3.7.2 \(\int \frac {(a+b \sec (c+d x))^4}{\sqrt {\sec (c+d x)}} \, dx\) [602]

Optimal. Leaf size=209 \[ \frac {2 \left (5 a^4-30 a^2 b^2-3 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a b \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b^2 \left (29 a^2+3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {28 a b^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b^2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d} \]

[Out]

28/15*a*b^3*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*b^2*(29*a^2+3*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*b^2*(a+b*se
c(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(5*a^4-30*a^2*b^2-3*b^4)*(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+8/3*a*b*(3*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]
time = 0.25, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3927, 4161, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 b^2 \left (29 a^2+3 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a b \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 \left (5 a^4-30 a^2 b^2-3 b^4\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 {28 a b^3 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b^2 \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4/Sqrt[Sec[c + d*x]],x]

[Out]

(2*(5*a^4 - 30*a^2*b^2 - 3*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (8*a*
b*(3*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b^2*(29*a^2 + 3*b^
2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (28*a*b^3*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*b^2*Sqrt[Se
c[c + d*x]]*(a + b*Sec[c + d*x])^2*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 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]

Rule 4161

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

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^4}{\sqrt {\sec (c+d x)}} \, dx &=\frac {2 b^2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2}{5} \int \frac {(a+b \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)+7 a b^2 \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {28 a b^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b^2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {3}{4} a^2 \left (5 a^2-b^2\right )+5 a b \left (3 a^2+b^2\right ) \sec (c+d x)+\frac {3}{4} b^2 \left (29 a^2+3 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {28 a b^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b^2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {3}{4} a^2 \left (5 a^2-b^2\right )+\frac {3}{4} b^2 \left (29 a^2+3 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (4 a b \left (3 a^2+b^2\right )\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 b^2 \left (29 a^2+3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {28 a b^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b^2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (5 a^4-30 a^2 b^2-3 b^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (4 a b \left (3 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 {8 a b \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b^2 \left (29 a^2+3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {28 a b^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b^2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (\left (5 a^4-30 a^2 b^2-3 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (5 a^4-30 a^2 b^2-3 b^4\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 {8 a b \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b^2 \left (29 a^2+3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {28 a b^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b^2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.43, size = 146, normalized size = 0.70 \begin {gather*} \frac {\sec ^{\frac {5}{2}}(c+d x) \left (12 \left (5 a^4-30 a^2 b^2-3 b^4\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+b \left (80 a \left (3 a^2+b^2\right ) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 b \left (15 \left (6 a^2+b^2\right )+40 a b \cos (c+d x)+9 \left (10 a^2+b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )\right )}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4/Sqrt[Sec[c + d*x]],x]

[Out]

(Sec[c + d*x]^(5/2)*(12*(5*a^4 - 30*a^2*b^2 - 3*b^4)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + b*(80*a*(3
*a^2 + b^2)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*b*(15*(6*a^2 + b^2) + 40*a*b*Cos[c + d*x] + 9*(10
*a^2 + b^2)*Cos[2*(c + d*x)])*Sin[c + d*x])))/(30*d)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(879\) vs. \(2(237)=474\).
time = 0.30, size = 880, normalized size = 4.21

method	result	size
default	\(\text {Expression too large to display}\)	\(880\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^4/sec(d*x+c)^(1/2),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^4*(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))
-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*a^4*(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))+8*b*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)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))+8*b^3*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*si
n(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^4/(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*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)^
(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos
(1/2*d*x+1/2*c)-3*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+12*b^2*a^2/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*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^4/sqrt(sec(d*x + c)), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.86, size = 264, normalized size = 1.26 \begin {gather*} -\frac {20 \, \sqrt {2} {\left (3 i \, a^{3} b + i \, a b^{3}\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 ) + 20 \, \sqrt {2} {\left (-3 i \, a^{3} b - i \, a b^{3}\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^{4} + 30 i \, a^{2} b^{2} + 3 i \, b^{4}\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^{4} - 30 i \, a^{2} b^{2} - 3 i \, b^{4}\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 (20 \, a b^{3} \cos \left (d x + c\right ) + 3 \, b^{4} + 9 \, {\left (10 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/15*(20*sqrt(2)*(3*I*a^3*b + I*a*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
)) + 20*sqrt(2)*(-3*I*a^3*b - I*a*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)
) + 3*sqrt(2)*(-5*I*a^4 + 30*I*a^2*b^2 + 3*I*b^4)*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^4 - 30*I*a^2*b^2 - 3*I*b^4)*cos(d*x + c)^2*weierstrass
Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(20*a*b^3*cos(d*x + c) + 3*b^4 + 9*
(10*a^2*b^2 + b^4)*cos(d*x + c)^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**4/sec(d*x+c)**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 deep

________________________________________________________________________________________

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))^4/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^4/sqrt(sec(d*x + c)), x)

________________________________________________________________________________________

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 )}^4}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^4/(1/cos(c + d*x))^(1/2),x)

[Out]

int((a + b/cos(c + d*x))^4/(1/cos(c + d*x))^(1/2), x)

________________________________________________________________________________________