\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx\) [338]

Optimal result

Integrand size = 31, antiderivative size = 274 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \]

[Out]

(A*a^3+4*A*a*b^2-3*B*a^2*b-2*B*b^3)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2
)/d-1/3*a^2*(A*b-B*a)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^3+1/6*a*(A*a^2*b-6*A*b^3-4*B*a^3+9*B*a*b^2)*
tan(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^2+1/6*(A*a^4*b-10*A*a^2*b^3-6*A*b^5+2*B*a^5-5*B*a^3*b^2+18*B*a*b
^4)*tan(d*x+c)/b^2/(a^2-b^2)^3/d/(a+b*sec(d*x+c))

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4113, 4165, 4088, 12, 3916, 2738, 214} \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac {\left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac {\left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \]

[In]

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^4,x]

[Out]

((a^3*A + 4*a*A*b^2 - 3*a^2*b*B - 2*b^3*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)
*(a + b)^(7/2)*d) - (a^2*(A*b - a*B)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + (a*(a^2*A*b
- 6*A*b^3 - 4*a^3*B + 9*a*b^2*B)*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + ((a^4*A*b - 10
*a^2*A*b^3 - 6*A*b^5 + 2*a^5*B - 5*a^3*b^2*B + 18*a*b^4*B)*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^3*d*(a + b*Sec[c +
d*x]))

Rule 12

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

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4088

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a
*B)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1]

Rule 4113

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> Simp[(-a^2)*(A*b - a*B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2)
)), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[a*b*(A*b - a*B)*
(m + 1) - (A*b - a*B)*(a^2 + b^2*(m + 1))*Csc[e + f*x] + b*B*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4165

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e +
f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e
+ f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {\int \frac {\sec (c+d x) \left (-3 a b (A b-a B)-\left (a^2-3 b^2\right ) (A b-a B) \sec (c+d x)-3 b \left (a^2-b^2\right ) B \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b^2 \left (a^2-b^2\right )} \\ & = -\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x) \left (2 b^2 \left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right )+b \left (a^3 A b-6 a A b^3+2 a^4 B-3 a^2 b^2 B+6 b^4 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2} \\ & = -\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\int -\frac {3 b^3 \left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^3} \\ & = -\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = -\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3} \\ & = -\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right )^3 d} \\ & = \frac {\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.84 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {-\frac {6 \left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {2 (-A b+a B) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))^3}+\frac {\left (3 a^2 A+2 A b^2-5 a b B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (b+a \cos (c+d x))^2}+\frac {\left (-13 a^2 A b-2 A b^3+4 a^3 B+11 a b^2 B\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (b+a \cos (c+d x))}}{6 d} \]

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^4,x]

[Out]

((-6*(a^3*A + 4*a*A*b^2 - 3*a^2*b*B - 2*b^3*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^
2)^(7/2) + (2*(-(A*b) + a*B)*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])^3) + ((3*a^2*A + 2*A*b^2 - 5*
a*b*B)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])^2) + ((-13*a^2*A*b - 2*A*b^3 + 4*a^3*B + 11*a*b
^2*B)*Sin[c + d*x])/((a - b)^3*(a + b)^3*(b + a*Cos[c + d*x])))/(6*d)

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.37

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (a^{3} A +6 A \,a^{2} b +2 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-3 B \,a^{2} b -6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (7 A \,a^{2} b +3 A \,b^{3}-B \,a^{3}-9 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (a^{3} A -6 A \,a^{2} b +2 A a \,b^{2}-2 A \,b^{3}+2 B \,a^{3}-3 B \,a^{2} b +6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (a^{3} A +4 A a \,b^{2}-3 B \,a^{2} b -2 B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(375\)
default	\(\frac {-\frac {2 \left (-\frac {\left (a^{3} A +6 A \,a^{2} b +2 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-3 B \,a^{2} b -6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (7 A \,a^{2} b +3 A \,b^{3}-B \,a^{3}-9 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (a^{3} A -6 A \,a^{2} b +2 A a \,b^{2}-2 A \,b^{3}+2 B \,a^{3}-3 B \,a^{2} b +6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (a^{3} A +4 A a \,b^{2}-3 B \,a^{2} b -2 B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(375\)
risch	\(\text {Expression too large to display}\)	\(1180\)

