3.227 \(\int \frac{(a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=253 \[ \frac{4 a^3 (13 A+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}-\frac{4 a^3 (41 A-35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d}+\frac{2 (7 A+5 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{12 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{35 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(4*a^3*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(13*A + 35*
C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) - (4*a^3*(41*A - 35*C)*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/(105*d) + (2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (12*A*(a^2
+ a^2*Sec[c + d*x])^2*Sin[c + d*x])/(35*a*d*Sec[c + d*x]^(3/2)) + (2*(7*A + 5*C)*(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 = 0.572871, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4087, 4017, 3997, 3787, 3771, 2639, 2641} \[ -\frac{4 a^3 (41 A-35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d}+\frac{2 (7 A+5 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (13 A+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^3 (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{12 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{35 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(4*a^3*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(13*A + 35*
C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) - (4*a^3*(41*A - 35*C)*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/(105*d) + (2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (12*A*(a^2
+ a^2*Sec[c + d*x])^2*Sin[c + d*x])/(35*a*d*Sec[c + d*x]^(3/2)) + (2*(7*A + 5*C)*(a^3 + a^3*Sec[c + d*x])*Sin
[c + d*x])/(15*d*Sqrt[Sec[c + d*x]])

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 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 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^3 \left (3 a A-\frac{1}{2} a (A-7 C) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{12 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \sec (c+d x))^2 \left (\frac{7}{4} a^2 (7 A+5 C)-\frac{1}{4} a^2 (11 A-35 C) \sec (c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{12 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (7 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{8 \int \frac{(a+a \sec (c+d x)) \left (\frac{1}{2} a^3 (53 A+70 C)-\frac{1}{4} a^3 (41 A-35 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{105 a}\\ &=-\frac{4 a^3 (41 A-35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{12 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (7 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{16 \int \frac{\frac{21}{8} a^4 (7 A+5 C)+\frac{5}{8} a^4 (13 A+35 C) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{105 a}\\ &=-\frac{4 a^3 (41 A-35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{12 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (7 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{1}{5} \left (2 a^3 (7 A+5 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (2 a^3 (13 A+35 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=-\frac{4 a^3 (41 A-35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{12 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (7 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{1}{5} \left (2 a^3 (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (2 a^3 (13 A+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^3 (7 A+5 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^3 (13 A+35 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{4 a^3 (41 A-35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{12 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (7 A+5 C) \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 = 2.23135, size = 218, normalized size = 0.86 \[ \frac{a^3 e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (-112 i (7 A+5 C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+80 (13 A+35 C) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+126 A \sin (c+d x)+550 A \sin (2 (c+d x))+126 A \sin (3 (c+d x))+15 A \sin (4 (c+d x))+2352 i A \cos (c+d x)+840 C \sin (c+d x)+140 C \sin (2 (c+d x))+1680 i C \cos (c+d x)\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*((2352*I)*A*Cos[c + d*x] + (1680*I)*C*Cos[c + d*x] + 80*(13*A
+ 35*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (112*I)*(7*A + 5*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))] + 126*A*Sin[c + d*x] + 840*C*Sin[c + d*x] +
550*A*Sin[2*(c + d*x)] + 140*C*Sin[2*(c + d*x)] + 126*A*Sin[3*(c + d*x)] + 15*A*Sin[4*(c + d*x)]))/(420*d*E^(I
*d*x))

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Maple [B]  time = 2.406, size = 569, normalized size = 2.3 \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+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x)

[Out]

-4/105*a^3*(120*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)^8
-432*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+14*(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(43*A+5*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-4*(-2*sin(1/2
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(52*A+35*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+65*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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-147*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)*(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+175*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)-105*C*(-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+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),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^{3} \sec \left (d x + c\right )^{5} + 3 \, C a^{3} \sec \left (d x + c\right )^{4} +{\left (A + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{3} +{\left (3 \, A + C\right )} a^{3} \sec \left (d x + c\right )^{2} + 3 \, A a^{3} \sec \left (d x + c\right ) + A a^{3}}{\sec \left (d x + c\right )^{\frac{7}{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+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((C*a^3*sec(d*x + c)^5 + 3*C*a^3*sec(d*x + c)^4 + (A + 3*C)*a^3*sec(d*x + c)^3 + (3*A + C)*a^3*sec(d*x
+ c)^2 + 3*A*a^3*sec(d*x + c) + A*a^3)/sec(d*x + c)^(7/2), 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+C*sec(d*x+c)**2)/sec(d*x+c)**(7/2),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 )}^{3}}{\sec \left (d x + c\right )^{\frac{7}{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+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3/sec(d*x + c)^(7/2), x)