\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [364]

Optimal result

Integrand size = 33, antiderivative size = 469 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 b^3 (25 A-49 B)-6 a b^2 (60 A-19 B)+15 a^2 b (3 A-11 B)-10 a^3 B\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}+\frac {2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 A b-2 a B) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 B (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d} \]

[Out]

-2/315*(a-b)*(45*A*a^3*b+435*A*a*b^3-10*B*a^4+279*B*a^2*b^2+147*B*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(
1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/
2)/b^3/d-2/315*(a-b)*(3*b^3*(25*A-49*B)-6*a*b^2*(60*A-19*B)+15*a^2*b*(3*A-11*B)-10*B*a^3)*cot(d*x+c)*EllipticF
((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec
(d*x+c))/(a-b))^(1/2)/b^2/d+2/315*(45*A*a*b-10*B*a^2+49*B*b^2)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/63*(9*A
*b-2*B*a)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d+2/9*B*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b/d+2/315*(45*A*a^2*b+
75*A*b^3-10*B*a^3+114*B*a*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4095, 4087, 4090, 3917, 4089} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{315 b d}-\frac {2 (a-b) \sqrt {a+b} \left (-10 a^3 B+15 a^2 b (3 A-11 B)-6 a b^2 (60 A-19 B)+3 b^3 (25 A-49 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{315 b^2 d}+\frac {2 \left (-10 a^3 B+45 a^2 A b+114 a b^2 B+75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{315 b d}-\frac {2 (a-b) \sqrt {a+b} \left (-10 a^4 B+45 a^3 A b+279 a^2 b^2 B+435 a A b^3+147 b^4 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^3 d}+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{63 b d}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d} \]

[In]

Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(-2*(a - b)*Sqrt[a + b]*(45*a^3*A*b + 435*a*A*b^3 - 10*a^4*B + 279*a^2*b^2*B + 147*b^4*B)*Cot[c + d*x]*Ellipti
cE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-(
(b*(1 + Sec[c + d*x]))/(a - b))])/(315*b^3*d) - (2*(a - b)*Sqrt[a + b]*(3*b^3*(25*A - 49*B) - 6*a*b^2*(60*A -
19*B) + 15*a^2*b*(3*A - 11*B) - 10*a^3*B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]],
(a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(315*b^2*d) +
(2*(45*a^2*A*b + 75*A*b^3 - 10*a^3*B + 114*a*b^2*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(315*b*d) + (2*(45*
a*A*b - 10*a^2*B + 49*b^2*B)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(315*b*d) + (2*(9*A*b - 2*a*B)*(a + b*Se
c[c + d*x])^(5/2)*Tan[c + d*x])/(63*b*d) + (2*B*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d)

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4087

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

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

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

Rule 4095

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 B (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {2 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {7 b B}{2}+\frac {1}{2} (9 A b-2 a B) \sec (c+d x)\right ) \, dx}{9 b} \\ & = \frac {2 (9 A b-2 a B) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 B (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {4 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {3}{4} b (15 A b+13 a B)+\frac {1}{4} \left (45 a A b-10 a^2 B+49 b^2 B\right ) \sec (c+d x)\right ) \, dx}{63 b} \\ & = \frac {2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 A b-2 a B) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 B (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {8 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {3}{8} b \left (120 a A b+55 a^2 B+49 b^2 B\right )+\frac {3}{8} \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sec (c+d x)\right ) \, dx}{315 b} \\ & = \frac {2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 A b-2 a B) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 B (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {16 \int \frac {\sec (c+d x) \left (\frac {3}{16} b \left (405 a^2 A b+75 A b^3+155 a^3 B+261 a b^2 B\right )+\frac {3}{16} \left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{945 b} \\ & = \frac {2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 A b-2 a B) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 B (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {\left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b}+\frac {\left (16 \left (\frac {3}{16} b \left (405 a^2 A b+75 A b^3+155 a^3 B+261 a b^2 B\right )-\frac {3}{16} \left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right )\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{945 b} \\ & = -\frac {2 (a-b) \sqrt {a+b} \left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 b^3 (25 A-49 B)-6 a b^2 (60 A-19 B)-10 a^3 B+a^2 (45 A b-165 b B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}+\frac {2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 A b-2 a B) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 B (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3781\) vs. \(2(469)=938\).

Time = 24.95 (sec) , antiderivative size = 3781, normalized size of antiderivative = 8.06 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Result too large to show} \]

