\(\int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx\) [594]

Optimal result

Integrand size = 23, antiderivative size = 189 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx=-\frac {6 b \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {6 b \left (5 a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 b^2 \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d} \]

[Out]

8/5*a*b^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*b^2*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))*sin(d*x+c)/d+6/5*b*(5*a^2+b^
2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-6/5*b*(5*a^2+b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(s
in(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2*a*(a^2+b^2)*(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 = 0.23 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3927, 4132, 3853, 3856, 2719, 4131, 2720} \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx=\frac {6 b \left (5 a^2+b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 b \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d} \]

[In]

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

[Out]

(-6*b*(5*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(a^2 + b^2)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (6*b*(5*a^2 + b^2)*Sqrt[Sec[c + d*x]]*Sin
[c + d*x])/(5*d) + (8*a*b^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*b^2*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 3927

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-b^2)
*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Dist[1/(d*(m + n - 1)),
Int[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2)
+ 3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e,
f, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Int
egerQ[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^2 \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 \left (5 a^2+b^2\right )+\frac {3}{2} b \left (5 a^2+b^2\right ) \sec (c+d x)+6 a b^2 \sec ^2(c+d x)\right ) \, dx \\ & = \frac {2 b^2 \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 \left (5 a^2+b^2\right )+6 a b^2 \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (3 b \left (5 a^2+b^2\right )\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {6 b \left (5 a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 b^2 \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\left (a \left (a^2+b^2\right )\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{5} \left (3 b \left (5 a^2+b^2\right )\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {6 b \left (5 a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 b^2 \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\left (a \left (a^2+b^2\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 (3 b \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {6 b \left (5 a^2+b^2\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 a \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {6 b \left (5 a^2+b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 b^2 \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 = 1.68 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.71 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx=\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 b \left (5 a^2+b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 a \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {b \left (5 \left (3 a^2+b^2\right )+10 a b \cos (c+d x)+3 \left (5 a^2+b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}\right )}{5 d} \]

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-3*b*(5*a^2 + b^2)*EllipticE[(c + d*x)/2, 2] + 5*a*(a^2 + b^2)*Ellip
ticF[(c + d*x)/2, 2] + (b*(5*(3*a^2 + b^2) + 10*a*b*Cos[c + d*x] + 3*(5*a^2 + b^2)*Cos[2*(c + d*x)])*Sin[c + d
*x])/(2*Cos[c + d*x]^(5/2))))/(5*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(219)=438\).

Time = 37.53 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.76

method	result	size
default	\(\text {Expression too large to display}\)	\(711\)
parts	\(\text {Expression too large to display}\)	\(898\)

[In]

int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3,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^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))+
2/5*b^3/(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)+6*a*b^2*(-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)))+6*a^2*b/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)))/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 = 0.10 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.29 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx=-\frac {5 \, \sqrt {2} {\left (i \, a^{3} + i \, a b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{3} - i \, a b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, a^{2} b + i \, b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, a^{2} b - i \, b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (5 \, a b^{2} \cos \left (d x + c\right ) + b^{3} + 3 \, {\left (5 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{5 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx=\int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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