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

Optimal. Leaf size=196 \[ \frac{4 a^2 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 a^2 (15 A+17 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{8 C \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{15 d}-\frac{16 a^2 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 C \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d} \]

[Out]

(-16*a^2*C*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(3*A + C)*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^2*(15*A + 17*C)*Sqrt[Sec[c + d*x]]*Sin[c
+ d*x])/(15*d) + (2*C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + (8*C*Sqrt[Sec[c + d*x]]*
(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(15*d)

________________________________________________________________________________________

Rubi [A]  time = 0.389384, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4089, 4018, 3997, 3787, 3771, 2639, 2641} \[ \frac{2 a^2 (15 A+17 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^2 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{8 C \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{15 d}-\frac{16 a^2 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 C \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*a^2*C*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(3*A + C)*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^2*(15*A + 17*C)*Sqrt[Sec[c + d*x]]*Sin[c
+ d*x])/(15*d) + (2*C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + (8*C*Sqrt[Sec[c + d*x]]*
(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(15*d)

Rule 4089

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

Rule 4018

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

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx &=\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{2 \int \frac{(a+a \sec (c+d x))^2 \left (\frac{1}{2} a (5 A-C)+2 a C \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{5 a}\\ &=\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{4 \int \frac{(a+a \sec (c+d x)) \left (\frac{1}{4} a^2 (15 A-7 C)+\frac{1}{4} a^2 (15 A+17 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{15 a}\\ &=\frac{2 a^2 (15 A+17 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{8 \int \frac{-3 a^3 C+\frac{5}{4} a^3 (3 A+C) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a}\\ &=\frac{2 a^2 (15 A+17 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac{1}{5} \left (8 a^2 C\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (2 a^2 (3 A+C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a^2 (15 A+17 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac{1}{5} \left (8 a^2 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^2 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{16 a^2 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{4 a^2 (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^2 (15 A+17 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{8 C \sqrt{\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}\\ \end{align*}

Mathematica [C]  time = 5.88315, size = 312, normalized size = 1.59 \[ \frac{a^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 \left (A+C \sec ^2(c+d x)\right ) \left (\frac{-3 \csc (c) \cos (d x) (5 A \cos (2 c)-5 A-16 C)+30 A \cos (c) \sin (d x)+2 C \tan (c+d x) (3 \sec (c+d x)+10)}{2 d \sec ^{\frac{7}{2}}(c+d x)}-\frac{2 i \sqrt{2} \cos ^4(c+d x) \left (5 \left (-1+e^{2 i c}\right ) (3 A+C) e^{i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+12 \left (-1+e^{2 i c}\right ) C \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+12 C \sqrt{1+e^{2 i (c+d x)}}\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}}}\right )}{15 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*(((-2*I)*Sqrt[2]*Cos[c + d*x]^4*(12*C*Sqrt
[1 + E^((2*I)*(c + d*x))] + 12*C*(-1 + E^((2*I)*c))*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] +
5*(3*A + C)*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*(-1
+ E^((2*I)*c))*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]) + (-3*(-5*A - 1
6*C + 5*A*Cos[2*c])*Cos[d*x]*Csc[c] + 30*A*Cos[c]*Sin[d*x] + 2*C*(10 + 3*Sec[c + d*x])*Tan[c + d*x])/(2*d*Sec[
c + d*x]^(7/2))))/(15*(A + 2*C + A*Cos[2*(c + d*x)]))

________________________________________________________________________________________

Maple [B]  time = 6.369, size = 756, normalized size = 3.9 \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))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x)

[Out]

4/15*a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(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)^3*(60*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))*sin(1/2*d*x+1/2*c)^4-60*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^6+20*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*s
in(1/2*d*x+1/2*c)^4+48*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))*sin(1/2*d*x+1/2*c)^4-96*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*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))*sin(1/2*d*x+1/2*c)^2+60*A*sin(1/2*
d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-20*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)*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-48*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))*sin(1/2*d*x+1/2*c)^2+116*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+
15*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))-15*A*
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*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)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*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))-37*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*
x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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}}{\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))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),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)/sqrt(sec(d*x + c)), x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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