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

Optimal. Leaf size=517 \[ \frac {2 \tan (c+d x) \sec (c+d x) \left (-6 a^2 C+9 a b B+63 A b^2+49 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{315 b^2 d}-\frac {2 \tan (c+d x) \left (-8 a^3 C+12 a^2 b B-a b^2 (21 A+13 C)-75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{315 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-16 a^3 C+12 a^2 b (2 B-C)-6 a b^2 (7 A-3 B+6 C)-3 b^3 (63 A-25 B+49 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-16 a^4 C+24 a^3 b B-6 a^2 b^2 (7 A+4 C)+57 a b^3 B+21 b^4 (9 A+7 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^5 d}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{63 b d}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d} \]

[Out]

-2/315*(a-b)*(24*a^3*b*B+57*a*b^3*B-16*a^4*C-6*a^2*b^2*(7*A+4*C)+21*b^4*(9*A+7*C))*cot(d*x+c)*EllipticE((a+b*s
ec(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-2/315*(a-b)*(12*a^2*b*(2*B-C)-16*a^3*C-6*a*b^2*(7*A-3*B+6*C)-3*b^3*(63*A-25*B+49*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-2/315*(12*a^2*b*B-75*b^3*B-8*a^3*C-a*b^2*(21*A+13*C))*(a+b*sec(d*x
+c))^(1/2)*tan(d*x+c)/b^3/d+2/315*(63*A*b^2+9*B*a*b-6*C*a^2+49*C*b^2)*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*
x+c)/b^2/d+2/63*(9*B*b+C*a)*sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d+2/9*C*sec(d*x+c)^3*(a+b*sec(d*x
+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 1.56, antiderivative size = 517, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4096, 4102, 4092, 4082, 4005, 3832, 4004} \[ \frac {2 \tan (c+d x) \sec (c+d x) \left (-6 a^2 C+9 a b B+63 A b^2+49 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{315 b^2 d}-\frac {2 \tan (c+d x) \left (12 a^2 b B-8 a^3 C-a b^2 (21 A+13 C)-75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{315 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (12 a^2 b (2 B-C)-16 a^3 C-6 a b^2 (7 A-3 B+6 C)-3 b^3 (63 A-25 B+49 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-6 a^2 b^2 (7 A+4 C)+24 a^3 b B-16 a^4 C+57 a b^3 B+21 b^4 (9 A+7 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^5 d}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{63 b d}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - b)*Sqrt[a + b]*(24*a^3*b*B + 57*a*b^3*B - 16*a^4*C - 6*a^2*b^2*(7*A + 4*C) + 21*b^4*(9*A + 7*C))*Cot[
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))])/(315*b^5*d) - (2*(a - b)*Sqrt[a + b]*(12*a^2*b*(2*B - C) - 1
6*a^3*C - 6*a*b^2*(7*A - 3*B + 6*C) - 3*b^3*(63*A - 25*B + 49*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))])/(315*b^4*d) - (2*(12*a^2*b*B - 75*b^3*B - 8*a^3*C - a*b^2*(21*A + 13*C))*Sqrt[a + b*Sec[c + d*x]]*Ta
n[c + d*x])/(315*b^3*d) + (2*(63*A*b^2 + 9*a*b*B - 6*a^2*C + 49*b^2*C)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*T
an[c + d*x])/(315*b^2*d) + (2*(9*b*B + a*C)*Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(63*b*d) + (
2*C*Sec[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(9*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 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 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]

