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

Optimal. Leaf size=234 \[ \frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 a \left (5 a^2+9 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 a \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))}{7 d}+\frac {32 a b^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d} \]

[Out]

2/21*b*(21*a^2+5*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+32/35*a*b^2*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*b^2*sec(d*x+
c)^(5/2)*(a+b*sec(d*x+c))*sin(d*x+c)/d+2/5*a*(5*a^2+9*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*a*(5*a^2+9*b^2)*(
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+2/21*b*(21*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 = 0.24, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3842, 4047, 3768, 3771, 2641, 4046, 2639} \[ \frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 a \left (5 a^2+9 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 a \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))}{7 d}+\frac {32 a b^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(5*a^2 + 9*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*b*(21*a^2 +
5*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*(5*a^2 + 9*b^2)*Sqrt[Sec
[c + d*x]]*Sin[c + d*x])/(5*d) + (2*b*(21*a^2 + 5*b^2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (32*a*b^2*Sec
[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*b^2*Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])*Sin[c + d*x])/(7*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]

Rubi steps

\begin {align*} \int \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}+\frac {2}{7} \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a \left (7 a^2+3 b^2\right )+\frac {1}{2} b \left (21 a^2+5 b^2\right ) \sec (c+d x)+8 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}+\frac {2}{7} \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a \left (7 a^2+3 b^2\right )+8 a b^2 \sec ^2(c+d x)\right ) \, dx+\frac {1}{7} \left (b \left (21 a^2+5 b^2\right )\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx\\ &=\frac {2 b \left (21 a^2+5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {32 a b^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{21} \left (b \left (21 a^2+5 b^2\right )\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (a \left (5 a^2+9 b^2\right )\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 a \left (5 a^2+9 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {32 a b^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}-\frac {1}{5} \left (a \left (5 a^2+9 b^2\right )\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (b \left (21 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 {2 b \left (21 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 a \left (5 a^2+9 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {32 a b^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}-\frac {1}{5} \left (a \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 a \left (5 a^2+9 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 b \left (21 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 a \left (5 a^2+9 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {32 a b^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 3.74, size = 177, normalized size = 0.76 \[ \frac {\sec ^{\frac {7}{2}}(c+d x) \left (40 b \left (21 a^2+5 b^2\right ) \cos ^{\frac {7}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-168 a \left (5 a^2+9 b^2\right ) \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) \left (105 a^3 \cos (3 (c+d x))+10 \left (21 a^2 b+5 b^3\right ) \cos (2 (c+d x))+63 a \left (5 a^2+13 b^2\right ) \cos (c+d x)+210 a^2 b+189 a b^2 \cos (3 (c+d x))+110 b^3\right )\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^(7/2)*(-168*a*(5*a^2 + 9*b^2)*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2] + 40*b*(21*a^2 + 5*b^
2)*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2] + 2*(210*a^2*b + 110*b^3 + 63*a*(5*a^2 + 13*b^2)*Cos[c + d*x]
+ 10*(21*a^2*b + 5*b^3)*Cos[2*(c + d*x)] + 105*a^3*Cos[3*(c + d*x)] + 189*a*b^2*Cos[3*(c + d*x)])*Sin[c + d*x]
))/(420*d)

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fricas [F]  time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} \sec \left (d x + c\right )^{4} + 3 \, a b^{2} \sec \left (d x + c\right )^{3} + 3 \, a^{2} b \sec \left (d x + c\right )^{2} + a^{3} \sec \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^3*sec(d*x + c)^4 + 3*a*b^2*sec(d*x + c)^3 + 3*a^2*b*sec(d*x + c)^2 + a^3*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 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 12.80, size = 847, normalized size = 3.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*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))
)+6*a^2*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)))-6/5*b^2*a/(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^3*(-(-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)*EllipticE(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))/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))^3,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a + b/cos(c + d*x))^3*(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))**3,x)

[Out]

Timed out

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