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

Optimal. Leaf size=242 \[ \frac {2 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+4 A b^2\right )}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right )}{21 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{5 d}+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

2/35*a*(4*A*b+7*B*a)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/7*A*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/21
*(4*A*b^2+14*a*b*B+a^2*(5*A+7*C))*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*(6*A*a*b+3*B*a^2+5*B*b^2+10*C*a*b)*(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+2/21*(14*a*b*B+7*b^2*(A+3*C)+a^2*(5*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 = 0.52, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4094, 4074, 4047, 3771, 2639, 4045, 2641} \[ \frac {2 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+4 A b^2\right )}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right )}{21 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{5 d}+\frac {2 a (7 a B+4 A b) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^2}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*a*A*b + 3*a^2*B + 5*b^2*B + 10*a*b*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(
5*d) + (2*(14*a*b*B + 7*b^2*(A + 3*C) + a^2*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec
[c + d*x]])/(21*d) + (2*a*(4*A*b + 7*a*B)*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*(4*A*b^2 + 14*a*b*B + a
^2*(5*A + 7*C))*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c
+ d*x]^(5/2))

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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4047

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]

Rule 4074

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4094

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+b \sec (c+d x)) \left (\frac {1}{2} (4 A b+7 a B)+\frac {1}{2} (5 a A+7 b B+7 a C) \sec (c+d x)+\frac {1}{2} b (A+7 C) \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a (4 A b+7 a B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {4}{35} \int \frac {-\frac {5}{4} \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right )-\frac {7}{4} \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) \sec (c+d x)-\frac {5}{4} b^2 (A+7 C) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a (4 A b+7 a B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {4}{35} \int \frac {-\frac {5}{4} \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right )-\frac {5}{4} b^2 (A+7 C) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx-\frac {1}{5} \left (-6 a A b-3 a^2 B-5 b^2 B-10 a b C\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a (4 A b+7 a B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {1}{21} \left (-14 a b B-7 b^2 (A+3 C)-a^2 (5 A+7 C)\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{5} \left (\left (-6 a A b-3 a^2 B-5 b^2 B-10 a b C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) \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 {2 a (4 A b+7 a B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {1}{21} \left (\left (-14 a b B-7 b^2 (A+3 C)-a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) \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 {2 \left (14 a b B+7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \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 (4 A b+7 a B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 6.51, size = 251, normalized size = 1.04 \[ \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\sin (2 (c+d x)) \left (5 \left (3 a^2 A \cos (2 (c+d x))+a^2 (13 A+14 C)+28 a b B+14 A b^2\right )+42 a (a B+2 A b) \cos (c+d x)\right )+20 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right )+84 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+2 a b (3 A+5 C)+5 b^2 B\right )\right )}{105 d \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+b)^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(84*(3*a^2*B + 5*b^2*B + 2*a*b*(3*A + 5*C))*Sq
rt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*(14*a*b*B + 7*b^2*(A + 3*C) + a^2*(5*A + 7*C))*Sqrt[Cos[c + d*
x]]*EllipticF[(c + d*x)/2, 2] + (42*a*(2*A*b + a*B)*Cos[c + d*x] + 5*(14*A*b^2 + 28*a*b*B + a^2*(13*A + 14*C)
+ 3*a^2*A*Cos[2*(c + d*x)]))*Sin[2*(c + d*x)]))/(105*d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*
Cos[2*(c + d*x)])*Sec[c + d*x]^(7/2))

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{2} \sec \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )}{\sec \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 5.33, size = 706, normalized size = 2.92 \[ \text {result too large to display} \]

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

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*A*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8+(-360*A*a^2-336*A*a*b-168*B*a^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*A*a^2+336*A*a*b+140*A*b^2+16
8*B*a^2+280*B*a*b+140*C*a^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-80*A*a^2-84*A*a*b-70*A*b^2-42*B*a^2-140
*B*a*b-70*C*a^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*A*EllipticF(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)*a^2+35*A*EllipticF(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)*b^2-126*A*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)*a*b+70*B*EllipticF(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)*a*b-63*B*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)*a^2-105*B*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)*b^2+35*C*EllipticF(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)*a^2+105*C*EllipticF(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)*b^2-210*C*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)*a*b)/(-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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((a + b/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/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 \[ \int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{2} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(c + d*x))**2*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sec(c + d*x)**(7/2), x)

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