3.220 \(\int \frac{(a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x))}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=237 \[ \frac{4 a^2 (5 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a^2 (19 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+7 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{16 a^2 (2 A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{8 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

[Out]

(16*a^2*(2*A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*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^2*(19*A + 21*C)*Sin[c + d*x]
)/(105*d*Sec[c + d*x]^(3/2)) + (4*a^2*(5*A + 7*C)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c
+ d*x])^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (8*A*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(63*d*Sec[c + d
*x]^(5/2))

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Rubi [A]  time = 0.436035, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4087, 4017, 3996, 3787, 3769, 3771, 2641, 2639} \[ \frac{2 a^2 (19 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+7 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{4 a^2 (5 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{16 a^2 (2 A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{8 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(9/2),x]

[Out]

(16*a^2*(2*A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*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^2*(19*A + 21*C)*Sin[c + d*x]
)/(105*d*Sec[c + d*x]^(3/2)) + (4*a^2*(5*A + 7*C)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c
+ d*x])^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (8*A*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(63*d*Sec[c + d
*x]^(5/2))

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])

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 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 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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 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 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))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^2 \left (2 a A+\frac{3}{2} a (A+3 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \sec (c+d x)) \left (\frac{3}{4} a^2 (19 A+21 C)+\frac{3}{4} a^2 (11 A+21 C) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac{2 a^2 (19 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{8 \int \frac{-\frac{45}{4} a^3 (5 A+7 C)-21 a^3 (2 A+3 C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac{2 a^2 (19 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{15} \left (8 a^2 (2 A+3 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{7} \left (2 a^2 (5 A+7 C)\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (19 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+7 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} \left (2 a^2 (5 A+7 C)\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{15} \left (8 a^2 (2 A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{16 a^2 (2 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)}}{15 d}+\frac{2 a^2 (19 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+7 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} \left (2 a^2 (5 A+7 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 (2 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)}}{15 d}+\frac{4 a^2 (5 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 (19 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+7 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 2.91122, size = 206, normalized size = 0.87 \[ \frac{a^2 e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (-448 i (2 A+3 C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+240 (5 A+7 C) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\cos (c+d x) (60 (23 A+28 C) \sin (c+d x)+14 (37 A+18 C) \sin (2 (c+d x))+180 A \sin (3 (c+d x))+35 A \sin (4 (c+d x))+2688 i A+4032 i C)\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(9/2),x]

[Out]

(a^2*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(240*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]
- (448*I)*(2*A + 3*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]*((2688*I)*A + (4032*I)*C + 60*(23*A + 28*C)*Sin[c + d*x] + 14*(37*A + 18*C)*Sin[2*
(c + d*x)] + 180*A*Sin[3*(c + d*x)] + 35*A*Sin[4*(c + d*x)])))/(1260*d*E^(I*d*x))

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Maple [A]  time = 1.928, size = 408, normalized size = 1.7 \begin{align*} -{\frac{4\,{a}^{2}}{315\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -560\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}+1840\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -2368\,A-252\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 1568\,A+672\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -387\,A-273\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +75\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -168\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +105\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -252\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \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)^(9/2),x)

[Out]

-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-560*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^10+1840*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-2368*A-252*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(
1568*A+672*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-387*A-273*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+7
5*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))-168*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)*EllipticE(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-252*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(-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)^(9/2),x, algorithm="maxima")

[Out]

Timed out

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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}}{\sec \left (d x + c\right )^{\frac{9}{2}}}, 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)^(9/2),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)^(9/2), 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))**2*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(9/2),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} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac{9}{2}}}\,{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)^(9/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(9/2), x)