3.821 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=573 \[ -\frac {2 \left (-24 a^2 C+44 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{693 b^3 d}+\frac {2 \left (-48 a^3 C+88 a^2 b B-204 a b^2 C+539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{3465 b^3 d}+\frac {2 \left (-48 a^4 C+88 a^3 b B-144 a^2 b^2 C+429 a b^3 B+675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3465 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (-48 a^4 C+4 a^3 b (22 B-9 C)+6 a^2 b^2 (11 B-24 C)+3 a b^3 (143 B-471 C)-3 b^4 (539 B-225 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3465 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (-48 a^5 C+88 a^4 b B-108 a^3 b^2 C+363 a^2 b^3 B+2088 a b^4 C+1617 b^5 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3465 b^5 d}+\frac {2 (11 b B-6 a C) \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{99 b^2 d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2}}{11 b d} \]

[Out]

-2/3465*(a-b)*(88*B*a^4*b+363*B*a^2*b^3+1617*B*b^5-48*C*a^5-108*C*a^3*b^2+2088*C*a*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^5/d-2/3465*(a-b)*(3*a*b^3*(143*B-471*C)-3*b^4*(539*B-225*C)+6*a^2*b^2*(11*B-24*C)+4*a^3*
b*(22*B-9*C)-48*a^4*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^4/d+2/3465*(88*B*a^2*b+539*B*b^3-48*C*a^3-2
04*C*a*b^2)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b^3/d-2/693*(44*B*a*b-24*C*a^2-81*C*b^2)*(a+b*sec(d*x+c))^(5/2)*
tan(d*x+c)/b^3/d+2/99*(11*B*b-6*C*a)*sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d+2/11*C*sec(d*x+c)^2*(a
+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d+2/3465*(88*B*a^3*b+429*B*a*b^3-48*C*a^4-144*C*a^2*b^2+675*C*b^4)*(a+b*sec(
d*x+c))^(1/2)*tan(d*x+c)/b^3/d

________________________________________________________________________________________

Rubi [A]  time = 1.98, antiderivative size = 573, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4072, 4033, 4092, 4082, 4002, 4005, 3832, 4004} \[ -\frac {2 \left (-24 a^2 C+44 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{693 b^3 d}+\frac {2 \left (88 a^2 b B-48 a^3 C-204 a b^2 C+539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{3465 b^3 d}+\frac {2 \left (-144 a^2 b^2 C+88 a^3 b B-48 a^4 C+429 a b^3 B+675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3465 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (6 a^2 b^2 (11 B-24 C)+4 a^3 b (22 B-9 C)-48 a^4 C+3 a b^3 (143 B-471 C)-3 b^4 (539 B-225 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3465 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (363 a^2 b^3 B-108 a^3 b^2 C+88 a^4 b B-48 a^5 C+2088 a b^4 C+1617 b^5 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3465 b^5 d}+\frac {2 (11 b B-6 a C) \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{99 b^2 d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2}}{11 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(-2*(a - b)*Sqrt[a + b]*(88*a^4*b*B + 363*a^2*b^3*B + 1617*b^5*B - 48*a^5*C - 108*a^3*b^2*C + 2088*a*b^4*C)*Co
t[c + d*x]*EllipticE[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))])/(3465*b^5*d) - (2*(a - b)*Sqrt[a + b]*(3*a*b^3*(143*B - 47
1*C) - 3*b^4*(539*B - 225*C) + 6*a^2*b^2*(11*B - 24*C) + 4*a^3*b*(22*B - 9*C) - 48*a^4*C)*Cot[c + d*x]*Ellipti
cF[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))])/(3465*b^4*d) + (2*(88*a^3*b*B + 429*a*b^3*B - 48*a^4*C - 144*a^2*b^2*C + 675
*b^4*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3465*b^3*d) + (2*(88*a^2*b*B + 539*b^3*B - 48*a^3*C - 204*a*b^
2*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(3465*b^3*d) - (2*(44*a*b*B - 24*a^2*C - 81*b^2*C)*(a + b*Sec[c
+ d*x])^(5/2)*Tan[c + d*x])/(693*b^3*d) + (2*(11*b*B - 6*a*C)*Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*Tan[c +
d*x])/(99*b^2*d) + (2*C*Sec[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(11*b*d)

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 4002

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] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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]

Rule 4005

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 4033

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

Rule 4072

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

Rule 4082

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[(C*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*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4092

