3.208 \(\int \sqrt {\sec (c+d x)} (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=172 \[ \frac {2 a (5 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 a (5 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d} \]

[Out]

2/3*a*C*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*C*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/5*a*(5*A+3*C)*sin(d*x+c)*sec(d*x
+c)^(1/2)/d-2/5*a*(5*A+3*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+2/3*a*(3*A+C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
F(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 = 0.20, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4077, 4047, 3768, 3771, 2639, 4046, 2641} \[ \frac {2 a (5 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 a (5 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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 4046

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 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 4077

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

Rubi steps

\begin {align*} \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 a C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\sec (c+d x)} \left (\frac {5 a A}{2}+\frac {1}{2} a (5 A+3 C) \sec (c+d x)+\frac {5}{2} a C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 a C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\sec (c+d x)} \left (\frac {5 a A}{2}+\frac {5}{2} a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} (a (5 A+3 C)) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 a (5 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a C \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} (a (3 A+C)) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{5} (a (5 A+3 C)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a (5 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a C \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} \left (a (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (a (5 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 a (5 A+3 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 {2 a (3 A+C) \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 (5 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a C \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [C]  time = 2.37, size = 286, normalized size = 1.66 \[ \frac {2 a e^{-i c} \left (-1+e^{2 i c}\right ) \csc (c) \left (A+C \sec ^2(c+d x)\right ) \left ((5 A+3 C) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-5 i (3 A+C) \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-15 A e^{i (c+d x)}-30 A e^{3 i (c+d x)}-15 A e^{5 i (c+d x)}-3 C e^{i (c+d x)}-24 C e^{3 i (c+d x)}-5 C e^{4 i (c+d x)}-9 C e^{5 i (c+d x)}+5 C\right )}{15 d \left (1+e^{2 i (c+d x)}\right )^2 \sec ^{\frac {3}{2}}(c+d x) (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(-1 + E^((2*I)*c))*Csc[c]*(5*C - 15*A*E^(I*(c + d*x)) - 3*C*E^(I*(c + d*x)) - 30*A*E^((3*I)*(c + d*x)) -
24*C*E^((3*I)*(c + d*x)) - 5*C*E^((4*I)*(c + d*x)) - 15*A*E^((5*I)*(c + d*x)) - 9*C*E^((5*I)*(c + d*x)) - (5*I
)*(3*A + C)*(1 + E^((2*I)*(c + d*x)))^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (5*A + 3*C)*E^(I*(c + d
*x))*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*(A + C*Sec[c + d*
x]^2))/(15*d*E^(I*c)*(1 + E^((2*I)*(c + d*x)))^2*(A + 2*C + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(3/2))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a \sec \left (d x + c\right )^{3} + C a \sec \left (d x + c\right )^{2} + A a \sec \left (d x + c\right ) + A a\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 13.25, size = 729, normalized size = 4.24 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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*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)^(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*(-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)*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)-2/5*C/(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*cos(1/2*d*x+1/2*c)
*sin(1/2*d*x+1/2*c)^6-12*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*(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))-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*C*(-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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(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))))/s
in(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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

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))*(1/cos(c + d*x))^(1/2),x)

[Out]

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

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sympy [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+a*sec(d*x+c))*(A+C*sec(d*x+c)**2)*sec(d*x+c)**(1/2),x)

[Out]

Timed out

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