3.422 \(\int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=406 \[ \frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 b^2 d \left (a^2-b^2\right )}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac {\left (-5 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b^3 d \left (a^2-b^2\right )}-\frac {\left (-5 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}-\frac {a \left (-5 a^3 B+3 a^2 A b+7 a b^2 B-5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]

[Out]

-1/3*(3*A*a*b-5*B*a^2+2*B*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/b^2/(a^2-b^2)/d+a*(A*b-B*a)*sec(d*x+c)^(5/2)*sin(d*
x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))+(3*A*a^2*b-2*A*b^3-5*B*a^3+4*B*a*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/b^3/(a^2
-b^2)/d-(3*A*a^2*b-2*A*b^3-5*B*a^3+4*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/
2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/(a^2-b^2)/d-1/3*(3*A*a*b-5*B*a^2+2*B*b^2)*(cos(1/2
*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2
)/b^2/(a^2-b^2)/d-a*(3*A*a^2*b-5*A*b^3-5*B*a^3+7*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli
pticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/(a-b)/b^3/(a+b)^2/d

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Rubi [A]  time = 1.16, antiderivative size = 406, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4029, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 b^2 d \left (a^2-b^2\right )}+\frac {\left (3 a^2 A b-5 a^3 B+4 a b^2 B-2 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b^3 d \left (a^2-b^2\right )}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac {\left (3 a^2 A b-5 a^3 B+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}-\frac {a \left (3 a^2 A b-5 a^3 B+7 a b^2 B-5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^(7/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^2,x]

[Out]

-(((3*a^2*A*b - 2*A*b^3 - 5*a^3*B + 4*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]
)/(b^3*(a^2 - b^2)*d)) - ((3*a*A*b - 5*a^2*B + 2*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[
c + d*x]])/(3*b^2*(a^2 - b^2)*d) - (a*(3*a^2*A*b - 5*A*b^3 - 5*a^3*B + 7*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticP
i[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^3*(a + b)^2*d) + ((3*a^2*A*b - 2*A*b^3 - 5*a^3
*B + 4*a*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b^3*(a^2 - b^2)*d) - ((3*a*A*b - 5*a^2*B + 2*b^2*B)*Sec[c +
d*x]^(3/2)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d) + (a*(A*b - a*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(b*(a^2 - b^2
)*d*(a + b*Sec[c + d*x]))

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4029

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

Rule 4102

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[(C*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 0]

Rule 4106

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

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\frac {a (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3}{2} a (A b-a B)-b (A b-a B) \sec (c+d x)-\frac {1}{2} \left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {2 \int \frac {\sqrt {\sec (c+d x)} \left (-\frac {1}{4} a \left (3 a A b-5 a^2 B+2 b^2 B\right )+\frac {1}{2} b \left (3 a A b-2 a^2 B-b^2 B\right ) \sec (c+d x)+\frac {3}{4} \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {4 \int \frac {-\frac {3}{8} a \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )-\frac {1}{4} b \left (6 a^2 A b-3 A b^3-10 a^3 B+7 a b^2 B\right ) \sec (c+d x)-\frac {1}{8} \left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {4 \int \frac {-\frac {3}{8} a^2 \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )-\left (-\frac {3}{8} a b \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )+\frac {1}{4} a b \left (6 a^2 A b-3 A b^3-10 a^3 B+7 a b^2 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^2 b^3 \left (a^2-b^2\right )}-\frac {\left (a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 b^2 \left (a^2-b^2\right )}-\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}-\frac {\left (a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac {a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )}-\frac {\left (\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 7.32, size = 733, normalized size = 1.81 \[ \frac {\sqrt {\sec (c+d x)} \left (\frac {a^2 A b \sin (c+d x)-a^3 B \sin (c+d x)}{b^2 \left (b^2-a^2\right ) (a \cos (c+d x)+b)}+\frac {\left (5 a^3 B-3 a^2 A b-4 a b^2 B+2 A b^3\right ) \sin (c+d x)}{b^3 \left (b^2-a^2\right )}+\frac {2 B \tan (c+d x)}{3 b^2}\right )}{d}+\frac {\frac {2 \left (40 a^3 b B-24 a^2 A b^2-28 a b^3 B+12 A b^4\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )}{a \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}+\frac {\left (15 a^4 B-9 a^3 A b-12 a^2 b^2 B+6 a A b^3\right ) \sin (c+d x) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (2 a^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 b^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)-2 a (a-2 b) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a^2 b \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a \cos (c+d x)+b)}+\frac {2 \left (45 a^4 B-27 a^3 A b-44 a^2 b^2 B+30 a A b^3-4 b^4 B\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right )}{b \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}}{12 b^3 d (a-b) (a+b)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^(7/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^2,x]

[Out]

((2*(-27*a^3*A*b + 30*a*A*b^3 + 45*a^4*B - 44*a^2*b^2*B - 4*b^4*B)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c
+ d*x]]], -1] - EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x
]^2]*Sin[c + d*x])/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(-24*a^2*A*b^2 + 12*A*b^4 + 40*a^3*b*B -
28*a*b^3*B)*Cos[c + d*x]^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a + b*Sec[c + d*x])*Sqrt[1 - S
ec[c + d*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((-9*a^3*A*b + 6*a*A*b^3 + 15*a^4
*B - 12*a^2*b^2*B)*Cos[2*(c + d*x)]*(a + b*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcS
in[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*a*(a - 2*b)*EllipticF[ArcSin[Sqrt[
Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c +
d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*b^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]]
, -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]
^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(12*(a - b)*b^3*(a + b)*d) + (Sqrt[Sec[c + d*x]]*(((-3*a^2*A*b +
2*A*b^3 + 5*a^3*B - 4*a*b^2*B)*Sin[c + d*x])/(b^3*(-a^2 + b^2)) + (a^2*A*b*Sin[c + d*x] - a^3*B*Sin[c + d*x])
/(b^2*(-a^2 + b^2)*(b + a*Cos[c + d*x])) + (2*B*Tan[c + d*x])/(3*b^2)))/d

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^(7/2)/(b*sec(d*x + c) + a)^2, x)

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maple [B]  time = 19.50, size = 1024, normalized size = 2.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^2*(A*b-2*B*a)/b^3/(a^2-a*b)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(c
os(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*(A*b-2*B*a)/b^3*(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*
sin(1/2*d*x+1/2*c)^2-1)+2*B/b^2*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/
(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(A*b-B*a)*a/b^2*(a^2/b/(a^2-b
^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2
/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/
2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4
+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipti
cPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2
*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/
(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + b/cos(c + d*x))^2,x)

[Out]

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + b/cos(c + d*x))^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(7/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

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