3.221 \(\int \frac{A+B \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=228 \[ -\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{6 a^3 d}-\frac{(13 A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(49 A-9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(8 A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

((49*A - 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((13*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) - ((8*A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])
^2) - ((13*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 = 0.498514, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4020, 3787, 3771, 2639, 2641} \[ -\frac{(13 A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(49 A-9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(8 A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((49*A - 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((13*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) - ((8*A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])
^2) - ((13*A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Sec[c + d*x]))

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 \sec (c+d x)}{\sqrt{\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 (11 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{(8 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 (41 A-6 B)-\frac{3}{2} a^2 (8 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{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 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 (49 A-9 B)-\frac{5}{4} a^3 (13 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{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 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{(49 A-9 B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}-\frac{(13 A-3 B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 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 ((49 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{\left ((13 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{(49 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{(13 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{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 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 = 6.5179, size = 364, normalized size = 1.6 \[ -\frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) (A+B \sec (c+d x)) \left (i (49 A-9 B) e^{-\frac{3}{2} i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^5 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+160 (13 A-3 B) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+2 \cos (c+d x) \left (-30 i (49 A-9 B) \cos \left (\frac{1}{2} (c+d x)\right )-15 i (49 A-9 B) \cos \left (\frac{3}{2} (c+d x)\right )+142 A \sin \left (\frac{1}{2} (c+d x)\right )+205 A \sin \left (\frac{3}{2} (c+d x)\right )+87 A \sin \left (\frac{5}{2} (c+d x)\right )-147 i A \cos \left (\frac{5}{2} (c+d x)\right )-42 B \sin \left (\frac{1}{2} (c+d x)\right )-45 B \sin \left (\frac{3}{2} (c+d x)\right )-27 B \sin \left (\frac{5}{2} (c+d x)\right )+27 i B \cos \left (\frac{5}{2} (c+d x)\right )\right )\right )}{120 a^3 d (\sec (c+d x)+1)^3 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*(Cos[d*x] + I*Sin[d*x])*(160*(13*A - 3*B)*Cos[(c +
d*x)/2]^5*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (I*(49*A - 9*B)*(1 + E^(I*(c + d*x)))^5*Sqrt[1 + E^((
2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(((3*I)/2)*(c + d*x)) + 2*Cos[c + d
*x]*((-30*I)*(49*A - 9*B)*Cos[(c + d*x)/2] - (15*I)*(49*A - 9*B)*Cos[(3*(c + d*x))/2] - (147*I)*A*Cos[(5*(c +
d*x))/2] + (27*I)*B*Cos[(5*(c + d*x))/2] + 142*A*Sin[(c + d*x)/2] - 42*B*Sin[(c + d*x)/2] + 205*A*Sin[(3*(c +
d*x))/2] - 45*B*Sin[(3*(c + d*x))/2] + 87*A*Sin[(5*(c + d*x))/2] - 27*B*Sin[(5*(c + d*x))/2])))/(120*a^3*d*E^(
I*d*x)*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 1.895, 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*sec(d*x+c))/(a+a*sec(d*x+c))^3/sec(d*x+c)^(1/2),x)

[Out]

1/60/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(348*A*cos(1/2*d*x+1/2*c)^8+130*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
+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))-108*B*cos(1/2*d*x+1/2*c)^8-30*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-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))-578*A*cos(1/2*d*x+1/2*c)^6+198
*B*cos(1/2*d*x+1/2*c)^6+264*A*cos(1/2*d*x+1/2*c)^4-114*B*cos(1/2*d*x+1/2*c)^4-37*A*cos(1/2*d*x+1/2*c)^2+27*B*c
os(1/2*d*x+1/2*c)^2+3*A-3*B)/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*sec(d*x+c))/(a+a*sec(d*x+c))^3/sec(d*x+c)^(1/2),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{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{4} + 3 \, a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + a^{3} \sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a^3*sec(d*x + c)^4 + 3*a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x +
c)^2 + a^3*sec(d*x + c)), 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*sec(d*x+c))/(a+a*sec(d*x+c))**3/sec(d*x+c)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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