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

Optimal. Leaf size=263 \[ \frac {2 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}-\frac {2 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (7 a (a B+2 A b)+5 b^2 B\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 \left (7 a (a B+2 A b)+5 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b (9 a B+7 A b) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))}{7 d} \]

[Out]

2/21*(5*b^2*B+7*a*(2*A*b+B*a))*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/35*b*(7*A*b+9*B*a)*sec(d*x+c)^(5/2)*sin(d*x+c)/
d+2/7*b*B*sec(d*x+c)^(5/2)*(a+b*sec(d*x+c))*sin(d*x+c)/d+2/5*(5*A*a^2+3*A*b^2+6*B*a*b)*sin(d*x+c)*sec(d*x+c)^(
1/2)/d-2/5*(5*A*a^2+3*A*b^2+6*B*a*b)*(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*(5*b^2*B+7*a*(2*A*b+B*a))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c
os(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 = 0.37, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4026, 4047, 3768, 3771, 2641, 4046, 2639} \[ \frac {2 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}-\frac {2 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (7 a (a B+2 A b)+5 b^2 B\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 \left (7 a (a B+2 A b)+5 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b (9 a B+7 A b) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*
(5*b^2*B + 7*a*(2*A*b + a*B))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*(5*
a^2*A + 3*A*b^2 + 6*a*b*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*(5*b^2*B + 7*a*(2*A*b + a*B))*Sec[c + d
*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*b*(7*A*b + 9*a*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*b*B*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 4026

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(m + n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*Simp[a^2*A*(m + n) + a*b*B*n + (a
*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1))*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] &&  !
(IGtQ[n, 1] &&  !IntegerQ[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))^2 (A+B \sec (c+d x)) \, dx &=\frac {2 b B \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 (7 a A+3 b B)+\frac {1}{2} \left (5 b^2 B+7 a (2 A b+a B)\right ) \sec (c+d x)+\frac {1}{2} b (7 A b+9 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b B \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 (7 a A+3 b B)+\frac {1}{2} b (7 A b+9 a B) \sec ^2(c+d x)\right ) \, dx+\frac {1}{7} \left (5 b^2 B+7 a (2 A b+a B)\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx\\ &=\frac {2 \left (5 b^2 B+7 a (2 A b+a B)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b (7 A b+9 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{5} \left (5 a^2 A+3 A b^2+6 a b B\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{21} \left (5 b^2 B+7 a (2 A b+a B)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (5 b^2 B+7 a (2 A b+a B)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b (7 A b+9 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{5} \left (-5 a^2 A-3 A b^2-6 a b B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (\left (5 b^2 B+7 a (2 A b+a B)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (5 b^2 B+7 a (2 A b+a B)\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 (5 a^2 A+3 A b^2+6 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (5 b^2 B+7 a (2 A b+a B)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b (7 A b+9 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{5} \left (\left (-5 a^2 A-3 A b^2-6 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\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 \left (5 b^2 B+7 a (2 A b+a B)\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 (5 a^2 A+3 A b^2+6 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (5 b^2 B+7 a (2 A b+a B)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b (7 A b+9 a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b B \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 = 4.49, size = 221, normalized size = 0.84 \[ \frac {\sec ^{\frac {7}{2}}(c+d x) \left (40 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \cos ^{\frac {7}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-168 \left (5 a^2 A+6 a b B+3 A b^2\right ) \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) \left (21 \left (15 a^2 A+26 a b B+13 A b^2\right ) \cos (c+d x)+10 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \cos (2 (c+d x))+105 a^2 A \cos (3 (c+d x))+70 a^2 B+140 a A b+126 a b B \cos (3 (c+d x))+63 A b^2 \cos (3 (c+d x))+110 b^2 B\right )\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^(7/2)*(-168*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2] + 40*(14*
a*A*b + 7*a^2*B + 5*b^2*B)*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2] + 2*(140*a*A*b + 70*a^2*B + 110*b^2*B
+ 21*(15*a^2*A + 13*A*b^2 + 26*a*b*B)*Cos[c + d*x] + 10*(14*a*A*b + 7*a^2*B + 5*b^2*B)*Cos[2*(c + d*x)] + 105*
a^2*A*Cos[3*(c + d*x)] + 63*A*b^2*Cos[3*(c + d*x)] + 126*a*b*B*Cos[3*(c + d*x)])*Sin[c + d*x]))/(420*d)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b^{2} \sec \left (d x + c\right )^{4} + A a^{2} \sec \left (d x + c\right ) + {\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{3} + {\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )^{2}\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 16.16, size = 859, normalized size = 3.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),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^2*B*(-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
)))+2*a^2*A*(-(-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)+2*a*(2*A*b+B*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)/(-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)))-2/5*b*(A*b+2*B*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*El
lipticE(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))/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))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^2*(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))**2*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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