3.567 \(\int \frac{(a+b \sec (c+d x)) (A+B \sec (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=140 \[ \frac{2 (a B+A b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{2 (5 a A+3 b B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 a A+3 b B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 b B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
[Out]
(-2*(5*a*A + 3*b*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(A*b + a*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*b*
B*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(A*b + a*B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(5*a*A +
3*b*B)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])
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Rubi [A] time = 0.231375, antiderivative size = 140, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used =
{2954, 2968, 3021, 2748, 2636, 2641, 2639} \[ \frac{2 (a B+A b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 (5 a A+3 b B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 a A+3 b B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 b B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]
[Out]
(-2*(5*a*A + 3*b*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(A*b + a*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*b*
B*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(A*b + a*B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(5*a*A +
3*b*B)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])
Rule 2954
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3021
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Rule 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 2636
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]
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 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]
Rubi steps
\begin{align*} \int \frac{(a+b \sec (c+d x)) (A+B \sec (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\int \frac{(b+a \cos (c+d x)) (B+A \cos (c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\int \frac{b B+(A b+a B) \cos (c+d x)+a A \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 b B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{\frac{5}{2} (A b+a B)+\frac{1}{2} (5 a A+3 b B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+(A b+a B) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx+\frac{1}{5} (5 a A+3 b B) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 a A+3 b B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{1}{3} (A b+a B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} (-5 a A-3 b B) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 (5 a A+3 b B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (A b+a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 a A+3 b B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.842185, size = 134, normalized size = 0.96 \[ \frac{10 (a B+A b) \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-6 (5 a A+3 b B) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+15 a A \sin (2 (c+d x))+10 a B \sin (c+d x)+10 A b \sin (c+d x)+9 b B \sin (2 (c+d x))+6 b B \tan (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]
[Out]
(-6*(5*a*A + 3*b*B)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*(A*b + a*B)*Cos[c + d*x]^(3/2)*EllipticF
[(c + d*x)/2, 2] + 10*A*b*Sin[c + d*x] + 10*a*B*Sin[c + d*x] + 15*a*A*Sin[2*(c + d*x)] + 9*b*B*Sin[2*(c + d*x)
] + 6*b*B*Tan[c + d*x])/(15*d*Cos[c + d*x]^(3/2))
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Maple [B] time = 5.971, size = 663, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),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*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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*A*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(
1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(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)-2/5*B*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*sin(1/
2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-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)*(s
in(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*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)-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., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B b \sec \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \sec \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{3}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
integral((B*b*sec(d*x + c)^2 + A*a + (B*a + A*b)*sec(d*x + c))/cos(d*x + c)^(3/2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/cos(d*x+c)**(3/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)/cos(d*x + c)^(3/2), x)