3.199 \(\int \frac{(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=244 \[ \frac{4 a^3 (11 A+13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{4 a^3 (23 A+24 B) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (11 A+13 B) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (17 A+21 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a A \sin (c+d x) (a \sec (c+d x)+a)^2}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

[Out]

(4*a^3*(17*A + 21*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(11*A +
13*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^3*(23*A + 24*B)*Sin[c + d
*x])/(105*d*Sec[c + d*x]^(3/2)) + (4*a^3*(11*A + 13*B)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + a
*Sec[c + d*x])^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (2*(13*A + 9*B)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x
])/(63*d*Sec[c + d*x]^(5/2))

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Rubi [A]  time = 0.476664, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4017, 3996, 3787, 3769, 3771, 2641, 2639} \[ \frac{4 a^3 (23 A+24 B) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (11 A+13 B) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (11 A+13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^3 (17 A+21 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a A \sin (c+d x) (a \sec (c+d x)+a)^2}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(9/2),x]

[Out]

(4*a^3*(17*A + 21*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(11*A +
13*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^3*(23*A + 24*B)*Sin[c + d
*x])/(105*d*Sec[c + d*x]^(3/2)) + (4*a^3*(11*A + 13*B)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + a
*Sec[c + d*x])^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (2*(13*A + 9*B)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x
])/(63*d*Sec[c + d*x]^(5/2))

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))^3 (A+B \sec (c+d x))}{\sec ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2}{9} \int \frac{(a+a \sec (c+d x))^2 \left (\frac{1}{2} a (13 A+9 B)+\frac{3}{2} a (A+3 B) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (13 A+9 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4}{63} \int \frac{(a+a \sec (c+d x)) \left (\frac{3}{2} a^2 (23 A+24 B)+\frac{15}{2} a^2 (2 A+3 B) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{4 a^3 (23 A+24 B) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (13 A+9 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{8}{315} \int \frac{-\frac{45}{4} a^3 (11 A+13 B)-\frac{21}{4} a^3 (17 A+21 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^3 (23 A+24 B) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (13 A+9 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{7} \left (2 a^3 (11 A+13 B)\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{15} \left (2 a^3 (17 A+21 B)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{4 a^3 (23 A+24 B) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (11 A+13 B) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (13 A+9 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} \left (2 a^3 (11 A+13 B)\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{15} \left (2 a^3 (17 A+21 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^3 (17 A+21 B) \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^3 (23 A+24 B) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (11 A+13 B) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (13 A+9 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} \left (2 a^3 (11 A+13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^3 (17 A+21 B) \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^3 (11 A+13 B) \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{4 a^3 (23 A+24 B) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (11 A+13 B) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (13 A+9 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 2.78518, size = 196, normalized size = 0.8 \[ \frac{a^3 \sqrt{\sec (c+d x)} \left (-112 i (17 A+21 B) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+240 (11 A+13 B) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\cos (c+d x) (30 (97 A+107 B) \sin (c+d x)+14 (73 A+54 B) \sin (2 (c+d x))+270 A \sin (3 (c+d x))+35 A \sin (4 (c+d x))+5712 i A+90 B \sin (3 (c+d x))+7056 i B)\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(9/2),x]

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(240*(11*A + 13*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (112*I)*(17*A + 21*B
)*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]*((5712*I)*A + (7056*I)*B + 30*(97*A + 107*B)*Sin[c + d*x] + 14*(73*A + 54*B)*Sin[2*(c + d*x)] + 270*A*
Sin[3*(c + d*x)] + 90*B*Sin[3*(c + d*x)] + 35*A*Sin[4*(c + d*x)])))/(1260*d)

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Maple [A]  time = 1.717, size = 413, normalized size = 1.7 \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))^3*(A+B*sec(d*x+c))/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^3*(-560*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^10+(2200*A+360*B)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-3412*A-1296*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x
+1/2*c)+(2702*A+1806*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-738*A-624*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)+165*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))-357*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
+195*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-441
*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*si
n(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))^3*(A+B*sec(d*x+c))/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{B a^{3} \sec \left (d x + c\right )^{4} +{\left (A + 3 \, B\right )} a^{3} \sec \left (d x + c\right )^{3} + 3 \,{\left (A + B\right )} a^{3} \sec \left (d x + c\right )^{2} +{\left (3 \, A + B\right )} a^{3} \sec \left (d x + c\right ) + A a^{3}}{\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))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

integral((B*a^3*sec(d*x + c)^4 + (A + 3*B)*a^3*sec(d*x + c)^3 + 3*(A + B)*a^3*sec(d*x + c)^2 + (3*A + B)*a^3*s
ec(d*x + c) + A*a^3)/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))**3*(A+B*sec(d*x+c))/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 (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\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))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/sec(d*x + c)^(9/2), x)