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

Optimal. Leaf size=185 \[ \frac{2 (5 A+7 C) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{21 b^4 d}+\frac{2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}+\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}} \]

[Out]

(6*B*EllipticE[(c + d*x)/2, 2])/(5*b^3*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*(5*A + 7*C)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*b^4*d) + (2*B*Sin[c + d*x])/(5*b^2*d*(b*Sec[c + d*
x])^(3/2)) + (2*(5*A + 7*C)*Sin[c + d*x])/(21*b^3*d*Sqrt[b*Sec[c + d*x]]) + (2*A*Tan[c + d*x])/(7*d*(b*Sec[c +
d*x])^(7/2))

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Rubi [A]  time = 0.158653, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4047, 3769, 3771, 2639, 4045, 2641} \[ \frac{2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt{b \sec (c+d x)}}+\frac{2 (5 A+7 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{21 b^4 d}+\frac{2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}+\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*B*EllipticE[(c + d*x)/2, 2])/(5*b^3*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*(5*A + 7*C)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*b^4*d) + (2*B*Sin[c + d*x])/(5*b^2*d*(b*Sec[c + d*
x])^(3/2)) + (2*(5*A + 7*C)*Sin[c + d*x])/(21*b^3*d*Sqrt[b*Sec[c + d*x]]) + (2*A*Tan[c + d*x])/(7*d*(b*Sec[c +
d*x])^(7/2))

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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 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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx &=\frac{B \int \frac{1}{(b \sec (c+d x))^{5/2}} \, dx}{b}+\int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx\\ &=\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}+\frac{(3 B) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx}{5 b^3}+\frac{(5 A+7 C) \int \frac{1}{(b \sec (c+d x))^{3/2}} \, dx}{7 b^2}\\ &=\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}+\frac{(5 A+7 C) \int \sqrt{b \sec (c+d x)} \, dx}{21 b^4}+\frac{(3 B) \int \sqrt{\cos (c+d x)} \, dx}{5 b^3 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}+\frac{\left ((5 A+7 C) \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 b^4}\\ &=\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 (5 A+7 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{21 b^4 d}+\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}\\ \end{align*}

Mathematica [C]  time = 3.52956, size = 313, normalized size = 1.69 \[ \frac{\sec ^{\frac{3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{2 \tan (c+d x) (15 A \cos (2 (c+d x))+65 A+42 B \cos (c+d x)+70 C)-252 B \cot (c) \sec (c+d x)}{d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 i \sqrt{2} \left (-5 \left (-1+e^{2 i c}\right ) (5 A+7 C) e^{i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+63 B \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+63 B \sqrt{1+e^{2 i (c+d x)}}\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}}}\right )}{105 (b \sec (c+d x))^{7/2} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(((4*I)*Sqrt[2]*(63*B*Sqrt[1 + E^((2*I)*(c + d*x))
] + 63*B*(-1 + E^((2*I)*c))*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - 5*(5*A + 7*C)*E^(I*(c +
d*x))*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*(-1 + E^((2*I)*c))*Sqrt[E
^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]) + (-252*B*Cot[c]*Sec[c + d*x] + 2*(65
*A + 70*C + 42*B*Cos[c + d*x] + 15*A*Cos[2*(c + d*x)])*Tan[c + d*x])/(d*Sec[c + d*x]^(3/2))))/(105*(A + 2*C +
2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(b*Sec[c + d*x])^(7/2))

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Maple [C]  time = 0.25, size = 645, 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((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x)

[Out]

2/105/d*(63*I*B*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos
(d*x+c)+1))^(1/2)+35*I*C*cos(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1
/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+63*I*B*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*(1/(cos(d*x+
c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+35*I*C*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin
(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-63*I*B*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+
c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-15*A*cos(d*x+c)^5-63*I*B*EllipticE
(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)+25*I*A*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+
c)+1))^(1/2)+25*I*A*sin(d*x+c)*cos(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-21*B*cos(d*x+c)^4-10*A*cos(d*x+c)^3-35*C*cos(d*x+c)^3-42*B*cos(d*x+c)^2+25*A*
cos(d*x+c)+63*B*cos(d*x+c)+35*C*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^4/(b/cos(d*x+c))^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right )}}{b^{4} \sec \left (d x + c\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c))/(b^4*sec(d*x + c)^4), 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)+C*sec(d*x+c)**2)/(b*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{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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