3.1082 \(\int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=80 \[ \frac{2 (7 A+5 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]

[Out]

(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*(7*A + 5*C
)*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2))

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Rubi [A]  time = 0.0755043, 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, 2641} \[ \frac{2 (7 A+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 C \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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*(7*A + 5*C
)*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2))

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 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+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{C+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{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 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}-\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 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.573954, size = 73, normalized size = 0.91 \[ \frac{2 (7 A+5 C) \cos ^{\frac{5}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+(7 A+5 C) \sin (2 (c+d x))+6 C \tan (c+d x)}{21 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*A + 5*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + (7*A + 5*C)*Sin[2*(c + d*x)] + 6*C*Tan[c + d*x])
/(21*d*Cos[c + d*x]^(5/2))

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Maple [B]  time = 4.45, size = 376, normalized size = 4.7 \begin{align*} -{\frac{1}{d}\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 2\,C \left ( -{\frac{\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{56\, \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{4}}}-{\frac{5\,\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{42\, \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{2}}}+{\frac{5\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) }{21\,\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}} \right ) +2\,A \left ( -1/6\,{\frac{\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{ \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{2}}}+1/3\,{\frac{\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) }{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}} \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \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)^(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*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*
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))))/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} + 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+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + 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} + 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+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + 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+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} + 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+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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