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

Optimal. Leaf size=219 \[ \frac{4 a^2 (6 A+7 B+14 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a^2 (33 A+49 B+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{4 a^2 (3 A+4 B+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (4 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(4*a^2*(3*A + 4*B + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(6*A
+ 7*B + 14*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^2*(33*A + 49*B +
35*C)*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(
5/2)) + (2*(4*A + 7*B)*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2))

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Rubi [A]  time = 0.496946, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4086, 4017, 3996, 3787, 3771, 2639, 2641} \[ \frac{2 a^2 (33 A+49 B+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{4 a^2 (6 A+7 B+14 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^2 (3 A+4 B+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (4 A+7 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(3*A + 4*B + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(6*A
+ 7*B + 14*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^2*(33*A + 49*B +
35*C)*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(
5/2)) + (2*(4*A + 7*B)*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2))

Rule 4086

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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 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+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^2 \left (\frac{1}{2} a (4 A+7 B)+\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))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \sec (c+d x)) \left (\frac{1}{4} a^2 (33 A+49 B+35 C)+\frac{1}{4} a^2 (9 A+7 B+35 C) \sec (c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac{2 a^2 (33 A+49 B+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{8 \int \frac{-\frac{21}{4} a^3 (3 A+4 B+5 C)-\frac{5}{4} a^3 (6 A+7 B+14 C) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{105 a}\\ &=\frac{2 a^2 (33 A+49 B+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (2 a^2 (3 A+4 B+5 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (2 a^2 (6 A+7 B+14 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a^2 (33 A+49 B+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (2 a^2 (3 A+4 B+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^2 (6 A+7 B+14 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^2 (3 A+4 B+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^2 (6 A+7 B+14 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 (33 A+49 B+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 2.25534, size = 189, normalized size = 0.86 \[ \frac{a^2 \sqrt{\sec (c+d x)} \left (-112 i (3 A+4 B+5 C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+80 (6 A+7 (B+2 C)) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+2 \cos (c+d x) (5 (51 A+28 (2 B+C)) \sin (c+d x)+42 (2 A+B) \sin (2 (c+d x))+15 A \sin (3 (c+d x))+504 i A+672 i B+840 i C)\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[Sec[c + d*x]]*(80*(6*A + 7*(B + 2*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (112*I)*(3*A +
4*B + 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))
] + 2*Cos[c + d*x]*((504*I)*A + (672*I)*B + (840*I)*C + 5*(51*A + 28*(2*B + C))*Sin[c + d*x] + 42*(2*A + B)*Si
n[2*(c + d*x)] + 15*A*Sin[3*(c + d*x)])))/(420*d)

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Maple [A]  time = 2.378, size = 483, normalized size = 2.2 \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+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x)

[Out]

-4/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(120*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8+(-348*A-84*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(378*A+224*B+70*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+
1/2*c)+(-117*A-91*B-35*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+30*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))-63*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))+35*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))-84*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+70*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))-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)*E
llipticE(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+B*sec(d*x+c)+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^{2} \sec \left (d x + c\right )^{4} +{\left (B + 2 \, C\right )} a^{2} \sec \left (d x + c\right )^{3} +{\left (A + 2 \, B + C\right )} a^{2} \sec \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} a^{2} \sec \left (d x + c\right ) + A a^{2}}{\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))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

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

[Out]

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