3.7.4 \(\int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [604]

Optimal. Leaf size=207 \[ \frac {2 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {28 a^3 b \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 b^2 \left (a^2-5 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/5*a^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2)+28/15*a^3*b*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/5*b^2*(a^
2-5*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(3*a^4+30*a^2*b^2-5*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*(a^2+3*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

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Rubi [A]
time = 0.26, antiderivative size = 207, 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 = {3926, 4159, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {28 a^3 b \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 b^2 \left (a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a b \left (a^2+3 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 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (3 a^4+30 a^2 b^2-5 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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4/Sec[c + d*x]^(5/2),x]

[Out]

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

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 3926

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^2*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

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]

Rubi steps

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

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Mathematica [A]
time = 0.39, size = 138, normalized size = 0.67 \begin {gather*} \frac {\sqrt {\sec (c+d x)} \left (12 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+80 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (3 a^4+30 b^4+40 a^3 b \cos (c+d x)+3 a^4 \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4/Sec[c + d*x]^(5/2),x]

[Out]

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

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Maple [A]
time = 0.16, size = 432, normalized size = 2.09

method	result	size
default	\(\frac {\frac {16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{5}-\frac {16 a^{4} \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {32 a^{3} \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{3}+\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{5}+\frac {16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3} b}{3}+4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{4}-\frac {8 b \,a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3}-8 b^{3} a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+\frac {6 a^{4} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{5}+12 b^{2} a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{4}}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(432\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^4/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/15*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a^4-24*a^4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-80*a^3*sin
(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)*b+6*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^4+40*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^2*a^3*b+30*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^4-20*b*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))-60*b^3*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))+9*a^4*(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))+90*b^2*a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*(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^4)/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]
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)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^4/sec(d*x + c)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.99, size = 215, normalized size = 1.04 \begin {gather*} -\frac {20 \, \sqrt {2} {\left (i \, a^{3} b + 3 i \, a b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 20 \, \sqrt {2} {\left (-i \, a^{3} b - 3 i \, a b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-3 i \, a^{4} - 30 i \, a^{2} b^{2} + 5 i \, b^{4}\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 ) + 3 \, \sqrt {2} {\left (3 i \, a^{4} + 30 i \, a^{2} b^{2} - 5 i \, b^{4}\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 (3 \, a^{4} \cos \left (d x + c\right )^{2} + 20 \, a^{3} b \cos \left (d x + c\right ) + 15 \, b^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \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)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(20*sqrt(2)*(I*a^3*b + 3*I*a*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 20*sqrt(2)
*(-I*a^3*b - 3*I*a*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(-3*I*a^4 - 30*I
*a^2*b^2 + 5*I*b^4)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt
(2)*(3*I*a^4 + 30*I*a^2*b^2 - 5*I*b^4)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(
d*x + c))) - 2*(3*a^4*cos(d*x + c)^2 + 20*a^3*b*cos(d*x + c) + 15*b^4)*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 )^{4}}{\sec ^{\frac {5}{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))**4/sec(d*x+c)**(5/2),x)

[Out]

Integral((a + b*sec(c + d*x))**4/sec(c + d*x)**(5/2), 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))^4/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^4/sec(d*x + c)^(5/2), 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 )}^4}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,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))^(5/2),x)

[Out]

int((a + b/cos(c + d*x))^4/(1/cos(c + d*x))^(5/2), x)

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