Rule 4096

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

Rubi steps

\begin {align*} \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {2}{9} \int \frac {\sec ^3(c+d x) \left (\frac {3}{2} a (3 A+2 C)+\frac {1}{2} (9 A b+9 a B+7 b C) \sec (c+d x)+\frac {1}{2} (9 b B+a C) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 (9 b B+a C) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{63 b d}+\frac {2 C \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {4 \int \frac {\sec ^2(c+d x) \left (a (9 b B+a C)+\frac {1}{4} b (63 a A+45 b B+47 a C) \sec (c+d x)+\frac {1}{4} \left (63 A b^2+9 a b B-6 a^2 C+49 b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{63 b}\\ &=\frac {2 \left (63 A b^2+9 a b B-6 a^2 C+49 b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 (9 b B+a C) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{63 b d}+\frac {2 C \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {8 \int \frac {\sec (c+d x) \left (\frac {1}{4} a \left (63 A b^2+9 a b B-6 a^2 C+49 b^2 C\right )+\frac {1}{8} b \left (189 A b^2+207 a b B+2 a^2 C+147 b^2 C\right ) \sec (c+d x)-\frac {3}{8} \left (12 a^2 b B-75 b^3 B-8 a^3 C-a b^2 (21 A+13 C)\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b^2}\\ &=-\frac {2 \left (12 a^2 b B-75 b^3 B-8 a^3 C-a b^2 (21 A+13 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^3 d}+\frac {2 \left (63 A b^2+9 a b B-6 a^2 C+49 b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 (9 b B+a C) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{63 b d}+\frac {2 C \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {16 \int \frac {\sec (c+d x) \left (\frac {3}{16} b \left (6 a^2 b B+75 b^3 B-4 a^3 C+3 a b^2 (49 A+37 C)\right )+\frac {3}{16} \left (24 a^3 b B+57 a b^3 B-16 a^4 C-6 a^2 b^2 (7 A+4 C)+21 b^4 (9 A+7 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{945 b^3}\\ &=-\frac {2 \left (12 a^2 b B-75 b^3 B-8 a^3 C-a b^2 (21 A+13 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^3 d}+\frac {2 \left (63 A b^2+9 a b B-6 a^2 C+49 b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 (9 b B+a C) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{63 b d}+\frac {2 C \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {\left (24 a^3 b B+57 a b^3 B-16 a^4 C-6 a^2 b^2 (7 A+4 C)+21 b^4 (9 A+7 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b^3}-\frac {\left ((a-b) \left (12 a^2 b (2 B-C)-16 a^3 C-6 a b^2 (7 A-3 B+6 C)-3 b^3 (63 A-25 B+49 C)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b^3}\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (24 a^3 b B+57 a b^3 B-16 a^4 C-6 a^2 b^2 (7 A+4 C)+21 b^4 (9 A+7 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}}}{315 b^5 d}-\frac {2 (a-b) \sqrt {a+b} \left (12 a^2 b (2 B-C)-16 a^3 C-6 a b^2 (7 A-3 B+6 C)-3 b^3 (63 A-25 B+49 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}}}{315 b^4 d}-\frac {2 \left (12 a^2 b B-75 b^3 B-8 a^3 C-a b^2 (21 A+13 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^3 d}+\frac {2 \left (63 A b^2+9 a b B-6 a^2 C+49 b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 (9 b B+a C) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{63 b d}+\frac {2 C \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 20.38, size = 920, normalized size = 1.78 \[ \frac {\sqrt {a+b \sec (c+d x)} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac {4 (9 b B \sin (c+d x)+a C \sin (c+d x)) \sec ^3(c+d x)}{63 b}+\frac {4}{9} C \tan (c+d x) \sec ^3(c+d x)+\frac {4 \left (-6 C \sin (c+d x) a^2+9 b B \sin (c+d x) a+63 A b^2 \sin (c+d x)+49 b^2 C \sin (c+d x)\right ) \sec ^2(c+d x)}{315 b^2}+\frac {4 \left (8 C \sin (c+d x) a^3-12 b B \sin (c+d x) a^2+21 A b^2 \sin (c+d x) a+13 b^2 C \sin (c+d x) a+75 b^3 B \sin (c+d x)\right ) \sec (c+d x)}{315 b^3}+\frac {4 \left (-16 C a^4+24 b B a^3-42 A b^2 a^2-24 b^2 C a^2+57 b^3 B a+189 A b^4+147 b^4 C\right ) \sin (c+d x)}{315 b^4}\right ) \cos ^2(c+d x)}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac {4 \sqrt {a+b \sec (c+d x)} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left ((a+b) \left (16 C a^4-24 b B a^3+6 b^2 (7 A+4 C) a^2-57 b^3 B a-21 b^4 (9 A+7 C)\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 (-16 C a^3+12 b (2 B+C) a^2-6 b^2 (7 A+3 B+6 C) a+3 b^3 (63 A+25 B+49 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 (16 C a^4-24 b B a^3+6 b^2 (7 A+4 C) a^2-57 b^3 B a-21 b^4 (9 A+7 C)\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 )}{315 b^4 d \sqrt {b+a \cos (c+d x)} (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac {5}{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}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((a +
b)*(-24*a^3*b*B - 57*a*b^3*B + 16*a^4*C + 6*a^2*b^2*(7*A + 4*C) - 21*b^4*(9*A + 7*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)*(-16*a^3*C + 12*a^2*b*(2*B + C) - 6*a*b^2*(7*A + 3*B + 6*
C) + 3*b^3*(63*A + 25*B + 49*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)] + (-24*a^3*b
*B - 57*a*b^3*B + 16*a^4*C + 6*a^2*b^2*(7*A + 4*C) - 21*b^4*(9*A + 7*C))*Tan[(c + d*x)/2]*(b - b*Tan[(c + d*x)
/2]^4 + a*(-1 + Tan[(c + d*x)/2]^2)^2)))/(315*b^4*d*Sqrt[b + a*Cos[c + d*x]]*(A + 2*C + 2*B*Cos[c + d*x] + A*C
os[2*c + 2*d*x])*Sec[c + d*x]^(5/2)*(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]^2*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*
Sec[c + d*x]^2)*((4*(-42*a^2*A*b^2 + 189*A*b^4 + 24*a^3*b*B + 57*a*b^3*B - 16*a^4*C - 24*a^2*b^2*C + 147*b^4*C
)*Sin[c + d*x])/(315*b^4) + (4*Sec[c + d*x]^3*(9*b*B*Sin[c + d*x] + a*C*Sin[c + d*x]))/(63*b) + (4*Sec[c + d*x
]^2*(63*A*b^2*Sin[c + d*x] + 9*a*b*B*Sin[c + d*x] - 6*a^2*C*Sin[c + d*x] + 49*b^2*C*Sin[c + d*x]))/(315*b^2) +
(4*Sec[c + d*x]*(21*a*A*b^2*Sin[c + d*x] - 12*a^2*b*B*Sin[c + d*x] + 75*b^3*B*Sin[c + d*x] + 8*a^3*C*Sin[c +
d*x] + 13*a*b^2*C*Sin[c + d*x]))/(315*b^3) + (4*C*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(A + 2*C + 2*B*Cos[c + d
*x] + A*Cos[2*c + 2*d*x]))

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{5} + B \sec \left (d x + c\right )^{4} + A \sec \left (d x + c\right )^{3}\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 4.01, size = 5961, normalized size = 11.53 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),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)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^3,x)

[Out]

int(((a + b/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+b*sec(d*x+c))**(1/2),x)

[Out]

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

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