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

Optimal. Leaf size=115 \[ \frac {2 (a A+3 b B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 (a B+A b) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

2/3*a*A*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2*(A*b+B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(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+2/3*(A*a+3*B*b)*(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 = 0.15, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3996, 3787, 3771, 2639, 2641} \[ \frac {2 (a A+3 b B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 (a B+A b) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 90, normalized size = 0.78 \[ \frac {\sqrt {\sec (c+d x)} \left (2 (a A+3 b B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 (a B+A b) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+a A \sin (2 (c+d x))\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(6*(A*b + a*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 2*(a*A + 3*b*B)*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2] + a*A*Sin[2*(c + d*x)]))/(3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 4.17, size = 326, normalized size = 2.83 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 A a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b -2 A a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 B b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*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))*b-2*A*a*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+3*B*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*(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))*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*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 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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