3.268 \(\int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=231 \[ \frac {a b^2}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac {2 a b \left (a^2+b^2\right )}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {\csc ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac {a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}+\frac {(2 a-b) \log (1-\cos (c+d x))}{4 d (a+b)^4}+\frac {(2 a+b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

1/2*a*b^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))^2-2*a*b*(a^2+b^2)/(a^2-b^2)^3/d/(b+a*cos(d*x+c))-1/2*(a*(a^2+3*b^2)-b
*(3*a^2+b^2)*cos(d*x+c))*csc(d*x+c)^2/(a^2-b^2)^3/d+1/4*(2*a-b)*ln(1-cos(d*x+c))/(a+b)^4/d+1/4*(2*a+b)*ln(1+co
s(d*x+c))/(a-b)^4/d-a*(a^4+8*a^2*b^2+3*b^4)*ln(b+a*cos(d*x+c))/(a^2-b^2)^4/d

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Rubi [A]  time = 0.62, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4397, 2837, 12, 1647, 1629} \[ \frac {a b^2}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac {2 a b \left (a^2+b^2\right )}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {a \left (8 a^2 b^2+a^4+3 b^4\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}-\frac {\csc ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}+\frac {(2 a-b) \log (1-\cos (c+d x))}{4 d (a+b)^4}+\frac {(2 a+b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]

[Out]

(a*b^2)/(2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) - (2*a*b*(a^2 + b^2))/((a^2 - b^2)^3*d*(b + a*Cos[c + d*x])
) - ((a*(a^2 + 3*b^2) - b*(3*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)^3*d) + ((2*a - b)*Log[1 -
Cos[c + d*x]])/(4*(a + b)^4*d) + ((2*a + b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) - (a*(a^4 + 8*a^2*b^2 + 3*
b^4)*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^4*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

\begin {align*} \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx &=\int \frac {\cot ^2(c+d x) \csc (c+d x)}{(b+a \cos (c+d x))^3} \, dx\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \frac {x^2}{a^2 (b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac {a \operatorname {Subst}\left (\int \frac {x^2}{(b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac {\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac {\operatorname {Subst}\left (\int \frac {\frac {a^2 b^4 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3}+\frac {a^2 b^3 \left (7 a^2-3 b^2\right ) x}{\left (a^2-b^2\right )^3}+\frac {a^2 \left (2 a^4-3 a^2 b^2-3 b^4\right ) x^2}{\left (a^2-b^2\right )^3}-\frac {a^2 b \left (3 a^2+b^2\right ) x^3}{\left (a^2-b^2\right )^3}}{(b+x)^3 \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{2 a d}\\ &=-\frac {\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac {\operatorname {Subst}\left (\int \left (\frac {a (2 a-b)}{2 (a+b)^4 (a-x)}-\frac {a (2 a+b)}{2 (a-b)^4 (a+x)}+\frac {2 a^2 b^2}{\left (a^2-b^2\right )^2 (b+x)^3}-\frac {4 a^2 b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 (b+x)^2}+\frac {2 \left (a^6+8 a^4 b^2+3 a^2 b^4\right )}{\left (a^2-b^2\right )^4 (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{2 a d}\\ &=\frac {a b^2}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}-\frac {2 a b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac {\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}+\frac {(2 a-b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}+\frac {(2 a+b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}-\frac {a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d}\\ \end {align*}

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Mathematica [C]  time = 6.38, size = 703, normalized size = 3.04 \[ \frac {2 i \left (a^5+8 a^3 b^2+3 a b^4\right ) (c+d x) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{d (a-b)^4 (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {\left (-a^5-8 a^3 b^2-3 a b^4\right ) \tan ^3(c+d x) (a \cos (c+d x)+b)^3 \log (a \cos (c+d x)+b)}{d \left (b^2-a^2\right )^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {a b^2 \tan ^3(c+d x) (a \cos (c+d x)+b)}{2 d (b-a)^2 (a+b)^2 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {2 a b (b-i a) (b+i a) \tan ^3(c+d x) (a \cos (c+d x)+b)^2}{d (b-a)^3 (a+b)^3 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (2 a+b) \tan ^{-1}(\tan (c+d x)) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{2 d (b-a)^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (2 a-b) \tan ^{-1}(\tan (c+d x)) \tan ^3(c+d x) (a \cos (c+d x)+b)^3}{2 d (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(2 a+b) \tan ^3(c+d x) \log \left (\cos ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}{4 d (b-a)^4 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(2 a-b) \tan ^3(c+d x) \log \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}{4 d (a+b)^4 (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {\tan ^3(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^3}{8 d (a+b)^3 (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {\tan ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^3}{8 d (b-a)^3 (a \sin (c+d x)+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]