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*(45*a^3*A*b + 435*a*A*b^3 - 10*a^4*B + 279*a^2*b^2*B + 147*b^4*
B)*Sin[c + d*x])/(315*b^2) + (2*Sec[c + d*x]^3*(9*A*b^2*Sin[c + d*x] + 19*a*b*B*Sin[c + d*x]))/63 + (2*Sec[c +
d*x]^2*(135*a*A*b*Sin[c + d*x] + 75*a^2*B*Sin[c + d*x] + 49*b^2*B*Sin[c + d*x]))/315 + (2*Sec[c + d*x]*(135*a
^2*A*b*Sin[c + d*x] + 75*A*b^3*Sin[c + d*x] + 5*a^3*B*Sin[c + d*x] + 163*a*b^2*B*Sin[c + d*x]))/(315*b) + (2*b
^2*B*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])^2) + (2*(-1/7*(a^3*A)/(Sqrt[b + a*Cos[c + d*x]]*
Sqrt[Sec[c + d*x]]) - (29*a*A*b^2)/(21*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a^4*B)/(63*b*Sqrt[b +
a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (31*a^2*b*B)/(35*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (7*b^3*
B)/(15*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (a^4*A*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]
) - (2*a^2*A*b*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) + (5*A*b^3*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a
*Cos[c + d*x]]) - (124*a^3*B*Sqrt[Sec[c + d*x]])/(315*Sqrt[b + a*Cos[c + d*x]]) + (2*a^5*B*Sqrt[Sec[c + d*x]])
/(63*b^2*Sqrt[b + a*Cos[c + d*x]]) + (38*a*b^2*B*Sqrt[Sec[c + d*x]])/(105*Sqrt[b + a*Cos[c + d*x]]) - (a^4*A*C
os[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (29*a^2*A*b*Cos[2*(c + d*x)]*Sqrt[Sec[c +
d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (31*a^3*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(35*Sqrt[b + a*Cos[c +
d*x]]) + (2*a^5*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(63*b^2*Sqrt[b + a*Cos[c + d*x]]) - (7*a*b^2*B*Cos[2*(c
+ d*x)]*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c
+ d*x])^(5/2)*(2*(a + b)*(-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Sqrt[Cos[c + d*x]
/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]
], (a - b)/(a + b)] + 2*b*(a + b)*(-10*a^3*B + 15*a^2*b*(3*A + 11*B) + 6*a*b^2*(60*A + 19*B) + 3*b^3*(25*A + 4
9*B))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[
ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B
)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(315*b^2*d*(b + a*Cos[c + d*x])^3*Sq
rt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(5/2)*((a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*(2*(a + b)*(-
45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b
+ a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a
+ b)*(-10*a^3*B + 15*a^2*b*(3*A + 11*B) + 6*a*b^2*(60*A + 19*B) + 3*b^3*(25*A + 49*B))*Sqrt[Cos[c + d*x]/(1 +
Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a
- b)/(a + b)] + (-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Cos[c + d*x]*(b + a*Cos[c
+ d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(315*b^2*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2]^2]) -
(Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(c + d*x)/2]*(2*(a + b)*(-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279
*a^2*b^2*B - 147*b^4*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c +
d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-10*a^3*B + 15*a^2*b*(3*A + 11*B)
+ 6*a*b^2*(60*A + 19*B) + 3*b^3*(25*A + 49*B))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])
/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-45*a^3*A*b - 435*a*A*b
^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/
2]))/(315*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*((
(-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c +
d*x)/2]^4)/2 + ((a + b)*(-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Sqrt[(b + a*Cos[c +
d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*((Cos[c + d*x]*Sin[c
+ d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + (b*(
a + b)*(-10*a^3*B + 15*a^2*b*(3*A + 11*B) + 6*a*b^2*(60*A + 19*B) + 3*b^3*(25*A + 49*B))*Sqrt[(b + a*Cos[c + d
*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*((Cos[c + d*x]*Sin[c +
d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + ((a +
b)*(-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*El
lipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b + a
*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c +
d*x]))] + (b*(a + b)*(-10*a^3*B + 15*a^2*b*(3*A + 11*B) + 6*a*b^2*(60*A + 19*B) + 3*b^3*(25*A + 49*B))*Sqrt[Co
s[c + d*x]/(1 + Cos[c + d*x])]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/((a +
b)*(1 + Cos[c + d*x]))) + ((b + a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b + a*Cos
[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] - a*(-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B
)*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] - (-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*
a^2*b^2*B - 147*b^4*B)*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] + (-45*a^3*A*b -
435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(
c + d*x)/2]^2 + (b*(a + b)*(-10*a^3*B + 15*a^2*b*(3*A + 11*B) + 6*a*b^2*(60*A + 19*B) + 3*b^3*(25*A + 49*B))*S
qrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Sec[(c + d*x)/2]^
2)/(Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]) + ((a + b)*(-45*a^3*A*b - 435
*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x
])/((a + b)*(1 + Cos[c + d*x]))]*Sec[(c + d*x)/2]^2*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])/Sqrt[1 - T
an[(c + d*x)/2]^2]))/(315*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + ((2*(a + b)*(-45*a^3*A*b -
435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c +
d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-10*a^
3*B + 15*a^2*b*(3*A + 11*B) + 6*a*b^2*(60*A + 19*B) + 3*b^3*(25*A + 49*B))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x]
)]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)
] + (-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(
c + d*x)/2]^2*Tan[(c + d*x)/2])*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c
+ d*x]*Tan[c + d*x]))/(315*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c
+ d*x]])))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5583\) vs. \(2(431)=862\).

Time = 63.30 (sec) , antiderivative size = 5584, normalized size of antiderivative = 11.91

method	result	size
parts	\(\text {Expression too large to display}\)	\(5584\)
default	\(\text {Expression too large to display}\)	\(5649\)

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*b^2*sec(d*x + c)^5 + A*a^2*sec(d*x + c)^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^4 + (B*a^2 + 2*A*a*b)*s
ec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), x)

Sympy [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{2}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**(5/2)*sec(c + d*x)**2, x)

Maxima [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]

[In]

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(5/2))/cos(c + d*x)^2,x)

[Out]

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(5/2))/cos(c + d*x)^2, x)