3.600 \(\int \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=287 \[ \frac {14 b^2 \left (7 a^2+b^2\right ) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{45 d}+\frac {8 a b \left (7 a^2+5 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {8 a b \left (7 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (15 a^4+54 a^2 b^2+7 b^4\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}-\frac {2 \left (15 a^4+54 a^2 b^2+7 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {44 a b^3 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}{9 d} \]

[Out]

8/21*a*b*(7*a^2+5*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+14/45*b^2*(7*a^2+b^2)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+44/63
*a*b^3*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/9*b^2*sec(d*x+c)^(5/2)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+2/15*(15*a^4+54*
a^2*b^2+7*b^4)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/15*(15*a^4+54*a^2*b^2+7*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+8/21*a*b*(7*a^2+5*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 = 0.41, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3842, 4076, 4047, 3768, 3771, 2641, 4046, 2639} \[ \frac {14 b^2 \left (7 a^2+b^2\right ) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{45 d}+\frac {8 a b \left (7 a^2+5 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 \left (54 a^2 b^2+15 a^4+7 b^4\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {8 a b \left (7 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 \left (54 a^2 b^2+15 a^4+7 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {44 a b^3 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^4,x]

[Out]

(-2*(15*a^4 + 54*a^2*b^2 + 7*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (8
*a*b*(7*a^2 + 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*(15*a^4 + 54
*a^2*b^2 + 7*b^4)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (8*a*b*(7*a^2 + 5*b^2)*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(21*d) + (14*b^2*(7*a^2 + b^2)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(45*d) + (44*a*b^3*Sec[c + d*x]^(7/2)*Si
n[c + d*x])/(63*d) + (2*b^2*Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(9*d)

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3768

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 3771

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 3842

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)), In
t[(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] &&  !Integ
erQ[m])

Rule 4046

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 4047

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 4076

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 \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2}{9} \int \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \left (\frac {3}{2} a \left (3 a^2+b^2\right )+\frac {1}{2} b \left (27 a^2+7 b^2\right ) \sec (c+d x)+11 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {44 a b^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {4}{63} \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {21}{4} a^2 \left (3 a^2+b^2\right )+9 a b \left (7 a^2+5 b^2\right ) \sec (c+d x)+\frac {49}{4} b^2 \left (7 a^2+b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {44 a b^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {4}{63} \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {21}{4} a^2 \left (3 a^2+b^2\right )+\frac {49}{4} b^2 \left (7 a^2+b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{7} \left (4 a b \left (7 a^2+5 b^2\right )\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx\\ &=\frac {8 a b \left (7 a^2+5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {14 b^2 \left (7 a^2+b^2\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {44 a b^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{21} \left (4 a b \left (7 a^2+5 b^2\right )\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{15} \left (15 a^4+54 a^2 b^2+7 b^4\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (15 a^4+54 a^2 b^2+7 b^4\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 a b \left (7 a^2+5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {14 b^2 \left (7 a^2+b^2\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {44 a b^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{15} \left (-15 a^4-54 a^2 b^2-7 b^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (4 a b \left (7 a^2+5 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 (7 a^2+5 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)}}{21 d}+\frac {2 \left (15 a^4+54 a^2 b^2+7 b^4\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 a b \left (7 a^2+5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {14 b^2 \left (7 a^2+b^2\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {44 a b^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{15} \left (\left (-15 a^4-54 a^2 b^2-7 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (15 a^4+54 a^2 b^2+7 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)}}{15 d}+\frac {8 a b \left (7 a^2+5 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)}}{21 d}+\frac {2 \left (15 a^4+54 a^2 b^2+7 b^4\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 a b \left (7 a^2+5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {14 b^2 \left (7 a^2+b^2\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {44 a b^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 2.43, size = 256, normalized size = 0.89 \[ -\frac {2 (a+b \sec (c+d x))^4 \left (-315 a^4 \sin (c+d x)-420 a^3 b \tan (c+d x)-1134 a^2 b^2 \sin (c+d x)-60 a b \left (7 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-378 a^2 b^2 \tan (c+d x) \sec (c+d x)+21 \left (15 a^4+54 a^2 b^2+7 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-300 a b^3 \tan (c+d x)-180 a b^3 \tan (c+d x) \sec ^2(c+d x)-147 b^4 \sin (c+d x)-35 b^4 \tan (c+d x) \sec ^3(c+d x)-49 b^4 \tan (c+d x) \sec (c+d x)\right )}{315 d \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+b)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^4,x]

[Out]

(-2*(a + b*Sec[c + d*x])^4*(21*(15*a^4 + 54*a^2*b^2 + 7*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] - 60
*a*b*(7*a^2 + 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - 315*a^4*Sin[c + d*x] - 1134*a^2*b^2*Sin[c
+ d*x] - 147*b^4*Sin[c + d*x] - 420*a^3*b*Tan[c + d*x] - 300*a*b^3*Tan[c + d*x] - 378*a^2*b^2*Sec[c + d*x]*Tan
[c + d*x] - 49*b^4*Sec[c + d*x]*Tan[c + d*x] - 180*a*b^3*Sec[c + d*x]^2*Tan[c + d*x] - 35*b^4*Sec[c + d*x]^3*T
an[c + d*x]))/(315*d*(b + a*Cos[c + d*x])^4*Sec[c + d*x]^(7/2))

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{4} \sec \left (d x + c\right )^{5} + 4 \, a b^{3} \sec \left (d x + c\right )^{4} + 6 \, a^{2} b^{2} \sec \left (d x + c\right )^{3} + 4 \, a^{3} b \sec \left (d x + c\right )^{2} + a^{4} \sec \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

integral((b^4*sec(d*x + c)^5 + 4*a*b^3*sec(d*x + c)^4 + 6*a^2*b^2*sec(d*x + c)^3 + 4*a^3*b*sec(d*x + c)^2 + a^
4*sec(d*x + c))*sqrt(sec(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^4*sec(d*x + c)^(3/2), x)

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maple [B]  time = 15.93, size = 1174, normalized size = 4.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^4,x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(8*a*b^3*(-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)/(-1/2+cos(1/2*d*x+1/2*c)^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)/(-1/2+cos(1/2*d*x+1/2*c)^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
)))-12/5*a^2*b^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)/sin(1/2*d*x+1/2*c)^
2*(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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*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)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+
1/2*c)+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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-8
*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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a^4*(-(-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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-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)*
sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)+8*a^3*b*(-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)/(-1/2+cos(1/2*d*x+1/2*c)^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^4*(-1/144*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)/(-1/2+cos
(1/2*d*x+1/2*c)^2)^5-7/180*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)/(-1/2+cos(1
/2*d*x+1/2*c)^2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2
*c)^2)^(1/2)+7/15*(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))-7/15*(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)))))/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(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^4\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^4*(1/cos(c + d*x))^(3/2),x)

[Out]

int((a + b/cos(c + d*x))^4*(1/cos(c + d*x))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(3/2)*(a+b*sec(d*x+c))**4,x)

[Out]

Timed out

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