[Out]

(a*b^2*(b + a*Cos[c + d*x])*Tan[c + d*x]^3)/(2*(-a + b)^2*(a + b)^2*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + (
2*a*b*((-I)*a + b)*(I*a + b)*(b + a*Cos[c + d*x])^2*Tan[c + d*x]^3)/((-a + b)^3*(a + b)^3*d*(a*Sin[c + d*x] +
b*Tan[c + d*x])^3) + ((2*I)*(a^5 + 8*a^3*b^2 + 3*a*b^4)*(c + d*x)*(b + a*Cos[c + d*x])^3*Tan[c + d*x]^3)/((a -
b)^4*(a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) - ((I/2)*(2*a - b)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c +
d*x])^3*Tan[c + d*x]^3)/((a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) - ((I/2)*(2*a + b)*ArcTan[Tan[c + d*
x]]*(b + a*Cos[c + d*x])^3*Tan[c + d*x]^3)/((-a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) - ((b + a*Cos[c
+ d*x])^3*Csc[(c + d*x)/2]^2*Tan[c + d*x]^3)/(8*(a + b)^3*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + ((2*a + b)*
(b + a*Cos[c + d*x])^3*Log[Cos[(c + d*x)/2]^2]*Tan[c + d*x]^3)/(4*(-a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x
])^3) + ((-a^5 - 8*a^3*b^2 - 3*a*b^4)*(b + a*Cos[c + d*x])^3*Log[b + a*Cos[c + d*x]]*Tan[c + d*x]^3)/((-a^2 +
b^2)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + ((2*a - b)*(b + a*Cos[c + d*x])^3*Log[Sin[(c + d*x)/2]^2]*Tan[
c + d*x]^3)/(4*(a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + ((b + a*Cos[c + d*x])^3*Sec[(c + d*x)/2]^2*T
an[c + d*x]^3)/(8*(-a + b)^3*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3)

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fricas [B]  time = 0.84, size = 1076, normalized size = 4.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(8*a^5*b^2 + 8*a^3*b^4 - 16*a*b^6 - 2*(7*a^6*b - 2*a^4*b^3 - 5*a^2*b^5)*cos(d*x + c)^3 + 2*(a^7 - 7*a^5*b^
2 - a^3*b^4 + 7*a*b^6)*cos(d*x + c)^2 + 2*(6*a^6*b + a^4*b^3 - 8*a^2*b^5 + b^7)*cos(d*x + c) + 4*(a^5*b^2 + 8*
a^3*b^4 + 3*a*b^6 - (a^7 + 8*a^5*b^2 + 3*a^3*b^4)*cos(d*x + c)^4 - 2*(a^6*b + 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x +
c)^3 + (a^7 + 7*a^5*b^2 - 5*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^2 + 2*(a^6*b + 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c
))*log(a*cos(d*x + c) + b) - (2*a^5*b^2 + 9*a^4*b^3 + 16*a^3*b^4 + 14*a^2*b^5 + 6*a*b^6 + b^7 - (2*a^7 + 9*a^6
*b + 16*a^5*b^2 + 14*a^4*b^3 + 6*a^3*b^4 + a^2*b^5)*cos(d*x + c)^4 - 2*(2*a^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*
a^3*b^4 + 6*a^2*b^5 + a*b^6)*cos(d*x + c)^3 + (2*a^7 + 9*a^6*b + 14*a^5*b^2 + 5*a^4*b^3 - 10*a^3*b^4 - 13*a^2*
b^5 - 6*a*b^6 - b^7)*cos(d*x + c)^2 + 2*(2*a^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*a^3*b^4 + 6*a^2*b^5 + a*b^6)*co
s(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - (2*a^5*b^2 - 9*a^4*b^3 + 16*a^3*b^4 - 14*a^2*b^5 + 6*a*b^6 - b^7 - (
2*a^7 - 9*a^6*b + 16*a^5*b^2 - 14*a^4*b^3 + 6*a^3*b^4 - a^2*b^5)*cos(d*x + c)^4 - 2*(2*a^6*b - 9*a^5*b^2 + 16*
a^4*b^3 - 14*a^3*b^4 + 6*a^2*b^5 - a*b^6)*cos(d*x + c)^3 + (2*a^7 - 9*a^6*b + 14*a^5*b^2 - 5*a^4*b^3 - 10*a^3*
b^4 + 13*a^2*b^5 - 6*a*b^6 + b^7)*cos(d*x + c)^2 + 2*(2*a^6*b - 9*a^5*b^2 + 16*a^4*b^3 - 14*a^3*b^4 + 6*a^2*b^
5 - a*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^8)*d
*cos(d*x + c)^4 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)^3 - (a^10 - 5*a^8*b^2 +
10*a^6*b^4 - 10*a^4*b^6 + 5*a^2*b^8 - b^10)*d*cos(d*x + c)^2 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 +
a*b^9)*d*cos(d*x + c) - (a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d)

