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

Optimal. Leaf size=172 \[ \frac{2 \left (3 a^2 A+2 a b B+A b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

[Out]

(-2*(10*a*A*b + 5*a^2*B + 3*b^2*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(3*a^2*A + A*b^2 + 2*a*b*B)*EllipticF
[(c + d*x)/2, 2])/(3*d) + (2*b^2*B*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*b*(A*b + 2*a*B)*Sin[c + d*x])/(
3*d*Cos[c + d*x]^(3/2)) + (2*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.374806, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2954, 2988, 3021, 2748, 2636, 2639, 2641} \[ \frac{2 \left (3 a^2 A+2 a b B+A b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(10*a*A*b + 5*a^2*B + 3*b^2*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(3*a^2*A + A*b^2 + 2*a*b*B)*EllipticF
[(c + d*x)/2, 2])/(3*d) + (2*b^2*B*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*b*(A*b + 2*a*B)*Sin[c + d*x])/(
3*d*Cos[c + d*x]^(3/2)) + (2*(10*a*A*b + 5*a^2*B + 3*b^2*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 2988

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/
(f*d^2*(n + 1)*(c^2 - d^2)), x] - Dist[1/(d^2*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*Simp[d*(n
+ 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n +
1))) + 2*a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x
]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && LtQ[n, -1]

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 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]

Rubi steps

\begin{align*} \int \frac{(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt{\cos (c+d x)}} \, dx &=\int \frac{(b+a \cos (c+d x))^2 (B+A \cos (c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2}{5} \int \frac{-\frac{5}{2} b (A b+2 a B)-\frac{1}{2} \left (10 a A b+5 a^2 B+3 b^2 B\right ) \cos (c+d x)-\frac{5}{2} a^2 A \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{4}{15} \int \frac{-\frac{3}{4} \left (10 a A b+5 a^2 B+3 b^2 B\right )-\frac{5}{4} \left (3 a^2 A+A b^2+2 a b B\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{1}{3} \left (-3 a^2 A-A b^2-2 a b B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} \left (-10 a A b-5 a^2 B-3 b^2 B\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 \left (3 a^2 A+A b^2+2 a b B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}-\frac{1}{5} \left (10 a A b+5 a^2 B+3 b^2 B\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (10 a A b+5 a^2 B+3 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (3 a^2 A+A b^2+2 a b B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.12825, size = 175, normalized size = 1.02 \[ \frac{10 \left (3 a^2 A+2 a b B+A b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-6 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+15 a^2 B \sin (2 (c+d x))+30 a A b \sin (2 (c+d x))+20 a b B \sin (c+d x)+10 A b^2 \sin (c+d x)+9 b^2 B \sin (2 (c+d x))+6 b^2 B \tan (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*(3*a^2*A + A*b^2 + 2*a*b*
B)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 10*A*b^2*Sin[c + d*x] + 20*a*b*B*Sin[c + d*x] + 30*a*A*b*Sin
[2*(c + d*x)] + 15*a^2*B*Sin[2*(c + d*x)] + 9*b^2*B*Sin[2*(c + d*x)] + 6*b^2*B*Tan[c + d*x])/(15*d*Cos[c + d*x
]^(3/2))

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Maple [B]  time = 6.644, size = 750, normalized size = 4.4 \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))^2*(A+B*sec(d*x+c))/cos(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^2*A*(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*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*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)*(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*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)+2*b*(A*b+2*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*(si
n(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*b+B*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*s
in(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., 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))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),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^{2} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))/sq
rt(cos(d*x + c)), 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))**2*(A+B*sec(d*x+c))/cos(d*x+c)**(1/2),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 )}^{2}}{\sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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