3.5.3 \(\int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [403]

Optimal. Leaf size=161 \[ -\frac {2 \left (b^2 B-a (2 A b+a B)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^2 A+3 A b^2+6 a b B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 b^2 B \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]

[Out]

2/3*a^2*A*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2*b^2*B*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2*(b^2*B-a*(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(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*(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)*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 = 0.17, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4109, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 \left (a^2 A+6 a b B+3 A b^2\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 a^2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 \left (b^2 B-a (a B+2 A b)\right ) \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 b^2 B \sin (c+d x) \sqrt {\sec (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

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 4109

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

Rule 4131

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 4132

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 \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 a^2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2}{3} \int \frac {-\frac {3}{2} a (2 A b+a B)+\left (A \left (-\frac {a^2}{2}-\frac {3 b^2}{2}\right )-3 a b B\right ) \sec (c+d x)-\frac {3}{2} b^2 B \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a^2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2}{3} \int \frac {-\frac {3}{2} a (2 A b+a B)-\frac {3}{2} b^2 B \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{3} \left (-a^2 A-3 A b^2-6 a b B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 a^2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 b^2 B \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\left (b^2 B-a (2 A b+a B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{3} \left (\left (-a^2 A-3 A b^2-6 a b 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 (a^2 A+3 A b^2+6 a b 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 a^2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 b^2 B \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\left (\left (b^2 B-a (2 A b+a B)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (b^2 B-a (2 A b+a B)\right ) \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 \left (a^2 A+3 A b^2+6 a b 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 a^2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 b^2 B \sqrt {\sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 124, normalized size = 0.77 \begin {gather*} \frac {\sqrt {\sec (c+d x)} \left (6 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (a^2 A+3 A b^2+6 a b B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (3 b^2 B+a^2 A \cos (c+d x)\right ) \sin (c+d x)\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(403\) vs. \(2(197)=394\).
time = 1.82, size = 404, normalized size = 2.51

method	result	size
default	\(-\frac {2 \left (4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}-2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}+a^{2} A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+6 B a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(404\)

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))/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2-2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^2+a^2*A*(s
in(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*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-6*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))*a*b-6*B*cos(1/2*d*x+
1/2*c)*sin(1/2*d*x+1/2*c)^2*b^2+6*B*a*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+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))*b^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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))/sec(d*x+c)^(3/2),x, algorithm="maxima")

[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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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.61, size = 208, normalized size = 1.29 \begin {gather*} \frac {\sqrt {2} {\left (-i \, A a^{2} - 6 i \, B a b - 3 i \, A b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A a^{2} + 6 i \, B a b + 3 i \, A b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-i \, B a^{2} - 2 i \, A a b + i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, A a b - i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (A a^{2} \cos \left (d x + c\right ) + 3 \, B b^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, d} \end {gather*}

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))/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(-I*A*a^2 - 6*I*B*a*b - 3*I*A*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sq
rt(2)*(I*A*a^2 + 6*I*B*a*b + 3*I*A*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*
(-I*B*a^2 - 2*I*A*a*b + I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c))) - 3*sqrt(2)*(I*B*a^2 + 2*I*A*a*b - I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c
) - I*sin(d*x + c))) + 2*(A*a^2*cos(d*x + c) + 3*B*b^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/d

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

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))/sec(d*x+c)**(3/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))/sec(d*x+c)^(3/2),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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\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 \end {gather*}

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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