[In]

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(-2*(-1/2*(A*a^3+6*A*a^2*b+2*A*a*b^2+2*A*b^3-2*B*a^3-3*B*a^2*b-6*B*a*b^2)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*
tan(1/2*d*x+1/2*c)^5+2/3*(7*A*a^2*b+3*A*b^3-B*a^3-9*B*a*b^2)/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c
)^3+1/2*(A*a^3-6*A*a^2*b+2*A*a*b^2-2*A*b^3+2*B*a^3-3*B*a^2*b+6*B*a*b^2)/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/
2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3+(A*a^3+4*A*a*b^2-3*B*a^2*b-2*B*b^3)/(a^6-3
*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 586 vs. \(2 (260) = 520\).

Time = 0.37 (sec) , antiderivative size = 1230, normalized size of antiderivative = 4.49 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[1/12*(3*(A*a^3*b^3 - 3*B*a^2*b^4 + 4*A*a*b^5 - 2*B*b^6 + (A*a^6 - 3*B*a^5*b + 4*A*a^4*b^2 - 2*B*a^3*b^3)*cos(
d*x + c)^3 + 3*(A*a^5*b - 3*B*a^4*b^2 + 4*A*a^3*b^3 - 2*B*a^2*b^4)*cos(d*x + c)^2 + 3*(A*a^4*b^2 - 3*B*a^3*b^3
+ 4*A*a^2*b^4 - 2*B*a*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)
^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x +
c) + b^2)) + 2*(2*B*a^7 + A*a^6*b - 7*B*a^5*b^2 - 11*A*a^4*b^3 + 23*B*a^3*b^4 + 4*A*a^2*b^5 - 18*B*a*b^6 + 6*A
*b^7 + (4*B*a^7 - 13*A*a^6*b + 7*B*a^5*b^2 + 11*A*a^4*b^3 - 11*B*a^3*b^4 + 2*A*a^2*b^5)*cos(d*x + c)^2 + 3*(A*
a^7 + B*a^6*b - 10*A*a^5*b^2 + 8*B*a^4*b^3 + 7*A*a^3*b^4 - 9*B*a^2*b^5 + 2*A*a*b^6)*cos(d*x + c))*sin(d*x + c)
)/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 -
4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x +
c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d), 1/6*(3*(A*a^3*b^3 - 3*B*a^2*b^4 + 4*A*a*b^5 - 2*
B*b^6 + (A*a^6 - 3*B*a^5*b + 4*A*a^4*b^2 - 2*B*a^3*b^3)*cos(d*x + c)^3 + 3*(A*a^5*b - 3*B*a^4*b^2 + 4*A*a^3*b^
3 - 2*B*a^2*b^4)*cos(d*x + c)^2 + 3*(A*a^4*b^2 - 3*B*a^3*b^3 + 4*A*a^2*b^4 - 2*B*a*b^5)*cos(d*x + c))*sqrt(-a^
2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (2*B*a^7 + A*a^6*b - 7*B*
a^5*b^2 - 11*A*a^4*b^3 + 23*B*a^3*b^4 + 4*A*a^2*b^5 - 18*B*a*b^6 + 6*A*b^7 + (4*B*a^7 - 13*A*a^6*b + 7*B*a^5*b
^2 + 11*A*a^4*b^3 - 11*B*a^3*b^4 + 2*A*a^2*b^5)*cos(d*x + c)^2 + 3*(A*a^7 + B*a^6*b - 10*A*a^5*b^2 + 8*B*a^4*b
^3 + 7*A*a^3*b^4 - 9*B*a^2*b^5 + 2*A*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5
*b^6 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 +
3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 -
4*a^2*b^9 + b^11)*d)]

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \]

[In]

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**4,x)

[Out]

Integral((A + B*sec(c + d*x))*sec(c + d*x)**3/(a + b*sec(c + d*x))**4, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (260) = 520\).