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giac [B]  time = 0.94, size = 801, normalized size = 3.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(2*(2*a - b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)
- 8*(a^5 + 8*a^3*b^2 + 3*a*b^4)*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) -
1)/(cos(d*x + c) + 1)))/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) + (a + b - 4*a*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4
*a*b^3 + b^4)*(cos(d*x + c) - 1)) + (cos(d*x + c) - 1)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(cos(d*x + c) + 1)) +
4*(3*a^7 - 2*a^6*b + 11*a^5*b^2 + 28*a^4*b^3 + 9*a^3*b^4 + 6*a^2*b^5 + 9*a*b^6 + 6*a^7*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) - 8*a^6*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 38*a^5*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) +
1) - 4*a^4*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 26*a^3*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*
a^2*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 18*a*b^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^7*(cos(d*
x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 6*a^6*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 27*a^5*b^2*(cos(d*x +
c) - 1)^2/(cos(d*x + c) + 1)^2 - 48*a^4*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 33*a^3*b^4*(cos(d*x +
c) - 1)^2/(cos(d*x + c) + 1)^2 - 18*a^2*b^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 9*a*b^6*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(a + b + a*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2))/d

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maple [A]  time = 0.23, size = 324, normalized size = 1.40 \[ -\frac {a^{5} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}-\frac {8 a^{3} b^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}-\frac {3 b^{4} a \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{2} a}{2 d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}-\frac {2 a^{3} b}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}-\frac {2 a \,b^{3}}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}+\frac {1}{4 d \left (a +b \right )^{3} \left (\cos \left (d x +c \right )-1\right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right ) a}{2 d \left (a +b \right )^{4}}-\frac {\ln \left (\cos \left (d x +c \right )-1\right ) b}{4 d \left (a +b \right )^{4}}-\frac {1}{4 d \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {a \ln \left (1+\cos \left (d x +c \right )\right )}{2 \left (a -b \right )^{4} d}+\frac {b \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)

[Out]

-1/d*a^5/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c))-8/d*a^3*b^2/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c))-3/d*b^4/(a+b)^4/(a-
b)^4*a*ln(b+a*cos(d*x+c))+1/2/d*b^2/(a+b)^2*a/(a-b)^2/(b+a*cos(d*x+c))^2-2/d*a^3*b/(a+b)^3/(a-b)^3/(b+a*cos(d*
x+c))-2/d*a*b^3/(a+b)^3/(a-b)^3/(b+a*cos(d*x+c))+1/4/d/(a+b)^3/(cos(d*x+c)-1)+1/2/d/(a+b)^4*ln(cos(d*x+c)-1)*a
-1/4/d/(a+b)^4*ln(cos(d*x+c)-1)*b-1/4/d/(a-b)^3/(1+cos(d*x+c))+1/2*a*ln(1+cos(d*x+c))/(a-b)^4/d+1/4*b*ln(1+cos
(d*x+c))/(a-b)^4/d

