\(\int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [401]

Optimal result

Integrand size = 33, antiderivative size = 221 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^2 A+A b^2+2 a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 A b+7 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d} \]

[Out]

2/15*b*(5*A*b+7*B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*b*B*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))*sin(d*x+c)/d+2/5*
(3*b^2*B+5*a*(2*A*b+B*a))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*(3*b^2*B+5*a*(2*A*b+B*a))*(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+2/3*(3*A*a^2
+A*b^2+2*B*a*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 = 0.35 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4111, 4132, 3853, 3856, 2719, 4131, 2720} \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2 \left (3 a^2 A+2 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \left (5 a (a B+2 A b)+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}-\frac {2 \left (5 a (a B+2 A b)+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d} \]

[In]

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

[Out]

(-2*(3*b^2*B + 5*a*(2*A*b + a*B))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*
(3*a^2*A + A*b^2 + 2*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*(3*b^2
*B + 5*a*(2*A*b + a*B))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*b*(5*A*b + 7*a*B)*Sec[c + d*x]^(3/2)*Sin[c
+ d*x])/(15*d) + (2*b*B*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])*Sin[c + d*x])/(5*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 4111

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\sec (c+d x)} \left (\frac {1}{2} a (5 a A+b B)+\frac {1}{2} \left (3 b^2 B+5 a (2 A b+a B)\right ) \sec (c+d x)+\frac {1}{2} b (5 A b+7 a B) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\sec (c+d x)} \left (\frac {1}{2} a (5 a A+b B)+\frac {1}{2} b (5 A b+7 a B) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (3 b^2 B+5 a (2 A b+a B)\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 A b+7 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{3} \left (3 a^2 A+A b^2+2 a b B\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (-3 b^2 B-5 a (2 A b+a B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 A b+7 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{3} \left (\left (3 a^2 A+A b^2+2 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (\left (-3 b^2 B-5 a (2 A b+a B)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {2 \left (3 b^2 B+5 a (2 A b+a B)\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 (3 a^2 A+A b^2+2 a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 A b+7 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.99 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.77 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\sec ^{\frac {5}{2}}(c+d x) \left (-12 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (3 a^2 A+A b^2+2 a b B\right ) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 \left (15 \left (2 a A b+a^2 B+b^2 B\right )+10 b (A b+2 a B) \cos (c+d x)+3 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{30 d} \]

[In]

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

[Out]

(Sec[c + d*x]^(5/2)*(-12*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 20*(3*a
^2*A + A*b^2 + 2*a*b*B)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*(15*(2*a*A*b + a^2*B + b^2*B) + 10*b*
(A*b + 2*a*B)*Cos[c + d*x] + 3*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(30*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(722\) vs. \(2(249)=498\).

Time = 40.82 (sec) , antiderivative size = 723, normalized size of antiderivative = 3.27

method	result	size
default	\(\text {Expression too large to display}\)	\(723\)
parts	\(\text {Expression too large to display}\)	\(914\)

[In]

int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*a^2*(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)
)+2/5*b^2*B/(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)/sin(1/2*d*x+1/2*c)^2*(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)+2*a*(2*A*b+B*a)/sin(
1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(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*b*(A*b+2*B*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))))/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 = 0.12 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.29 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, A a^{2} + 2 i \, B a b + i \, A b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, A a^{2} - 2 i \, B a b - i \, A b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{2} + 10 i \, A a b + 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{2} - 10 i \, A a b - 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, B b^{2} + 3 \, {\left (5 \, B a^{2} + 10 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-1/15*(5*sqrt(2)*(3*I*A*a^2 + 2*I*B*a*b + I*A*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*
sin(d*x + c)) + 5*sqrt(2)*(-3*I*A*a^2 - 2*I*B*a*b - I*A*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x
+ c) - I*sin(d*x + c)) + 3*sqrt(2)*(5*I*B*a^2 + 10*I*A*a*b + 3*I*B*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0,
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-5*I*B*a^2 - 10*I*A*a*b - 3*I*B*b^2)*
cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*B*b^2
+ 3*(5*B*a^2 + 10*A*a*b + 3*B*b^2)*cos(d*x + c)^2 + 5*(2*B*a*b + A*b^2)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*
x + c)))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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