\(\int (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [952]

Optimal result

Integrand size = 35, antiderivative size = 521 \[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}+\frac {2 \sqrt {a+b} \left (15 a^3 (7 B-C)+b^3 (35 A-63 B+25 C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b d}-\frac {2 a^2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \]

[Out]

-2/105*(a-b)*(161*B*a^2*b+63*B*b^3+15*a^3*C+5*a*b^2*(49*A+29*C))*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^2
/d+2/105*(15*a^3*(7*B-C)+b^3*(35*A-63*B+25*C)+a^2*b*(315*A-161*B+135*C)-a*b^2*(245*A-119*B+145*C))*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/d-2*a^2*A*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((
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)/d+2/35*(7*B*b+5*
C*a)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*C*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/105*(35*A*b^2+56*B*a*b+15
*C*a^2+25*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4141, 4143, 4006, 3869, 3917, 4089} \[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 a^2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{105 d}+\frac {2 \sqrt {a+b} \cot (c+d x) \left (15 a^3 (7 B-C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{105 b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (15 a^3 C+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{105 b^2 d}+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d} \]

[In]

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

[Out]

(-2*(a - b)*Sqrt[a + b]*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 5*a*b^2*(49*A + 29*C))*Cot[c + d*x]*EllipticE[Arc
Sin[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))])/(105*b^2*d) + (2*Sqrt[a + b]*(15*a^3*(7*B - C) + b^3*(35*A - 63*B + 25*C) + a^2*b*
(315*A - 161*B + 135*C) - a*b^2*(245*A - 119*B + 145*C))*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))])
/(105*b*d) - (2*a^2*A*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/d + (2*(35
*A*b^2 + 56*a*b*B + 15*a^2*C + 25*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(105*d) + (2*(7*b*B + 5*a*C)*(
a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*C*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

Rule 3869

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b
*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b)*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b
*Csc[c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

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 4006

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 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 4141

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

Rule 4143

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {2}{7} \int (a+b \sec (c+d x))^{3/2} \left (\frac {7 a A}{2}+\frac {1}{2} (7 A b+7 a B+5 b C) \sec (c+d x)+\frac {1}{2} (7 b B+5 a C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {a+b \sec (c+d x)} \left (\frac {35 a^2 A}{4}+\frac {1}{4} \left (70 a A b+35 a^2 B+21 b^2 B+40 a b C\right ) \sec (c+d x)+\frac {1}{4} \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {8}{105} \int \frac {\frac {105 a^3 A}{8}+\frac {1}{8} \left (105 a^3 B+119 a b^2 B+45 a^2 b (7 A+3 C)+5 b^3 (7 A+5 C)\right ) \sec (c+d x)+\frac {1}{8} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx \\ & = \frac {2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {8}{105} \int \frac {\frac {105 a^3 A}{8}+\left (\frac {1}{8} \left (105 a^3 B+119 a b^2 B+45 a^2 b (7 A+3 C)+5 b^3 (7 A+5 C)\right )+\frac {1}{8} \left (-161 a^2 b B-63 b^3 B-15 a^3 C-5 a b^2 (49 A+29 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{105} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx \\ & = -\frac {2 (a-b) \sqrt {a+b} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\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}}}{105 b^2 d}+\frac {2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\left (a^3 A\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{105} \left (15 a^3 (7 B-C)+b^3 (35 A-63 B+25 C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx \\ & = -\frac {2 (a-b) \sqrt {a+b} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\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}}}{105 b^2 d}+\frac {2 \sqrt {a+b} \left (15 a^3 (7 B-C)+b^3 (35 A-63 B+25 C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)\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}}}{105 b d}-\frac {2 a^2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\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}}}{d}+\frac {2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1401\) vs. \(2(521)=1042\).

