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

Optimal. Leaf size=222 \[ \frac{(A+3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(2 A+3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

-((A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((A + 3*B)*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/
(5*d*(a + a*Sec[c + d*x])^3) + ((2*A + 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) +
((A + 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.579967, antiderivative size = 222, 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 = {2960, 4019, 4020, 3787, 3771, 2639, 2641} \[ \frac{(A+3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(2 A+3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((A + 3*B)*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/
(5*d*(a + a*Sec[c + d*x])^3) + ((2*A + 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) +
((A + 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(6*d*(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 4019

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

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 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+B \cos (c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac{3}{2}}(c+d x)} \, dx &=\int \frac{\sqrt{\sec (c+d x)} (B+A \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\frac{1}{2} a (A-B)+\frac{5}{2} a (A+B) \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a^2 (A-6 B)+\frac{3}{2} a^2 (2 A+3 B) \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \frac{-\frac{3}{4} a^3 (A+9 B)+\frac{5}{4} a^3 (A+3 B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(A+3 B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}-\frac{(A+9 B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\left ((A+3 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{\left ((A+9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac{(A+9 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{(A+3 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{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(A+3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.95649, size = 793, normalized size = 3.57 \[ \frac{\cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} \left (\frac{2 \sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )-B \sin \left (\frac{d x}{2}\right )\right )}{5 d}-\frac{4 \sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (7 A \sin \left (\frac{d x}{2}\right )-12 B \sin \left (\frac{d x}{2}\right )\right )}{15 d}+\frac{2 (A-B) \tan \left (\frac{c}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}-\frac{4 (7 A-12 B) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )-9 B \sin \left (\frac{d x}{2}\right )\right )}{3 d}+\frac{2 (A+9 B) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos (d x)}{5 d}+\frac{4 (A-9 B) \tan \left (\frac{c}{2}\right )}{3 d}\right )}{(a \cos (c+d x)+a)^3}+\frac{\sqrt{2} A \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) 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 ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (a \cos (c+d x)+a)^3}+\frac{2 A \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d (a \cos (c+d x)+a)^3}+\frac{3 \sqrt{2} B \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) 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 ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d (a \cos (c+d x)+a)^3}+\frac{2 B \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*
Csc[c/2]*(-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))])*Sec[c/2])/(15*d*E^(I*d*x)*(a + a*Cos[c + d*x])^3) + (3*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/
(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-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))])*Sec[c/2]
)/(5*d*E^(I*d*x)*(a + a*Cos[c + d*x])^3) + (2*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c
+ d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])^3) + (2*B*Cos[c/2 + (d*x)/2]^6*Sqrt
[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(d*(a + a*Cos[c + d*x])^
3) + (Cos[c/2 + (d*x)/2]^6*Sqrt[Sec[c + d*x]]*((2*(A + 9*B)*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) - (4*Sec[c/2]*Se
c[c/2 + (d*x)/2]^3*(7*A*Sin[(d*x)/2] - 12*B*Sin[(d*x)/2]))/(15*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x
)/2] - 9*B*Sin[(d*x)/2]))/(3*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/(5*d) +
(4*(A - 9*B)*Tan[c/2])/(3*d) - (4*(7*A - 12*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (2*(A - B)*Sec[c/2 + (d
*x)/2]^4*Tan[c/2])/(5*d)))/(a + a*Cos[c + d*x])^3

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Maple [A]  time = 3.7, size = 451, normalized size = 2. \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)^(3/2),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)*(12*A*cos(1/2*d*x+1/2*c)^8+10*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+6*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))+108*B*cos(1/2*d*x+1/2*c)^8+30*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)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+54*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))-2*A*cos(1/2*d*x+1/2*c)^6-198*B*cos(1/
2*d*x+1/2*c)^6-24*A*cos(1/2*d*x+1/2*c)^4+114*B*cos(1/2*d*x+1/2*c)^4+17*A*cos(1/2*d*x+1/2*c)^2-27*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]  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{3}{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)^(3/2),x, algorithm="maxima")

[Out]

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

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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{3}{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)^(3/2),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)^(3/2)), 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)**(3/2),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{3}{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)^(3/2),x, algorithm="giac")

[Out]

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