3.197 \(\int \frac{(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=211 \[ \frac{4 a^3 (3 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{4 a^3 (6 A-5 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{2 (9 A+5 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (9 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(4*a^3*(9*A + 5*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(3*A + 5*B)
*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(6*A - 5*B)*Sqrt[Sec[c + d*x]
]*Sin[c + d*x])/(15*d) + (2*a*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(9*A + 5*B)
*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.413033, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4017, 3997, 3787, 3771, 2639, 2641} \[ -\frac{4 a^3 (6 A-5 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{2 (9 A+5 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (3 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^3 (9 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]

[Out]

(4*a^3*(9*A + 5*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(3*A + 5*B)
*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(6*A - 5*B)*Sqrt[Sec[c + d*x]
]*Sin[c + d*x])/(15*d) + (2*a*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(9*A + 5*B)
*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]])

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 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))^3 (A+B \sec (c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+a \sec (c+d x))^2 \left (\frac{1}{2} a (9 A+5 B)-\frac{1}{2} a (A-5 B) \sec (c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (9 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{4}{15} \int \frac{(a+a \sec (c+d x)) \left (\frac{1}{2} a^2 (21 A+20 B)-\frac{1}{2} a^2 (6 A-5 B) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{4 a^3 (6 A-5 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (9 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{8}{15} \int \frac{\frac{3}{4} a^3 (9 A+5 B)+\frac{5}{4} a^3 (3 A+5 B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{4 a^3 (6 A-5 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (9 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{1}{3} \left (2 a^3 (3 A+5 B)\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{5} \left (2 a^3 (9 A+5 B)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{4 a^3 (6 A-5 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (9 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{1}{3} \left (2 a^3 (3 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (2 a^3 (9 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^3 (9 A+5 B) \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^3 (3 A+5 B) \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{4 a^3 (6 A-5 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (9 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.6948, size = 207, normalized size = 0.98 \[ \frac{a^3 e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (-8 i (9 A+5 B) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+40 (3 A+5 B) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+3 A \sin (c+d x)+30 A \sin (2 (c+d x))+3 A \sin (3 (c+d x))+216 i A \cos (c+d x)+60 B \sin (c+d x)+10 B \sin (2 (c+d x))+120 i B \cos (c+d x)\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*((216*I)*A*Cos[c + d*x] + (120*I)*B*Cos[c + d*x] + 40*(3*A + 5
*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (8*I)*(9*A + 5*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))] + 3*A*Sin[c + d*x] + 60*B*Sin[c + d*x] + 30*A*Sin[
2*(c + d*x)] + 10*B*Sin[2*(c + d*x)] + 3*A*Sin[3*(c + d*x)]))/(30*d*E^(I*d*x))

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Maple [B]  time = 1.976, size = 519, normalized size = 2.5 \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)^(5/2),x)

[Out]

-4/15*a^3*(-12*A*(-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)^6+
2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(21*A+5*B)*sin(1/2*d*x+1/2*c)^4*cos(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)*(9*A+10*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15*A*(si
n(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))*(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-27*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))+25*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)-15*B*(-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*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))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),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{5}{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)^(5/2),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)^(5/2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \frac{A}{\sec ^{\frac{5}{2}}{\left (c + d x \right )}}\, dx + \int \frac{3 A}{\sec ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{3 A}{\sqrt{\sec{\left (c + d x \right )}}}\, dx + \int A \sqrt{\sec{\left (c + d x \right )}}\, dx + \int \frac{B}{\sec ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{3 B}{\sqrt{\sec{\left (c + d x \right )}}}\, dx + \int 3 B \sqrt{\sec{\left (c + d x \right )}}\, dx + \int B \sec ^{\frac{3}{2}}{\left (c + d x \right )}\, dx\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)**(5/2),x)

[Out]

a**3*(Integral(A/sec(c + d*x)**(5/2), x) + Integral(3*A/sec(c + d*x)**(3/2), x) + Integral(3*A/sqrt(sec(c + d*
x)), x) + Integral(A*sqrt(sec(c + d*x)), x) + Integral(B/sec(c + d*x)**(3/2), x) + Integral(3*B/sqrt(sec(c + d
*x)), x) + Integral(3*B*sqrt(sec(c + d*x)), x) + Integral(B*sec(c + d*x)**(3/2), x))

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/sec(d*x + c)^(5/2), x)