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

Optimal. Leaf size=143 \[ \frac{2 a (3 A+3 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 a (A-B-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a (B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{d}+\frac{2 a C \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.219038, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4076, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 a (3 A+3 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a (A-B-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a (B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{d}+\frac{2 a C \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4076

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[(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*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 1.84693, size = 208, normalized size = 1.45 \[ \frac{a e^{-i d x} \sec ^{\frac{3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-i (A-B-C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+2 (3 A+3 B+C) \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+3 i A \cos (2 (c+d x))+3 i A+3 B \sin (2 (c+d x))-3 i B \cos (2 (c+d x))-3 i B+2 C \sin (c+d x)+3 C \sin (2 (c+d x))-3 i C \cos (2 (c+d x))-3 i C\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])*((3*I)*A - (3*I)*B - (3*I)*C + (3*I)*A*Cos[2*(c + d*x)] - (3*I)*
B*Cos[2*(c + d*x)] - (3*I)*C*Cos[2*(c + d*x)] + 2*(3*A + 3*B + C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2]
- I*(A - B - C)*(1 + E^((2*I)*(c + d*x)))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + 2*C*
Sin[c + d*x] + 3*B*Sin[2*(c + d*x)] + 3*C*Sin[2*(c + d*x)]))/(3*d*E^(I*d*x))

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Maple [B]  time = 5.522, size = 516, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))*(A+B*sec(d*x+c)+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))
-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*B*(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*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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(c
os(1/2*d*x+1/2*c),2^(1/2)))+4*(1/2*C+1/2*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)*El
lipticE(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)+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))/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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C a \sec \left (d x + c\right )^{3} +{\left (B + C\right )} a \sec \left (d x + c\right )^{2} +{\left (A + B\right )} a \sec \left (d x + c\right ) + A a}{\sqrt{\sec \left (d x + c\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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