3.269 \(\int \frac {\sec ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=212 \[ \frac {3 a^2 \left (a^2+3 b^2\right )}{2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {3 a^2 b}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {6 a^2 b \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}+\frac {\csc ^2(c+d x) (b-a \cos (c+d x))}{2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {3 a \log (1-\cos (c+d x))}{4 d (a+b)^4}-\frac {3 a \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

-3/2*a^2*b/(a^2-b^2)^2/d/(b+a*cos(d*x+c))^2+3/2*a^2*(a^2+3*b^2)/(a^2-b^2)^3/d/(b+a*cos(d*x+c))+1/2*(b-a*cos(d*
x+c))*csc(d*x+c)^2/(a^2-b^2)/d/(b+a*cos(d*x+c))^2+3/4*a*ln(1-cos(d*x+c))/(a+b)^4/d-3/4*a*ln(1+cos(d*x+c))/(a-b
)^4/d+6*a^2*b*(a^2+b^2)*ln(b+a*cos(d*x+c))/(a^2-b^2)^4/d

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Rubi [A]  time = 0.36, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4397, 2837, 12, 823, 801} \[ \frac {3 a^2 \left (a^2+3 b^2\right )}{2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {3 a^2 b}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {6 a^2 b \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}+\frac {\csc ^2(c+d x) (b-a \cos (c+d x))}{2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {3 a \log (1-\cos (c+d x))}{4 d (a+b)^4}-\frac {3 a \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]

[Out]

(-3*a^2*b)/(2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) + (3*a^2*(a^2 + 3*b^2))/(2*(a^2 - b^2)^3*d*(b + a*Cos[c
+ d*x])) + ((b - a*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + (3*a*Log[1 - Cos[c
+ d*x]])/(4*(a + b)^4*d) - (3*a*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) + (6*a^2*b*(a^2 + b^2)*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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx &=\int \frac {\cot (c+d x) \csc ^2(c+d x)}{(b+a \cos (c+d x))^3} \, dx\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \frac {x}{a (b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac {a^2 \operatorname {Subst}\left (\int \frac {x}{(b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=\frac {(b-a \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {\operatorname {Subst}\left (\int \frac {-3 a^2 b+3 a^2 x}{(b+x)^3 \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=\frac {(b-a \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {\operatorname {Subst}\left (\int \left (\frac {3 a (a-b)}{2 (a+b)^3 (a-x)}+\frac {3 a (a+b)}{2 (a-b)^3 (a+x)}-\frac {6 a^2 b}{\left (a^2-b^2\right ) (b+x)^3}+\frac {3 a^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^2 (b+x)^2}-\frac {12 a^2 b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac {3 a^2 b}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac {3 a^2 \left (a^2+3 b^2\right )}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac {(b-a \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {3 a \log (1-\cos (c+d x))}{4 (a+b)^4 d}-\frac {3 a \log (1+\cos (c+d x))}{4 (a-b)^4 d}+\frac {6 a^2 b \left (a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d}\\ \end {align*}

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Mathematica [A]  time = 6.27, size = 217, normalized size = 1.02 \[ -\frac {a^2 \left (a^2+3 b^2\right )}{d (b-a)^3 (a+b)^3 (a \cos (c+d x)+b)}-\frac {a^2 b}{2 d (b-a)^2 (a+b)^2 (a \cos (c+d x)+b)^2}+\frac {6 \left (a^4 b+a^2 b^3\right ) \log (a \cos (c+d x)+b)}{d \left (b^2-a^2\right )^4}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d (a+b)^3}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d (b-a)^3}+\frac {3 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (a+b)^4}-\frac {3 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (b-a)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]

[Out]

-1/2*(a^2*b)/((-a + b)^2*(a + b)^2*d*(b + a*Cos[c + d*x])^2) - (a^2*(a^2 + 3*b^2))/((-a + b)^3*(a + b)^3*d*(b
+ a*Cos[c + d*x])) - Csc[(c + d*x)/2]^2/(8*(a + b)^3*d) - (3*a*Log[Cos[(c + d*x)/2]])/(2*(-a + b)^4*d) + (6*(a
^4*b + a^2*b^3)*Log[b + a*Cos[c + d*x]])/((-a^2 + b^2)^4*d) + (3*a*Log[Sin[(c + d*x)/2]])/(2*(a + b)^4*d) - Se
c[(c + d*x)/2]^2/(8*(-a + b)^3*d)

