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

Optimal. Leaf size=259 \[ -\frac{(13 A-33 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}+\frac{7 (7 A-17 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(13 A-33 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{7 (7 A-17 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A-2 B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2}+\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^3} \]

[Out]

(7*(7*A - 17*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((13*A - 33*B)*S
qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((13*A - 33*B)*Sin[c + d*x])/(6*a^
3*d*Sqrt[Sec[c + d*x]]) + ((A - B)*Sin[c + d*x])/(5*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3) + ((A - 2*B)*
Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2) + (7*(7*A - 17*B)*Sin[c + d*x])/(30*d*Sqrt[Sec
[c + d*x]]*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.623547, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2960, 4020, 3787, 3769, 3771, 2641, 2639} \[ -\frac{(13 A-33 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}+\frac{7 (7 A-17 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(13 A-33 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{7 (7 A-17 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A-2 B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2}+\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(7*A - 17*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((13*A - 33*B)*S
qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((13*A - 33*B)*Sin[c + d*x])/(6*a^
3*d*Sqrt[Sec[c + d*x]]) + ((A - B)*Sin[c + d*x])/(5*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3) + ((A - 2*B)*
Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2) + (7*(7*A - 17*B)*Sin[c + d*x])/(30*d*Sqrt[Sec
[c + d*x]]*(a^3 + a^3*Sec[c + d*x]))

Rule 2960

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

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+B \cos (c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac{7}{2}}(c+d x)} \, dx &=\int \frac{B+A \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\\ &=\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}+\frac{\int \frac{-\frac{1}{2} a (3 A-13 B)+\frac{7}{2} a (A-B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\int \frac{-\frac{3}{2} a^2 (8 A-23 B)+\frac{25}{2} a^2 (A-2 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \frac{-\frac{15}{4} a^3 (13 A-33 B)+\frac{21}{4} a^3 (7 A-17 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(13 A-33 B) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{4 a^3}+\frac{(7 (7 A-17 B)) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac{(13 A-33 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}+\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(13 A-33 B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}+\frac{\left (7 (7 A-17 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{7 (7 A-17 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{(13 A-33 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}+\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left ((13 A-33 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac{7 (7 A-17 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{(13 A-33 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}-\frac{(13 A-33 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}+\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 4.55882, size = 589, normalized size = 2.27 \[ \frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left ((806 A-1961 B) \cos \left (\frac{1}{2} (c-d x)\right )+(664 A-1609 B) \cos \left (\frac{1}{2} (3 c+d x)\right )+470 A \cos \left (\frac{1}{2} (c+3 d x)\right )+265 A \cos \left (\frac{1}{2} (5 c+3 d x)\right )+117 A \cos \left (\frac{1}{2} (3 c+5 d x)\right )+30 A \cos \left (\frac{1}{2} (7 c+5 d x)\right )-1165 B \cos \left (\frac{1}{2} (c+3 d x)\right )-620 B \cos \left (\frac{1}{2} (5 c+3 d x)\right )-292 B \cos \left (\frac{1}{2} (3 c+5 d x)\right )-65 B \cos \left (\frac{1}{2} (7 c+5 d x)\right )-5 B \cos \left (\frac{1}{2} (5 c+7 d x)\right )+5 B \cos \left (\frac{1}{2} (9 c+7 d x)\right )\right )}{8 \sqrt{\sec (c+d x)}}-98 \sqrt{2} A \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )-260 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+238 \sqrt{2} B \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+660 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 a^3 d (\cos (c+d x)+1)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[(c + d*x)/2]^6*((-98*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x
))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/
4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (238*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 +
E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometr
ic2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - (((806*A - 1961*B)*Cos[(c - d*x)/2] + (664*A - 1609*B
)*Cos[(3*c + d*x)/2] + 470*A*Cos[(c + 3*d*x)/2] - 1165*B*Cos[(c + 3*d*x)/2] + 265*A*Cos[(5*c + 3*d*x)/2] - 620
*B*Cos[(5*c + 3*d*x)/2] + 117*A*Cos[(3*c + 5*d*x)/2] - 292*B*Cos[(3*c + 5*d*x)/2] + 30*A*Cos[(7*c + 5*d*x)/2]
- 65*B*Cos[(7*c + 5*d*x)/2] - 5*B*Cos[(5*c + 7*d*x)/2] + 5*B*Cos[(9*c + 7*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c +
d*x)/2]^5)/(8*Sqrt[Sec[c + d*x]]) - 260*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + 66
0*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]))/(15*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 3.979, size = 465, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^3/sec(d*x+c)^(7/2),x)

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-160*B*cos(1/2*d*x+1/2*c)^10+348*A*cos(1/2*d*x+1
/2*c)^8+130*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))*cos(1/2*d*x+1/2*c)^5+294*A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^
(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-468*B*cos(1/2*d*x+1/2*c)^8-330*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*
cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-714*B*cos(1/2*d*x+1/2
*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-578
*A*cos(1/2*d*x+1/2*c)^6+1058*B*cos(1/2*d*x+1/2*c)^6+264*A*cos(1/2*d*x+1/2*c)^4-474*B*cos(1/2*d*x+1/2*c)^4-37*A
*cos(1/2*d*x+1/2*c)^2+47*B*cos(1/2*d*x+1/2*c)^2+3*A-3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-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+B*cos(d*x+c))/(a+a*cos(d*x+c))^3/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{B \cos \left (d x + c\right ) + A}{{\left (a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \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+B*cos(d*x+c))/(a+a*cos(d*x+c))^3/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

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

[Out]

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