Time = 0.37 (sec) , antiderivative size = 693, normalized size of antiderivative = 2.53 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {\frac {3 \, {\left (A a^{3} - 3 \, B a^{2} b + 4 \, A a b^{2} - 2 \, B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {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 )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {3 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 27 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(A*a^3 - 3*B*a^2*b + 4*A*a*b^2 - 2*B*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(
a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(
-a^2 + b^2)) + (3*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^5*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^4*b*tan(1/2*d*x + 1/2
*c)^5 + 3*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^3*b^2*tan(1/2*d*x + 1/2
*c)^5 + 12*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*A*a*b^4*tan(1/2*d*x + 1/
2*c)^5 - 18*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^5*tan(1/2*d*x + 1/2*c)^5 + 4*B*a^5*tan(1/2*d*x + 1/2*c)^3 -
28*A*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 32*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 16*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3
- 36*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^5*tan(1/2*d*x + 1/2*c)^3 - 3*A*a^5*tan(1/2*d*x + 1/2*c) - 6*B*a^
5*tan(1/2*d*x + 1/2*c) + 12*A*a^4*b*tan(1/2*d*x + 1/2*c) - 3*B*a^4*b*tan(1/2*d*x + 1/2*c) + 27*A*a^3*b^2*tan(1
/2*d*x + 1/2*c) - 6*B*a^3*b^2*tan(1/2*d*x + 1/2*c) + 12*A*a^2*b^3*tan(1/2*d*x + 1/2*c) - 27*B*a^2*b^3*tan(1/2*
d*x + 1/2*c) + 6*A*a*b^4*tan(1/2*d*x + 1/2*c) - 18*B*a*b^4*tan(1/2*d*x + 1/2*c) + 6*A*b^5*tan(1/2*d*x + 1/2*c)
)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^3))/d

Mupad [B] (verification not implemented)

Time = 19.60 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-B\,a^3+7\,A\,a^2\,b-9\,B\,a\,b^2+3\,A\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A\,a^3+2\,A\,b^3-2\,B\,a^3+2\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2-3\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a^3-2\,A\,b^3+2\,B\,a^3+2\,A\,a\,b^2-6\,A\,a^2\,b+6\,B\,a\,b^2-3\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}+\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (A\,a^3-3\,B\,a^2\,b+4\,A\,a\,b^2-2\,B\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^3*(a + b/cos(c + d*x))^4),x)

[Out]

((4*tan(c/2 + (d*x)/2)^3*(3*A*b^3 - B*a^3 + 7*A*a^2*b - 9*B*a*b^2))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) - (tan(c
/2 + (d*x)/2)^5*(A*a^3 + 2*A*b^3 - 2*B*a^3 + 2*A*a*b^2 + 6*A*a^2*b - 6*B*a*b^2 - 3*B*a^2*b))/((a + b)^3*(a - b
)) + (tan(c/2 + (d*x)/2)*(A*a^3 - 2*A*b^3 + 2*B*a^3 + 2*A*a*b^2 - 6*A*a^2*b + 6*B*a*b^2 - 3*B*a^2*b))/((a + b)
*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/(d*(tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) - tan(c/2 + (d
*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) + 3*a*b^2 + 3*a^2*b + a^3 + b^3 - tan(c/2 + (d*x)/2)^6*(3*a*b^2 -
3*a^2*b + a^3 - b^3))) + (atanh((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(2*(a + b)^(
1/2)*(a - b)^(7/2)))*(A*a^3 - 2*B*b^3 + 4*A*a*b^2 - 3*B*a^2*b))/(d*(a + b)^(7/2)*(a - b)^(7/2))