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maxima [B]  time = 0.38, size = 602, normalized size = 2.61 \[ -\frac {\frac {8 \, {\left (a^{5} + 8 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac {4 \, {\left (2 \, a - b\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {a^{6} - 2 \, a^{5} b - a^{4} b^{2} + 4 \, a^{3} b^{3} - a^{2} b^{4} - 2 \, a b^{5} + b^{6} - \frac {2 \, {\left (a^{6} - 20 \, a^{5} b - 11 \, a^{4} b^{2} - 24 \, a^{3} b^{3} - 29 \, a^{2} b^{4} + 4 \, a b^{5} - b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (a^{6} - 38 \, a^{5} b + 31 \, a^{4} b^{2} - 52 \, a^{3} b^{3} + 63 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{\frac {{\left (a^{9} + a^{8} b - 4 \, a^{7} b^{2} - 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} + 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} - 4 \, a^{2} b^{7} + a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, {\left (a^{9} - a^{8} b - 4 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} + 4 \, a^{2} b^{7} + a b^{8} - b^{9}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (a^{9} - 3 \, a^{8} b + 8 \, a^{6} b^{3} - 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} + 8 \, a^{3} b^{6} - 3 \, a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\sin \left (d x + c\right )^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/8*(8*(a^5 + 8*a^3*b^2 + 3*a*b^4)*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^8 - 4*a^6*b^2
+ 6*a^4*b^4 - 4*a^2*b^6 + b^8) - 4*(2*a - b)*log(sin(d*x + c)/(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 +
4*a*b^3 + b^4) + (a^6 - 2*a^5*b - a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + b^6 - 2*(a^6 - 20*a^5*b - 11*a^4*
b^2 - 24*a^3*b^3 - 29*a^2*b^4 + 4*a*b^5 - b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (a^6 - 38*a^5*b + 31*a^4*
b^2 - 52*a^3*b^3 + 63*a^2*b^4 - 6*a*b^5 + b^6)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/((a^9 + a^8*b - 4*a^7*b^2
- 4*a^6*b^3 + 6*a^5*b^4 + 6*a^4*b^5 - 4*a^3*b^6 - 4*a^2*b^7 + a*b^8 + b^9)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
- 2*(a^9 - a^8*b - 4*a^7*b^2 + 4*a^6*b^3 + 6*a^5*b^4 - 6*a^4*b^5 - 4*a^3*b^6 + 4*a^2*b^7 + a*b^8 - b^9)*sin(d
*x + c)^4/(cos(d*x + c) + 1)^4 + (a^9 - 3*a^8*b + 8*a^6*b^3 - 6*a^5*b^4 - 6*a^4*b^5 + 8*a^3*b^6 - 3*a*b^8 + b^
9)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + sin(d*x + c)^2/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(cos(d*x + c) + 1)^2
))/d

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mupad [B]  time = 1.12, size = 496, normalized size = 2.15 \[ \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-a^5+37\,a^4\,b+6\,a^3\,b^2+58\,a^2\,b^3-5\,a\,b^4+b^5\right )}{2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}-\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-21\,a^4\,b+10\,a^3\,b^2-34\,a^2\,b^3+5\,a\,b^4-b^5\right )}{\left (a-b\right )\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left (\left (4\,a^5-20\,a^4\,b+40\,a^3\,b^2-40\,a^2\,b^3+20\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (-8\,a^5+24\,a^4\,b-16\,a^3\,b^2-16\,a^2\,b^3+24\,a\,b^4-8\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,a^5-4\,a^4\,b-8\,a^3\,b^2+8\,a^2\,b^3+4\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a-b\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (a^5+8\,a^3\,b^2+3\,a\,b^4\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a*sin(c + d*x) + b*tan(c + d*x))^3),x)

[Out]

((tan(c/2 + (d*x)/2)^4*(37*a^4*b - 5*a*b^4 - a^5 + b^5 + 58*a^2*b^3 + 6*a^3*b^2))/(2*(a + b)*(2*a*b + a^2 + b^
2)) - (3*a*b^2 - 3*a^2*b + a^3 - b^3)/(2*(a + b)) + (tan(c/2 + (d*x)/2)^2*(5*a*b^4 - 21*a^4*b + a^5 - b^5 - 34
*a^2*b^3 + 10*a^3*b^2))/((a - b)*(2*a*b + a^2 + b^2)))/(d*(tan(c/2 + (d*x)/2)^2*(4*a*b^4 - 4*a^4*b + 4*a^5 - 4
*b^5 + 8*a^2*b^3 - 8*a^3*b^2) - tan(c/2 + (d*x)/2)^4*(8*a^5 - 24*a^4*b - 24*a*b^4 + 8*b^5 + 16*a^2*b^3 + 16*a^
3*b^2) + tan(c/2 + (d*x)/2)^6*(20*a*b^4 - 20*a^4*b + 4*a^5 - 4*b^5 - 40*a^2*b^3 + 40*a^3*b^2))) - tan(c/2 + (d
*x)/2)^2/(8*d*(a - b)^3) + (log(tan(c/2 + (d*x)/2))*(2*a - b))/(d*(8*a*b^3 + 8*a^3*b + 2*a^4 + 2*b^4 + 12*a^2*
b^2)) - (log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(3*a*b^4 + a^5 + 8*a^3*b^2))/(d*(a^8 + b
^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\left (c + d x \right )}}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)/(a*sin(c + d*x) + b*tan(c + d*x))**3, x)

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