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

Optimal. Leaf size=521 \[ \frac{2 \sqrt{a+b} \cot (c+d x) \left (a^2 b (315 A-161 B+135 C)+15 a^3 (7 B-C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{105 b d}+\frac{2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{105 d}-\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (161 a^2 b B+15 a^3 C+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b^2 d}-\frac{2 a^2 A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d}+\frac{2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.971456, antiderivative size = 521, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4056, 4058, 3921, 3784, 3832, 4004} \[ \frac{2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{105 d}+\frac{2 \sqrt{a+b} \cot (c+d x) \left (a^2 b (315 A-161 B+135 C)+15 a^3 (7 B-C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b d}-\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (161 a^2 b B+15 a^3 C+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b^2 d}-\frac{2 a^2 A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d}+\frac{2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac{2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[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 4056

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), Int
[(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 4058

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
]

Rule 3921

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 3784

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

Rule 3832

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

Rule 4004

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 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \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 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 (\sin ^{-1}\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 (\sin ^{-1}\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) F\left (\sin ^{-1}\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) \Pi \left (\frac{a+b}{a};\sin ^{-1}\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]  time = 21.2756, size = 1405, normalized size = 2.7 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[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 -
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)] + 210*a^3*A*b*Elliptic
Pi[-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[
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)]))/(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])/(
105*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]  time = 1.259, size = 5138, normalized size = 9.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \sec \left (d x + c\right )^{4} +{\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)