3.1081 \(\int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=80 \[ -\frac{2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

[Out]

(-2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*C*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(5*A + 3*C
)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.0747611, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4066, 3012, 2636, 2639} \[ -\frac{2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*C*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(5*A + 3*C
)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

Rule 4066

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

Rule 3012

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

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]

Rubi steps

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

Mathematica [A]  time = 0.363833, size = 73, normalized size = 0.91 \[ \frac{(5 A+3 C) \sin (2 (c+d x))-2 (5 A+3 C) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 C \tan (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*A + 3*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + (5*A + 3*C)*Sin[2*(c + d*x)] + 2*C*Tan[c + d*x]
)/(5*d*Cos[c + d*x]^(3/2))

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Maple [B]  time = 5.565, size = 593, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x)

[Out]

2/5*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(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)^3*(20*A*(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)^4-40*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+1
2*C*(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*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-20*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(co
s(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+40*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^4-12*C*(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*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A
+3*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)*EllipticE(cos(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)*C)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(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} + A}{\cos \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+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)/cos(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{C \sec \left (d x + c\right )^{2} + A}{\cos \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+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + A)/cos(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+C*sec(d*x+c)**2)/cos(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{C \sec \left (d x + c\right )^{2} + A}{\cos \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+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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