3.357 \(\int \frac{a+a \sec (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=111 \[ \frac{2 a \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{6 a \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}} \]

[Out]

(-6*a*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*Sin[c + d*x])/(5*d*Cos[c
+ d*x]^(5/2)) + (2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (6*a*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.0907566, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4225, 2748, 2636, 2639, 2641} \[ \frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{6 a \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*Sin[c + d*x])/(5*d*Cos[c
+ d*x]^(5/2)) + (2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (6*a*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

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

Mathematica [C]  time = 6.17843, size = 477, normalized size = 4.3 \[ a \left (\frac{3 \csc (c) (\cos (c+d x)+1) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\tan ^2(c)+1} \sqrt{1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt{\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac{\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt{\tan ^2(c)+1}}+\frac{2 \cos ^2(c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{10 d}-\frac{\csc (c) (\cos (c+d x)+1) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin (c) \left (-\sqrt{\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{3 d \sqrt{\cot ^2(c)+1}}+\sqrt{\cos (c+d x)} (\cos (c+d x)+1) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (\frac{\sec (c) \sin (d x) \sec ^3(c+d x)}{5 d}+\frac{\sec (c) (3 \sin (c)+5 \sin (d x)) \sec ^2(c+d x)}{15 d}+\frac{\sec (c) (5 \sin (c)+9 \sin (d x)) \sec (c+d x)}{15 d}+\frac{3 \csc (c) \sec (c)}{5 d}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*((3*Csc[c]*Sec[c])/(5*d) + (Sec[c]*Sec[c + d*x]^
3*Sin[d*x])/(5*d) + (Sec[c]*Sec[c + d*x]^2*(3*Sin[c] + 5*Sin[d*x]))/(15*d) + (Sec[c]*Sec[c + d*x]*(5*Sin[c] +
9*Sin[d*x]))/(15*d)) - ((1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]
]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c
]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) + (3*(1
+ Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]
^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]
*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*T
an[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/S
qrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(10*d))

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Maple [B]  time = 2.505, size = 384, normalized size = 3.5 \begin{align*} -4\,{\frac{\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}a}{\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}d} \left ( -1/12\,{\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}}}+{\frac{7\,\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 ) }{15\,\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}-1/40\,{\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 ) ^{3}}}-3/5\,{\frac{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) }{\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}-3/10\,{\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} \left ({\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))/cos(d*x+c)^(5/2),x)

[Out]

-4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(-1/12*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+7/15*(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))-
1/40*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)^3-3/5*
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-3/10*(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))-EllipticE(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{a \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+a*sec(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*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{a \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+a*sec(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((a*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+a*sec(d*x+c))/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{a \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+a*sec(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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