3.219 \(\int \frac {(a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=204 \[ \frac {2 a^2 (33 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {8 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^2 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

2/7*A*(a+a*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(5/2)+8/35*A*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(3/
2)+2/105*a^2*(33*A+35*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4/5*a^2*(3*A+5*C)*(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+8/21*a^2*(3*A+7*C)*(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 = 0.42, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4087, 4017, 3996, 3787, 3771, 2639, 2641} \[ \frac {2 a^2 (33 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {8 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^2 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(4*a^2*(3*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (8*a^2*(3*A + 7*C)
*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^2*(33*A + 35*C)*Sin[c + d*x])/
(105*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (8*A*(a^2 +
a^2*Sec[c + d*x])*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2))

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]

Rule 4017

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

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^2 \left (2 a A+\frac {1}{2} a (A+7 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \sec (c+d x)) \left (\frac {1}{4} a^2 (33 A+35 C)+\frac {1}{4} a^2 (9 A+35 C) \sec (c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac {2 a^2 (33 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {8 \int \frac {-\frac {21}{4} a^3 (3 A+5 C)-\frac {5}{2} a^3 (3 A+7 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{105 a}\\ &=\frac {2 a^2 (33 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (2 a^2 (3 A+5 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (4 a^2 (3 A+7 C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 a^2 (33 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (2 a^2 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (4 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^2 (3 A+5 C) \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 {8 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a^2 (33 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [C]  time = 2.43, size = 189, normalized size = 0.93 \[ \frac {a^2 e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (-56 i (3 A+5 C) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+\cos (c+d x) (5 (51 A+28 C) \sin (c+d x)+84 A \sin (2 (c+d x))+15 A \sin (3 (c+d x))+504 i A+840 i C)+80 (3 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(a^2*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(80*(3*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] -
(56*I)*(3*A + 5*C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(
c + d*x))] + Cos[c + d*x]*((504*I)*A + (840*I)*C + 5*(51*A + 28*C)*Sin[c + d*x] + 84*A*Sin[2*(c + d*x)] + 15*A
*Sin[3*(c + d*x)])))/(210*d*E^(I*d*x))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{2} \sec \left (d x + c\right )^{4} + 2 \, C a^{2} \sec \left (d x + c\right )^{3} + {\left (A + C\right )} a^{2} \sec \left (d x + c\right )^{2} + 2 \, A a^{2} \sec \left (d x + c\right ) + A a^{2}}{\sec \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((C*a^2*sec(d*x + c)^4 + 2*C*a^2*sec(d*x + c)^3 + (A + C)*a^2*sec(d*x + c)^2 + 2*A*a^2*sec(d*x + c) +
A*a^2)/sec(d*x + c)^(7/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(7/2), x)

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maple [A]  time = 4.53, size = 380, normalized size = 1.86 \[ -\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{2} \left (120 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-348 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (378 A +70 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-117 A -35 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+30 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+70 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x)

[Out]

-4/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(120*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8-348*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(378*A+70*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-117*A
-35*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+30*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))-63*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)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))+70*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-105*C*(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)/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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(7/2),x)

[Out]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {A}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 C}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int C \sqrt {\sec {\left (c + d x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(7/2),x)

[Out]

a**2*(Integral(A/sec(c + d*x)**(7/2), x) + Integral(2*A/sec(c + d*x)**(5/2), x) + Integral(A/sec(c + d*x)**(3/
2), x) + Integral(C/sec(c + d*x)**(3/2), x) + Integral(2*C/sqrt(sec(c + d*x)), x) + Integral(C*sqrt(sec(c + d*
x)), x))

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