3.11.1 \(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [1001]

Optimal. Leaf size=317 \[ \frac {2 \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}-\frac {2 b^2 (11 A b+7 a B-35 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

2/35*(6*A*b+7*B*a)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/7*A*(a+b*sec(d*x+c))^3*sin(d*x+c)/d/sec(
d*x+c)^(5/2)+2/105*a*(24*A*b^2+63*a*b*B+5*a^2*(5*A+7*C))*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/35*b^2*(11*A*b+7*B*a-
35*C*b)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(3*a^3*B+15*a*b^2*B+5*b^3*(A-C)+3*a^2*b*(3*A+5*C))*(cos(1/2*d*x+1/2*
c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/21*
(21*a^2*b*B+21*b^3*B+21*a*b^2*(A+3*C)+a^3*(5*A+7*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF
(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.57, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4179, 4159, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 a \sin (c+d x) \left (5 a^2 (5 A+7 C)+63 a b B+24 A b^2\right )}{105 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )}{21 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^3 B+3 a^2 b (3 A+5 C)+15 a b^2 B+5 b^3 (A-C)\right )}{5 d}-\frac {2 b^2 \sin (c+d x) \sqrt {\sec (c+d x)} (7 a B+11 A b-35 b C)}{35 d}+\frac {2 (7 a B+6 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{7 d \sec ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(2*(3*a^3*B + 15*a*b^2*B + 5*b^3*(A - C) + 3*a^2*b*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(5*d) + (2*(21*a^2*b*B + 21*b^3*B + 21*a*b^2*(A + 3*C) + a^3*(5*A + 7*C))*Sqrt[Cos[c + d*x]
]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*(24*A*b^2 + 63*a*b*B + 5*a^2*(5*A + 7*C))*Sin[c
+ d*x])/(105*d*Sqrt[Sec[c + d*x]]) - (2*b^2*(11*A*b + 7*a*B - 35*b*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(35*d)
+ (2*(6*A*b + 7*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + b*Sec[c + d*x]
)^3*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4131

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

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4159

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_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4179

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

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+b \sec (c+d x))^2 \left (\frac {1}{2} (6 A b+7 a B)+\frac {1}{2} (5 a A+7 b B+7 a C) \sec (c+d x)-\frac {1}{2} b (A-7 C) \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {(a+b \sec (c+d x)) \left (\frac {1}{4} \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right )+\frac {1}{4} \left (38 a A b+21 a^2 B+35 b^2 B+70 a b C\right ) \sec (c+d x)-\frac {1}{4} b (11 A b+7 a B-35 b C) \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {8}{105} \int \frac {-\frac {3}{8} \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right )-\frac {5}{8} \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \sec (c+d x)+\frac {3}{8} b^2 (11 A b+7 a B-35 b C) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {8}{105} \int \frac {-\frac {3}{8} \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right )+\frac {3}{8} b^2 (11 A b+7 a B-35 b C) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{21} \left (-21 a^2 b B-21 b^3 B-21 a b^2 (A+3 C)-a^3 (5 A+7 C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}-\frac {2 b^2 (11 A b+7 a B-35 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} \left (-3 a^3 B-15 a b^2 B-5 b^3 (A-C)-3 a^2 b (3 A+5 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{21} \left (\left (-21 a^2 b B-21 b^3 B-21 a b^2 (A+3 C)-a^3 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}-\frac {2 b^2 (11 A b+7 a B-35 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} \left (\left (-3 a^3 B-15 a b^2 B-5 b^3 (A-C)-3 a^2 b (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}-\frac {2 b^2 (11 A b+7 a B-35 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 (6 A b+7 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 5.22, size = 234, normalized size = 0.74 \begin {gather*} \frac {\sqrt {\sec (c+d x)} \left (168 \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+40 \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (42 \left (3 a^2 A b+a^3 B+10 b^3 C\right )+5 a \left (84 A b^2+84 a b B+a^2 (29 A+28 C)\right ) \cos (c+d x)+42 a^2 (3 A b+a B) \cos (2 (c+d x))+15 a^3 A \cos (3 (c+d x))\right ) \sin (c+d x)\right )}{420 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(Sqrt[Sec[c + d*x]]*(168*(3*a^3*B + 15*a*b^2*B + 5*b^3*(A - C) + 3*a^2*b*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*Ellip
ticE[(c + d*x)/2, 2] + 40*(21*a^2*b*B + 21*b^3*B + 21*a*b^2*(A + 3*C) + a^3*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2] + 2*(42*(3*a^2*A*b + a^3*B + 10*b^3*C) + 5*a*(84*A*b^2 + 84*a*b*B + a^2*(29*A + 28*C))
*Cos[c + d*x] + 42*a^2*(3*A*b + a*B)*Cos[2*(c + d*x)] + 15*a^3*A*Cos[3*(c + d*x)])*Sin[c + d*x]))/(420*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1277\) vs. \(2(341)=682\).
time = 0.14, size = 1278, normalized size = 4.03