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

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^4(c+d x) (a+b \sec (c+d x))^{3/2} (B+C \sec (c+d x)) \, dx\\ &=\frac {2 C \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{11 b d}+\frac {2 \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (2 a C+\frac {9}{2} b C \sec (c+d x)+\frac {1}{2} (11 b B-6 a C) \sec ^2(c+d x)\right ) \, dx}{11 b}\\ &=\frac {2 (11 b B-6 a C) \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{11 b d}+\frac {4 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {1}{2} a (11 b B-6 a C)+\frac {1}{4} b (77 b B-6 a C) \sec (c+d x)-\frac {1}{4} \left (44 a b B-24 a^2 C-81 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx}{99 b^2}\\ &=-\frac {2 \left (44 a b B-24 a^2 C-81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^3 d}+\frac {2 (11 b B-6 a C) \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{11 b d}+\frac {8 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (-\frac {3}{8} b \left (22 a b B-12 a^2 C-135 b^2 C\right )+\frac {1}{8} \left (88 a^2 b B+539 b^3 B-48 a^3 C-204 a b^2 C\right ) \sec (c+d x)\right ) \, dx}{693 b^3}\\ &=\frac {2 \left (88 a^2 b B+539 b^3 B-48 a^3 C-204 a b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^3 d}-\frac {2 \left (44 a b B-24 a^2 C-81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^3 d}+\frac {2 (11 b B-6 a C) \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{11 b d}+\frac {16 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (-\frac {3}{16} b \left (22 a^2 b B-539 b^3 B-12 a^3 C-471 a b^2 C\right )+\frac {3}{16} \left (88 a^3 b B+429 a b^3 B-48 a^4 C-144 a^2 b^2 C+675 b^4 C\right ) \sec (c+d x)\right ) \, dx}{3465 b^3}\\ &=\frac {2 \left (88 a^3 b B+429 a b^3 B-48 a^4 C-144 a^2 b^2 C+675 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^3 d}+\frac {2 \left (88 a^2 b B+539 b^3 B-48 a^3 C-204 a b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^3 d}-\frac {2 \left (44 a b B-24 a^2 C-81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^3 d}+\frac {2 (11 b B-6 a C) \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{11 b d}+\frac {32 \int \frac {\sec (c+d x) \left (\frac {3}{32} b \left (22 a^3 b B+2046 a b^3 B-12 a^4 C+1269 a^2 b^2 C+675 b^4 C\right )+\frac {3}{32} \left (88 a^4 b B+363 a^2 b^3 B+1617 b^5 B-48 a^5 C-108 a^3 b^2 C+2088 a b^4 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{10395 b^3}\\ &=\frac {2 \left (88 a^3 b B+429 a b^3 B-48 a^4 C-144 a^2 b^2 C+675 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^3 d}+\frac {2 \left (88 a^2 b B+539 b^3 B-48 a^3 C-204 a b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^3 d}-\frac {2 \left (44 a b B-24 a^2 C-81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^3 d}+\frac {2 (11 b B-6 a C) \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{11 b d}+\frac {\left (88 a^4 b B+363 a^2 b^3 B+1617 b^5 B-48 a^5 C-108 a^3 b^2 C+2088 a b^4 C\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3465 b^3}+\frac {\left (32 \left (\frac {3}{32} b \left (22 a^3 b B+2046 a b^3 B-12 a^4 C+1269 a^2 b^2 C+675 b^4 C\right )-\frac {3}{32} \left (88 a^4 b B+363 a^2 b^3 B+1617 b^5 B-48 a^5 C-108 a^3 b^2 C+2088 a b^4 C\right )\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{10395 b^3}\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (88 a^4 b B+363 a^2 b^3 B+1617 b^5 B-48 a^5 C-108 a^3 b^2 C+2088 a b^4 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}}}{3465 b^5 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 a b^3 (143 B-471 C)-3 b^4 (539 B-225 C)+a^3 b (88 B-36 C)+6 a^2 b^2 (11 B-24 C)-48 a^4 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}}}{3465 b^4 d}+\frac {2 \left (88 a^3 b B+429 a b^3 B-48 a^4 C-144 a^2 b^2 C+675 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^3 d}+\frac {2 \left (88 a^2 b B+539 b^3 B-48 a^3 C-204 a b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^3 d}-\frac {2 \left (44 a b B-24 a^2 C-81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^3 d}+\frac {2 (11 b B-6 a C) \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{11 b d}\\ \end {align*}

