3.11.81 \(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [1081]

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

[Out]

-2/5*(5*A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/5*C*s
in(d*x+c)/d/cos(d*x+c)^(5/2)+2/5*(5*A+3*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 = {4151, 3091, 2716, 2719} \begin {gather*} -\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)} \end {gather*}

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 2716

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 3091

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 4151

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]

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*}

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Mathematica [A]
time = 0.41, size = 73, normalized size = 0.91 \begin {gather*} \frac {-2 (5 A+3 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(5 A+3 C) \sin (2 (c+d x))+2 C \tan (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(592\) vs. \(2(96)=192\).
time = 0.16, size = 593, normalized size = 7.41

method	result	size
default	\(-\frac {2 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (40 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 A \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 C \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 A \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 C \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A -5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C -3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{5 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(593\)

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,method=_RETURNVERBOSE)

[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*(40*A*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)*(sin(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)^4+
24*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)^4-40*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+20*A
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)^2-24*C*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)*(sin(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+10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1
/2*c)*A-5*A*(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-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)))*(-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.04, size = 128, normalized size = 1.60 \begin {gather*} \frac {\sqrt {2} {\left (-5 i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (5 i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left ({\left (5 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

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]

1/5*(sqrt(2)*(-5*I*A - 3*I*C)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) +
I*sin(d*x + c))) + sqrt(2)*(5*I*A + 3*I*C)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co
s(d*x + c) - I*sin(d*x + c))) + 2*((5*A + 3*C)*cos(d*x + c)^2 + C)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x
+ c)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 4.93, size = 87, normalized size = 1.09 \begin {gather*} \frac {2\,A\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) +
(2*C*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2
))

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