method	result	size
default	\(\text {Expression too large to display}\)	\(1278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*(240*A*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^8*a^3
-24*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(15*A*a+21*A*b+7*B*a)*sin(1/2*d*x+1/2*c)^6*cos(1/
2*d*x+1/2*c)+28*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(10*A*a^2+18*A*a*b+15*A*b^2+6*B*a^2+15*
B*a*b+5*C*a^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
(40*A*a^3+63*A*a^2*b+105*A*a*b^2+21*B*a^3+105*B*a^2*b+35*C*a^3+105*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2
*c)+25*A*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+105*a*A*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*
x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
-189*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)
^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-105*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(
1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^3+1
05*a^2*b*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+105*b^3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-63*
B*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-315*B*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2+35*a^3
*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+315*C*b^2*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2
-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-315*C*(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b+105*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^3)/(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.77, size = 346, normalized size = 1.09 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (i \, {\left (5 \, A + 7 \, C\right )} a^{3} + 21 i \, B a^{2} b + 21 i \, {\left (A + 3 \, C\right )} a b^{2} + 21 i \, B b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 7 \, C\right )} a^{3} - 21 i \, B a^{2} b - 21 i \, {\left (A + 3 \, C\right )} a b^{2} - 21 i \, B b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (-3 i \, B a^{3} - 3 i \, {\left (3 \, A + 5 \, C\right )} a^{2} b - 15 i \, B a b^{2} - 5 i \, {\left (A - C\right )} b^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (3 i \, B a^{3} + 3 i \, {\left (3 \, A + 5 \, C\right )} a^{2} b + 15 i \, B a b^{2} + 5 i \, {\left (A - C\right )} b^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (15 \, A a^{3} \cos \left (d x + c\right )^{3} + 105 \, C b^{3} + 21 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left ({\left (5 \, A + 7 \, C\right )} a^{3} + 21 \, B a^{2} b + 21 \, A a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/105*(5*sqrt(2)*(I*(5*A + 7*C)*a^3 + 21*I*B*a^2*b + 21*I*(A + 3*C)*a*b^2 + 21*I*B*b^3)*weierstrassPInverse(-
4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*(5*A + 7*C)*a^3 - 21*I*B*a^2*b - 21*I*(A + 3*C)*a*b^2 - 2
1*I*B*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(-3*I*B*a^3 - 3*I*(3*A + 5*C
)*a^2*b - 15*I*B*a*b^2 - 5*I*(A - C)*b^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*s
in(d*x + c))) + 21*sqrt(2)*(3*I*B*a^3 + 3*I*(3*A + 5*C)*a^2*b + 15*I*B*a*b^2 + 5*I*(A - C)*b^3)*weierstrassZet
a(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(15*A*a^3*cos(d*x + c)^3 + 105*C*b^3 +
21*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)^2 + 5*((5*A + 7*C)*a^3 + 21*B*a^2*b + 21*A*a*b^2)*cos(d*x + c))*sin(d*x +
c)/sqrt(cos(d*x + c)))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{3} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(7/2),x)

[Out]

Integral((a + b*sec(c + d*x))**3*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sec(c + d*x)**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3/sec(d*x + c)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(7/2),x)

[Out]

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(7/2), x)

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