3.1186 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=147 \[ \frac{2 (7 A+5 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{6 B \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.152271, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4064, 3021, 2748, 2636, 2639, 2641} \[ \frac{2 (7 A+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{6 B \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Cos[c + d*x]^(5/2),x]

[Out]

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

Rule 4064

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)
]^2), x_Symbol] :> Dist[b^2, Int[(b*Cos[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; F
reeQ[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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)+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2}{7} \int \frac{\frac{7 B}{2}+\frac{1}{2} (7 A+5 C) \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+B \int \frac{1}{\cos ^{\frac{7}{2}}(c+d x)} \, dx+\frac{1}{7} (7 A+5 C) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} (3 B) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{21} (7 A+5 C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (7 A+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{6 B \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}-\frac{1}{5} (3 B) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (7 A+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{6 B \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.560619, size = 129, normalized size = 0.88 \[ \frac{10 (7 A+5 C) \cos ^{\frac{5}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+35 A \sin (2 (c+d x))+42 B \sin (c+d x)+126 B \sin (c+d x) \cos ^2(c+d x)-126 B \cos ^{\frac{5}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+25 C \sin (2 (c+d x))+30 C \tan (c+d x)}{105 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Cos[c + d*x]^(5/2),x]

[Out]

(-126*B*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 10*(7*A + 5*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2
, 2] + 42*B*Sin[c + d*x] + 126*B*Cos[c + d*x]^2*Sin[c + d*x] + 35*A*Sin[2*(c + d*x)] + 25*C*Sin[2*(c + d*x)] +
30*C*Tan[c + d*x])/(105*d*Cos[c + d*x]^(5/2))

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Maple [B]  time = 7.17, size = 684, normalized size = 4.7 \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)/cos(d*x+c)^(5/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(-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*C*(-1/56*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)^
4-5/42*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+5/
21*(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*(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))*(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*(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))*(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*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)-8*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))/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{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac{5}{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)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/cos(d*x + c)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac{5}{2}}}, 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)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/cos(d*x + c)^(5/2), 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*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2),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}{\cos \left (d x + c\right )^{\frac{5}{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)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/cos(d*x + c)^(5/2), x)