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

Optimal. Leaf size=244 \[ \frac {2 b \left (14 a^2 B+15 a A b+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d} \]

[Out]

2/15*b^2*(5*A*b+9*B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*b*(15*A*a*b+14*B*a^2+3*B*b^2)*sin(d*x+c)*sec(d*x+c)^(
1/2)/d+2/5*b*B*(a+b*sec(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(5*A*a^3-15*A*a*b^2-15*B*a^2*b-3*B*b^3)*(c
os(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+2/3*(9*A*a^2*b+A*b^3+3*B*a^3+3*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(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 = 0.48, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4026, 4076, 4047, 3771, 2641, 4046, 2639} \[ \frac {2 b \left (14 a^2 B+15 a A b+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 A b+3 a^3 B+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a^3*A - 15*a*A*b^2 - 15*a^2*b*B - 3*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x
]])/(5*d) + (2*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec
[c + d*x]])/(3*d) + (2*b*(15*a*A*b + 14*a^2*B + 3*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*b^2*(5*A*
b + 9*a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*b*B*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c +
d*x])/(5*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 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]

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 \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx &=\frac {2 b B \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 (5 a A-b B)+\frac {1}{2} \left (3 b^2 B+5 a (2 A b+a B)\right ) \sec (c+d x)+\frac {1}{2} b (5 A b+9 a B) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \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 (5 a A-b B)+\frac {5}{4} \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sec (c+d x)+\frac {3}{4} b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \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 (5 a A-b B)+\frac {3}{4} b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 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 (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\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 \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \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^3 A-15 a A b^2-15 a^2 b B-3 b^3 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^3 A-15 a A b^2-15 a^2 b B-3 b^3 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 (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\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 \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 2.55, size = 190, normalized size = 0.78 \[ \frac {\sec ^{\frac {5}{2}}(c+d x) \left (2 b \sin (c+d x) \left (9 \left (5 a^2 B+5 a A b+b^2 B\right ) \cos (2 (c+d x))+15 \left (3 a^2 B+3 a A b+b^2 B\right )+10 b (3 a B+A b) \cos (c+d x)\right )+20 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+12 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^(5/2)*(12*(5*a^3*A - 15*a*A*b^2 - 15*a^2*b*B - 3*b^3*B)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2
, 2] + 20*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^2*B)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*b*(15*(3*
a*A*b + 3*a^2*B + b^2*B) + 10*b*(A*b + 3*a*B)*Cos[c + d*x] + 9*(5*a*A*b + 5*a^2*B + b^2*B)*Cos[2*(c + d*x)])*S
in[c + d*x]))/(30*d)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B b^{3} \sec \left (d x + c\right )^{4} + A a^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} \sec \left (d x + c\right )^{3} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} \sec \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \sec \left (d x + c\right )}{\sqrt {\sec \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 13.16, size = 997, normalized size = 4.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),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*A*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)*(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*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*A*a^2*b*(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^3*B*(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*b*(A*b+B*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*b^2*(A*b+3*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^3*B/(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))/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 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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