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fricas [B]  time = 1.01, size = 939, normalized size = 4.43 \[ -\frac {2 \, a^{6} b + 18 \, a^{4} b^{3} - 18 \, a^{2} b^{5} - 2 \, b^{7} - 6 \, {\left (a^{7} + 2 \, a^{5} b^{2} - 3 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - 24 \, {\left (a^{4} b^{3} - a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{7} + 9 \, a^{5} b^{2} - 12 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) + 24 \, {\left (a^{4} b^{3} + a^{2} b^{5} - {\left (a^{6} b + a^{4} b^{3}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{6} b - a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - 3 \, {\left (a^{5} b^{2} + 4 \, a^{4} b^{3} + 6 \, a^{3} b^{4} + 4 \, a^{2} b^{5} + a b^{6} - {\left (a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} - 5 \, a^{3} b^{4} - 4 \, a^{2} b^{5} - a b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (a^{5} b^{2} - 4 \, a^{4} b^{3} + 6 \, a^{3} b^{4} - 4 \, a^{2} b^{5} + a b^{6} - {\left (a^{7} - 4 \, a^{6} b + 6 \, a^{5} b^{2} - 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b - 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} - 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{7} - 4 \, a^{6} b + 5 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 4 \, a^{2} b^{5} - a b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b - 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} - 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{10} - 4 \, a^{8} b^{2} + 6 \, a^{6} b^{4} - 4 \, a^{4} b^{6} + a^{2} b^{8}\right )} d \cos \left (d x + c\right )^{4} + 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right )^{3} - {\left (a^{10} - 5 \, a^{8} b^{2} + 10 \, a^{6} b^{4} - 10 \, a^{4} b^{6} + 5 \, a^{2} b^{8} - b^{10}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) - {\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(2*a^6*b + 18*a^4*b^3 - 18*a^2*b^5 - 2*b^7 - 6*(a^7 + 2*a^5*b^2 - 3*a^3*b^4)*cos(d*x + c)^3 - 24*(a^4*b^3
- a^2*b^5)*cos(d*x + c)^2 + 2*(2*a^7 + 9*a^5*b^2 - 12*a^3*b^4 + a*b^6)*cos(d*x + c) + 24*(a^4*b^3 + a^2*b^5 -
(a^6*b + a^4*b^3)*cos(d*x + c)^4 - 2*(a^5*b^2 + a^3*b^4)*cos(d*x + c)^3 + (a^6*b - a^2*b^5)*cos(d*x + c)^2 +
2*(a^5*b^2 + a^3*b^4)*cos(d*x + c))*log(a*cos(d*x + c) + b) - 3*(a^5*b^2 + 4*a^4*b^3 + 6*a^3*b^4 + 4*a^2*b^5 +
a*b^6 - (a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*cos(d*x + c)^4 - 2*(a^6*b + 4*a^5*b^2 + 6*a^4*b^3 +
4*a^3*b^4 + a^2*b^5)*cos(d*x + c)^3 + (a^7 + 4*a^6*b + 5*a^5*b^2 - 5*a^3*b^4 - 4*a^2*b^5 - a*b^6)*cos(d*x + c
)^2 + 2*(a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 3*(a
^5*b^2 - 4*a^4*b^3 + 6*a^3*b^4 - 4*a^2*b^5 + a*b^6 - (a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*cos(d*x
+ c)^4 - 2*(a^6*b - 4*a^5*b^2 + 6*a^4*b^3 - 4*a^3*b^4 + a^2*b^5)*cos(d*x + c)^3 + (a^7 - 4*a^6*b + 5*a^5*b^2
- 5*a^3*b^4 + 4*a^2*b^5 - a*b^6)*cos(d*x + c)^2 + 2*(a^6*b - 4*a^5*b^2 + 6*a^4*b^3 - 4*a^3*b^4 + a^2*b^5)*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 - 1
0*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 = 1.11, size = 689, normalized size = 3.25 \[ \frac {\frac {6 \, a \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {48 \, {\left (a^{4} b + a^{2} b^{3}\right )} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} + \frac {{\left (a + b - \frac {6 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {\cos \left (d x + c\right ) - 1}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {8 \, {\left (2 \, a^{7} - 5 \, a^{6} b - 8 \, a^{5} b^{2} - 2 \, a^{4} b^{3} - 10 \, a^{3} b^{4} - 9 \, a^{2} b^{5} + \frac {2 \, a^{7} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {16 \, a^{6} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {6 \, a^{5} b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a^{4} b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {8 \, a^{3} b^{4} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {18 \, a^{2} b^{5} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {9 \, a^{6} b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 \, a^{5} b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {18 \, a^{4} b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 \, a^{3} b^{4} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {9 \, a^{2} b^{5} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (a + b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\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)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(6*a*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) + 48*(a
^4*b + a^2*b^3)*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 - 6*a*(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)) + 8*(2*a^7 - 5*a^6*b - 8*a^5*b^2 - 2*a^4*b^3 - 10*a^3*b^4 - 9*a
^2*b^5 + 2*a^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 16*a^6*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 6*a^5*
b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*a^4*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 8*a^3*b^4*(cos(d
*x + c) - 1)/(cos(d*x + c) + 1) + 18*a^2*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 9*a^6*b*(cos(d*x + c) - 1
)^2/(cos(d*x + c) + 1)^2 + 18*a^5*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 18*a^4*b^3*(cos(d*x + c) - 1
)^2/(cos(d*x + c) + 1)^2 + 18*a^3*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 9*a^2*b^5*(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 = 0.25, size = 251, normalized size = 1.18 \[ -\frac {b \,a^{2}}{2 d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}+\frac {a^{4}}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}+\frac {3 b^{2} a^{2}}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}+\frac {6 b \ln \left (b +a \cos \left (d x +c \right )\right ) a^{4}}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {6 b^{3} \ln \left (b +a \cos \left (d x +c \right )\right ) a^{2}}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {1}{4 d \left (a +b \right )^{3} \left (\cos \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\cos \left (d x +c \right )-1\right ) a}{4 d \left (a +b \right )^{4}}+\frac {1}{4 d \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}-\frac {3 a \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)

