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

Optimal. Leaf size=229 \[ \frac{5 (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a d}+\frac{(A-B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(5 A-7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 a d}+\frac{5 (A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a d}-\frac{3 (5 A-7 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 a d}+\frac{3 (5 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d} \]

[Out]

(3*(5*A - 7*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a*d) + (5*(A - B)*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) - (3*(5*A - 7*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*
x])/(5*a*d) + (5*(A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) - ((5*A - 7*B)*Sec[c + d*x]^(5/2)*Sin[c + d*
x])/(5*a*d) + ((A - B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

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Rubi [A]  time = 0.24767, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4019, 3787, 3768, 3771, 2641, 2639} \[ \frac{(A-B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(5 A-7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 a d}+\frac{5 (A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a d}-\frac{3 (5 A-7 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 a d}+\frac{5 (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{3 (5 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(5*A - 7*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a*d) + (5*(A - B)*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) - (3*(5*A - 7*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*
x])/(5*a*d) + (5*(A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) - ((5*A - 7*B)*Sec[c + d*x]^(5/2)*Sin[c + d*
x])/(5*a*d) + ((A - B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

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 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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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{\sec ^{\frac{7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx &=\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{\int \sec ^{\frac{5}{2}}(c+d x) \left (\frac{5}{2} a (A-B)-\frac{1}{2} a (5 A-7 B) \sec (c+d x)\right ) \, dx}{a^2}\\ &=\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac{(5 A-7 B) \int \sec ^{\frac{7}{2}}(c+d x) \, dx}{2 a}+\frac{(5 (A-B)) \int \sec ^{\frac{5}{2}}(c+d x) \, dx}{2 a}\\ &=\frac{5 (A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac{(5 A-7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac{(3 (5 A-7 B)) \int \sec ^{\frac{3}{2}}(c+d x) \, dx}{10 a}+\frac{(5 (A-B)) \int \sqrt{\sec (c+d x)} \, dx}{6 a}\\ &=-\frac{3 (5 A-7 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 a d}+\frac{5 (A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac{(5 A-7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{(3 (5 A-7 B)) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{10 a}+\frac{\left (5 (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a}\\ &=\frac{5 (A-B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a d}-\frac{3 (5 A-7 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 a d}+\frac{5 (A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac{(5 A-7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{\left (3 (5 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{10 a}\\ &=\frac{3 (5 A-7 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 a d}+\frac{5 (A-B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a d}-\frac{3 (5 A-7 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 a d}+\frac{5 (A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac{(5 A-7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac{(A-B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 7.51226, size = 814, normalized size = 3.55 \[ -\frac{A 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)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\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 ) \sec \left (\frac{c}{2}\right ) (A+B \sec (c+d x)) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{\sqrt{2} d (B+A \cos (c+d x)) (\sec (c+d x) a+a)}+\frac{7 B 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)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\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 ) \sec \left (\frac{c}{2}\right ) (A+B \sec (c+d x)) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 \sqrt{2} d (B+A \cos (c+d x)) (\sec (c+d x) a+a)}+\frac{5 A \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\sec (c+d x)} (A+B \sec (c+d x)) \sin (c) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)}-\frac{5 B \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\sec (c+d x)} (A+B \sec (c+d x)) \sin (c) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)}+\frac{\sqrt{\sec (c+d x)} (A+B \sec (c+d x)) \left (\frac{4 B \sec (c) \sin (d x) \sec ^2(c+d x)}{5 d}+\frac{4 \sec (c) (3 B \sin (c)+5 A \sin (d x)-5 B \sin (d x)) \sec (c+d x)}{15 d}+\frac{3 (7 B-5 A) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{5 d}-\frac{(B-A) \sec \left (\frac{c}{2}\right ) \sec (c) \left (5 \sin \left (\frac{3 c}{2}\right )-\sin \left (\frac{c}{2}\right )\right )}{3 d}-\frac{2 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (B \sin \left (\frac{d x}{2}\right )-A \sin \left (\frac{d x}{2}\right )\right )}{d}\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{(B+A \cos (c+d x)) (\sec (c+d x) a+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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]^2*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]*(A + B*Sec[c + d*x]))/(Sqrt[2]*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x]
))) + (7*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]^2*
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]*(A + B*Sec[c + d*x]))/(5*Sqrt[2]*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(a + a*Sec[
c + d*x])) + (5*A*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec
[c + d*x]]*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) - (5*B*Cos[c/2 + (d*x)
/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x])*S
in[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) + (Cos[c/2 + (d*x)/2]^2*Sqrt[Sec[c + d*x]]*(A + B*Sec[c
+ d*x])*((3*(-5*A + 7*B)*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) - ((-A + B)*Sec[c/2]*Sec[c]*(-Sin[c/2] + 5*Sin[(3*
c)/2]))/(3*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/d + (4*B*Sec[c]*Sec[c + d
*x]^2*Sin[d*x])/(5*d) + (4*Sec[c]*Sec[c + d*x]*(3*B*Sin[c] + 5*A*Sin[d*x] - 5*B*Sin[d*x]))/(15*d)))/((B + A*Co
s[c + d*x])*(a + a*Sec[c + d*x]))

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Maple [B]  time = 6.423, size = 806, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((2*A-2*B)*(-1/6*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(
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)*EllipticF(cos(1/2*d*x+1/2*c),2^
(1/2)))-2/5*B/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(
12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2
*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2
*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-8*si
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(A-B)*(cos(1/2*d*x
+1/2*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))-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)/cos(1/2*d*x+1/2*c)/(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(-2*A+2*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(-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)^2)/sin(1/2*d*x+1/2*c)^2/(2
*sin(1/2*d*x+1/2*c)^2-1))/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(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),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 )^{4} + A \sec \left (d x + c\right )^{3}\right )} \sqrt{\sec \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*sqrt(sec(d*x + c))/(a*sec(d*x + c) + a), 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(sec(d*x+c)**(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{7}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

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