\(\int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [179]

Optimal result

Integrand size = 31, antiderivative size = 199 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {6 a (A+B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (7 A+5 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {6 a (A+B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a (A+B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

2/21*a*(7*A+5*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*(A+B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*a*B*sec(d*x+c)^(7
/2)*sin(d*x+c)/d+6/5*a*(A+B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-6/5*a*(A+B)*(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*a*(7*A+5*B)*(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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4082, 3872, 3853, 3856, 2720, 2719} \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 a (A+B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a (7 A+5 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {6 a (A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (7 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {6 a (A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d} \]

[In]

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

[Out]

(-6*a*(A + B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(7*A + 5*B)*Sqrt[C
os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (6*a*(A + B)*Sqrt[Sec[c + d*x]]*Sin[c + d*
x])/(5*d) + (2*a*(7*A + 5*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*a*(A + B)*Sec[c + d*x]^(5/2)*Sin[c +
d*x])/(5*d) + (2*a*B*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d)

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 3853

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

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 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4082

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2}{7} \int \sec ^{\frac {5}{2}}(c+d x) \left (\frac {1}{2} a (7 A+5 B)+\frac {7}{2} a (A+B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+(a (A+B)) \int \sec ^{\frac {7}{2}}(c+d x) \, dx+\frac {1}{7} (a (7 A+5 B)) \int \sec ^{\frac {5}{2}}(c+d x) \, dx \\ & = \frac {2 a (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a (A+B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{5} (3 a (A+B)) \int \sec ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{21} (a (7 A+5 B)) \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {6 a (A+B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a (A+B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac {1}{5} (3 a (A+B)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (a (7 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a (7 A+5 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {6 a (A+B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a (A+B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac {1}{5} \left (3 a (A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {6 a (A+B) \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 a (7 A+5 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {6 a (A+B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a (A+B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.26 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x)) (A+B \sec (c+d x)) \left (-63 (A+B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 (7 A+5 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+63 A \sin (c+d x)+63 B \sin (c+d x)+35 A \tan (c+d x)+25 B \tan (c+d x)+21 A \sec (c+d x) \tan (c+d x)+21 B \sec (c+d x) \tan (c+d x)+15 B \sec ^2(c+d x) \tan (c+d x)\right )}{105 d (B+A \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(a*Sec[(c + d*x)/2]^2*(1 + Sec[c + d*x])*(A + B*Sec[c + d*x])*(-63*(A + B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d
*x)/2, 2] + 5*(7*A + 5*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 63*A*Sin[c + d*x] + 63*B*Sin[c + d*x]
+ 35*A*Tan[c + d*x] + 25*B*Tan[c + d*x] + 21*A*Sec[c + d*x]*Tan[c + d*x] + 21*B*Sec[c + d*x]*Tan[c + d*x] + 1
5*B*Sec[c + d*x]^2*Tan[c + d*x]))/(105*d*(B + A*Cos[c + d*x])*Sec[c + d*x]^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(690\) vs. \(2(223)=446\).

Time = 41.64 (sec) , antiderivative size = 691, normalized size of antiderivative = 3.47

method	result	size
default	\(\text {Expression too large to display}\)	\(691\)
parts	\(\text {Expression too large to display}\)	\(824\)

[In]

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

[Out]

-a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*
x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
)+2*B*(-1/56*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)/(cos(1/2*d*x+1/2*c)^2-1/2
)^4-5/42*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+
5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+4/5*(1/2*A+1/2*B)/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)
^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4
-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))*(-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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.17 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {-5 i \, \sqrt {2} {\left (7 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 63 i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 63 i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (63 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{2} + 21 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 15 \, B a\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

1/105*(-5*I*sqrt(2)*(7*A + 5*B)*a*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5
*I*sqrt(2)*(7*A + 5*B)*a*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 63*I*sqrt(
2)*(A + B)*a*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)))
+ 63*I*sqrt(2)*(A + B)*a*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin
(d*x + c))) + 2*(63*(A + B)*a*cos(d*x + c)^3 + 5*(7*A + 5*B)*a*cos(d*x + c)^2 + 21*(A + B)*a*cos(d*x + c) + 15
*B*a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

[In]

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))*(1/cos(c + d*x))^(5/2),x)

[Out]

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))*(1/cos(c + d*x))^(5/2), x)