}+22 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}+97 C \,{\mathrm e}^{2 i \left (d x +c \right )}+18 \,{\mathrm e}^{i \left (d x +c \right )} A -48 B \,{\mathrm e}^{i \left (d x +c \right )}+75 C \,{\mathrm e}^{i \left (d x +c \right )}+8 A -20 B +32 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{2} d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{2} d}\)	\(394\)

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,method=_RETURNVERBOSE)

[Out]

1/2/d/a^2*(-1/3*A*tan(1/2*d*x+1/2*c)^3+1/3*B*tan(1/2*d*x+1/2*c)^3-1/3*C*tan(1/2*d*x+1/2*c)^3-3*A*tan(1/2*d*x+1
/2*c)+5*B*tan(1/2*d*x+1/2*c)-7*C*tan(1/2*d*x+1/2*c)-(2*B-5*C)/(tan(1/2*d*x+1/2*c)-1)+(-7*C+4*B-2*A)*ln(tan(1/2
*d*x+1/2*c)-1)+C/(tan(1/2*d*x+1/2*c)-1)^2-(2*B-5*C)/(tan(1/2*d*x+1/2*c)+1)+(2*A-4*B+7*C)*ln(tan(1/2*d*x+1/2*c)
+1)-C/(tan(1/2*d*x+1/2*c)+1)^2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (159) = 318\).
time = 0.28, size = 431, normalized size = 2.55 \begin {gather*} -\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 {gather*}

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 = 3.26, size = 252, normalized size = 1.49 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.53, size = 235, normalized size = 1.39 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 3.32, size = 170, normalized size = 1.01 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-2\,B+\frac {7\,C}{2}\right )}{a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B+C\right )}{2\,a^2}-\frac {2\,B-4\,C}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-5\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-3\,C\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*(A - 2*B + (7*C)/2))/(a^2*d) - (tan(c/2 + (d*x)/2)*((3*(A - B + C))/(2*a^2) - (2*
B - 4*C)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^3*(A - B + C))/(6*a^2*d) - (tan(c/2 + (d*x)/2)^3*(2*B - 5*C) - tan(
c/2 + (d*x)/2)*(2*B - 3*C))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c/2 + (d*x)/2)^2 + a^2))

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