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Mathematica [A]  time = 26.61, size = 890, normalized size = 1.55 \[ \frac {2 \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left ((a+b) \left (48 C a^5-88 b B a^4+108 b^2 C a^3-363 b^3 B a^2-2088 b^4 C a-1617 b^5 B\right ) E\left (\sin ^{-1}\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 \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 (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )+b (a+b) \left (-48 C a^4+4 b (22 B+9 C) a^3-6 b^2 (11 B+24 C) a^2+3 b^3 (143 B+471 C) a+3 b^4 (539 B+225 C)\right ) F\left (\sin ^{-1}\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 \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 (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )+\left (48 C a^5-88 b B a^4+108 b^2 C a^3-363 b^3 B a^2-2088 b^4 C a-1617 b^5 B\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (-b \tan ^4\left (\frac {1}{2} (c+d x)\right )+a \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right )^2+b\right )\right ) (a+b \sec (c+d x))^{3/2}}{3465 b^4 d (b+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}}}+\frac {\cos (c+d x) \left (\frac {2}{99} (11 b B \sin (c+d x)+12 a C \sin (c+d x)) \sec ^4(c+d x)+\frac {2}{11} b C \tan (c+d x) \sec ^4(c+d x)+\frac {2 \left (3 C \sin (c+d x) a^2+110 b B \sin (c+d x) a+81 b^2 C \sin (c+d x)\right ) \sec ^3(c+d x)}{693 b}+\frac {2 \left (-18 C \sin (c+d x) a^3+33 b B \sin (c+d x) a^2+606 b^2 C \sin (c+d x) a+539 b^3 B \sin (c+d x)\right ) \sec ^2(c+d x)}{3465 b^2}+\frac {2 \left (24 C \sin (c+d x) a^4-44 b B \sin (c+d x) a^3+57 b^2 C \sin (c+d x) a^2+968 b^3 B \sin (c+d x) a+675 b^4 C \sin (c+d x)\right ) \sec (c+d x)}{3465 b^3}-\frac {2 \left (48 C a^5-88 b B a^4+108 b^2 C a^3-363 b^3 B a^2-2088 b^4 C a-1617 b^5 B\right ) \sin (c+d x)}{3465 b^4}\right ) (a+b \sec (c+d x))^{3/2}}{d (b+a \cos (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*(a + b*Sec[c + d*x])^(3/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((a + b)*(-88*a^4*b*B - 363*a^2*b^3*B - 1617
*b^5*B + 48*a^5*C + 108*a^3*b^2*C - 2088*a*b^4*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*(a + b)*(-48*a^4*C + 4*a^3*b*(22*B + 9*C) - 6*a^2*b^2*(11*B + 24*C) + 3*b^4*(539*B + 225*C) + 3*a*b^3
*(143*B + 471*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)] + (-88*a^4*b*B - 363*a^2*b^
3*B - 1617*b^5*B + 48*a^5*C + 108*a^3*b^2*C - 2088*a*b^4*C)*Tan[(c + d*x)/2]*(b - b*Tan[(c + d*x)/2]^4 + a*(-1
+ Tan[(c + d*x)/2]^2)^2)))/(3465*b^4*d*(b + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)*(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]*(
a + b*Sec[c + d*x])^(3/2)*((-2*(-88*a^4*b*B - 363*a^2*b^3*B - 1617*b^5*B + 48*a^5*C + 108*a^3*b^2*C - 2088*a*b
^4*C)*Sin[c + d*x])/(3465*b^4) + (2*Sec[c + d*x]^4*(11*b*B*Sin[c + d*x] + 12*a*C*Sin[c + d*x]))/99 + (2*Sec[c
+ d*x]^3*(110*a*b*B*Sin[c + d*x] + 3*a^2*C*Sin[c + d*x] + 81*b^2*C*Sin[c + d*x]))/(693*b) + (2*Sec[c + d*x]^2*
(33*a^2*b*B*Sin[c + d*x] + 539*b^3*B*Sin[c + d*x] - 18*a^3*C*Sin[c + d*x] + 606*a*b^2*C*Sin[c + d*x]))/(3465*b
^2) + (2*Sec[c + d*x]*(-44*a^3*b*B*Sin[c + d*x] + 968*a*b^3*B*Sin[c + d*x] + 24*a^4*C*Sin[c + d*x] + 57*a^2*b^
2*C*Sin[c + d*x] + 675*b^4*C*Sin[c + d*x]))/(3465*b^3) + (2*b*C*Sec[c + d*x]^4*Tan[c + d*x])/11))/(d*(b + a*Co
s[c + d*x]))

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \sec \left (d x + c\right )^{6} + B a \sec \left (d x + c\right )^{4} + {\left (C a + B b\right )} \sec \left (d x + c\right )^{5}\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b*sec(d*x + c)^6 + B*a*sec(d*x + c)^4 + (C*a + B*b)*sec(d*x + c)^5)*sqrt(b*sec(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^(3/2)*(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))*(b*sec(d*x + c) + a)^(3/2)*sec(d*x + c)^3, x)

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maple [B]  time = 3.76, size = 5368, normalized size = 9.37 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

result too large to display

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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)^3*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*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 \[ \int \frac {\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(3/2))/cos(c + d*x)^3,x)

[Out]

int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(3/2))/cos(c + d*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((B + C*sec(c + d*x))*(a + b*sec(c + d*x))**(3/2)*sec(c + d*x)**4, x)

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