[Out]

-1/2/d*b*a^2/(a+b)^2/(a-b)^2/(b+a*cos(d*x+c))^2+1/d*a^4/(a+b)^3/(a-b)^3/(b+a*cos(d*x+c))+3/d*b^2/(a+b)^3/(a-b)
^3/(b+a*cos(d*x+c))*a^2+6/d*b/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c))*a^4+6/d*b^3/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c)
)*a^2+1/4/d/(a+b)^3/(cos(d*x+c)-1)+3/4/d/(a+b)^4*ln(cos(d*x+c)-1)*a+1/4/d/(a-b)^3/(1+cos(d*x+c))-3/4*a*ln(1+co
s(d*x+c))/(a-b)^4/d

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maxima [B]  time = 0.39, size = 596, normalized size = 2.81 \[ \frac {\frac {48 \, {\left (a^{4} b + a^{2} b^{3}\right )} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} + \frac {12 \, a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {a^{6} - 2 \, a^{5} b - a^{4} b^{2} + 4 \, a^{3} b^{3} - a^{2} b^{4} - 2 \, a b^{5} + b^{6} - \frac {2 \, {\left (9 \, a^{6} + 4 \, a^{5} b + 37 \, a^{4} b^{2} + 32 \, a^{3} b^{3} - 5 \, a^{2} b^{4} + 4 \, a b^{5} - b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (17 \, a^{6} - 6 \, a^{5} b + 63 \, a^{4} b^{2} - 84 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{\frac {{\left (a^{9} + a^{8} b - 4 \, a^{7} b^{2} - 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} + 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} - 4 \, a^{2} b^{7} + a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, {\left (a^{9} - a^{8} b - 4 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} + 4 \, a^{2} b^{7} + a b^{8} - b^{9}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (a^{9} - 3 \, a^{8} b + 8 \, a^{6} b^{3} - 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} + 8 \, a^{3} b^{6} - 3 \, a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\sin \left (d x + c\right )^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/8*(48*(a^4*b + a^2*b^3)*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) + 12*a*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*(9*a^6 + 4*a^5*b + 37*a^4*b^2 + 32*a^3*b^3
- 5*a^2*b^4 + 4*a*b^5 - b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (17*a^6 - 6*a^5*b + 63*a^4*b^2 - 84*a^3*b^
3 + 15*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 = 1.13, size = 492, normalized size = 2.32 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^3}-\frac {\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{2\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (9\,a^5-5\,a^4\,b+42\,a^3\,b^2-10\,a^2\,b^3+5\,a\,b^4-b^5\right )}{\left (a-b\right )\,\left (a^2+2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (17\,a^5+11\,a^4\,b+74\,a^3\,b^2-10\,a^2\,b^3+5\,a\,b^4-b^5\right )}{2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left (\left (4\,a^5-20\,a^4\,b+40\,a^3\,b^2-40\,a^2\,b^3+20\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (-8\,a^5+24\,a^4\,b-16\,a^3\,b^2-16\,a^2\,b^3+24\,a\,b^4-8\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,a^5-4\,a^4\,b-8\,a^3\,b^2+8\,a^2\,b^3+4\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (6\,a^4\,b+6\,a^2\,b^3\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )}+\frac {3\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a*sin(c + d*x) + b*tan(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*d*(a - b)^3) - ((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 - 5*a^4*b + 9*a^5 - b^5 - 10*a^2*b^3 + 42*a^3*b^2))/((a - b)*(2*a*b + a^2 + b^2)) + (tan(c/2 + (d*x)/2)
^4*(5*a*b^4 + 11*a^4*b + 17*a^5 - b^5 - 10*a^2*b^3 + 74*a^3*b^2))/(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))) + (log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(6*a^4*b + 6*a^2
*b^3))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2)) + (3*a*log(tan(c/2 + (d*x)/2)))/(d*(8*a*b^3 + 8*a^3
*b + 2*a^4 + 2*b^4 + 12*a^2*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**2/(a*sin(c + d*x) + b*tan(c + d*x))**3, x)

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