Time = 25.74 (sec) , antiderivative size = 1401, normalized size of antiderivative = 2.69 \[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (245 a^2 A b^2 \tan \left (\frac {1}{2} (c+d x)\right )+245 a A b^3 \tan \left (\frac {1}{2} (c+d x)\right )+161 a^3 b B \tan \left (\frac {1}{2} (c+d x)\right )+161 a^2 b^2 B \tan \left (\frac {1}{2} (c+d x)\right )+63 a b^3 B \tan \left (\frac {1}{2} (c+d x)\right )+63 b^4 B \tan \left (\frac {1}{2} (c+d x)\right )+15 a^4 C \tan \left (\frac {1}{2} (c+d x)\right )+15 a^3 b C \tan \left (\frac {1}{2} (c+d x)\right )+145 a^2 b^2 C \tan \left (\frac {1}{2} (c+d x)\right )+145 a b^3 C \tan \left (\frac {1}{2} (c+d x)\right )-490 a^2 A b^2 \tan ^3\left (\frac {1}{2} (c+d x)\right )-322 a^3 b B \tan ^3\left (\frac {1}{2} (c+d x)\right )-126 a b^3 B \tan ^3\left (\frac {1}{2} (c+d x)\right )-30 a^4 C \tan ^3\left (\frac {1}{2} (c+d x)\right )-290 a^2 b^2 C \tan ^3\left (\frac {1}{2} (c+d x)\right )+245 a^2 A b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )-245 a A b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )+161 a^3 b B \tan ^5\left (\frac {1}{2} (c+d x)\right )-161 a^2 b^2 B \tan ^5\left (\frac {1}{2} (c+d x)\right )+63 a b^3 B \tan ^5\left (\frac {1}{2} (c+d x)\right )-63 b^4 B \tan ^5\left (\frac {1}{2} (c+d x)\right )+15 a^4 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-15 a^3 b C \tan ^5\left (\frac {1}{2} (c+d x)\right )+145 a^2 b^2 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-145 a b^3 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-210 a^3 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-210 a^3 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+(a+b) \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-b \left (-15 a^3 (7 A-7 B-C)+b^3 (35 A+63 B+25 C)+a^2 b (315 A+161 B+135 C)+a b^2 (245 A+119 B+145 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{105 b d (b+a \cos (c+d x))^{5/2} (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {9}{2}}(c+d x) \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {\cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (245 a A b^2+161 a^2 b B+63 b^3 B+15 a^3 C+145 a b^2 C\right ) \sin (c+d x)}{105 b}+\frac {4}{35} \sec ^2(c+d x) \left (7 b^2 B \sin (c+d x)+15 a b C \sin (c+d x)\right )+\frac {4}{105} \sec (c+d x) \left (35 A b^2 \sin (c+d x)+77 a b B \sin (c+d x)+45 a^2 C \sin (c+d x)+25 b^2 C \sin (c+d x)\right )+\frac {4}{7} b^2 C \sec ^2(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))} \]

[In]

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

[Out]

(-4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(24
5*a^2*A*b^2*Tan[(c + d*x)/2] + 245*a*A*b^3*Tan[(c + d*x)/2] + 161*a^3*b*B*Tan[(c + d*x)/2] + 161*a^2*b^2*B*Tan
[(c + d*x)/2] + 63*a*b^3*B*Tan[(c + d*x)/2] + 63*b^4*B*Tan[(c + d*x)/2] + 15*a^4*C*Tan[(c + d*x)/2] + 15*a^3*b
*C*Tan[(c + d*x)/2] + 145*a^2*b^2*C*Tan[(c + d*x)/2] + 145*a*b^3*C*Tan[(c + d*x)/2] - 490*a^2*A*b^2*Tan[(c + d
*x)/2]^3 - 322*a^3*b*B*Tan[(c + d*x)/2]^3 - 126*a*b^3*B*Tan[(c + d*x)/2]^3 - 30*a^4*C*Tan[(c + d*x)/2]^3 - 290
*a^2*b^2*C*Tan[(c + d*x)/2]^3 + 245*a^2*A*b^2*Tan[(c + d*x)/2]^5 - 245*a*A*b^3*Tan[(c + d*x)/2]^5 + 161*a^3*b*
B*Tan[(c + d*x)/2]^5 - 161*a^2*b^2*B*Tan[(c + d*x)/2]^5 + 63*a*b^3*B*Tan[(c + d*x)/2]^5 - 63*b^4*B*Tan[(c + d*
x)/2]^5 + 15*a^4*C*Tan[(c + d*x)/2]^5 - 15*a^3*b*C*Tan[(c + d*x)/2]^5 + 145*a^2*b^2*C*Tan[(c + d*x)/2]^5 - 145
*a*b^3*C*Tan[(c + d*x)/2]^5 - 210*a^3*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - T
an[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 210*a^3*A*b*EllipticP
i[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b -
a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 5*a*b^2*
(49*A + 29*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c +
d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - b*(-15*a^3*(7*A - 7*B - C) +
b^3*(35*A + 63*B + 25*C) + a^2*b*(315*A + 161*B + 135*C) + a*b^2*(245*A + 119*B + 145*C))*EllipticF[ArcSin[Ta
n[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c
+ d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(105*b*d*(b + a*Cos[c + d*x])^(5/2)*(A + 2*C + 2*B*Cos[c + d*x
] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2)*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 +
b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + (Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c +
d*x] + C*Sec[c + d*x]^2)*((4*(245*a*A*b^2 + 161*a^2*b*B + 63*b^3*B + 15*a^3*C + 145*a*b^2*C)*Sin[c + d*x])/(10
5*b) + (4*Sec[c + d*x]^2*(7*b^2*B*Sin[c + d*x] + 15*a*b*C*Sin[c + d*x]))/35 + (4*Sec[c + d*x]*(35*A*b^2*Sin[c
+ d*x] + 77*a*b*B*Sin[c + d*x] + 45*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/105 + (4*b^2*C*Sec[c + d*x]^2
*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(6503\) vs. \(2(478)=956\).

Time = 14.65 (sec) , antiderivative size = 6504, normalized size of antiderivative = 12.48

method	result	size
parts	\(\text {Expression too large to display}\)	\(6504\)
default	\(\text {Expression too large to display}\)	\(6563\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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