3.1.43 \(\int \sqrt {b \cos (c+d x)} (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [43]

Optimal. Leaf size=110 \[ -\frac {2 (3 A+5 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}} \]

[Out]

2/5*A*b^3*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/5*b*(3*A+5*C)*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)-2/5*(3*A+5*C)*(c
os(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))*(b*cos(d*x+c))^(1/2)/d/cos
(d*x+c)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {16, 3091, 2716, 2721, 2719} \begin {gather*} \frac {2 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}-\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 2721

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

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]

Rubi steps

\begin {align*} \int \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=b^4 \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{7/2}} \, dx\\ &=\frac {2 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {1}{5} \left (b^2 (3 A+5 C)\right ) \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx\\ &=\frac {2 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}+\frac {1}{5} (-3 A-5 C) \int \sqrt {b \cos (c+d x)} \, dx\\ &=\frac {2 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}+\frac {\left ((-3 A-5 C) \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)}}\\ &=-\frac {2 (3 A+5 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 84, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {b \cos (c+d x)} \sec ^2(c+d x) \left (2 (3 A+5 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-(3 A+5 C) \sin (2 (c+d x))-2 A \tan (c+d x)\right )}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

-1/5*(Sqrt[b*Cos[c + d*x]]*Sec[c + d*x]^2*(2*(3*A + 5*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] - (3*A +
 5*C)*Sin[2*(c + d*x)] - 2*A*Tan[c + d*x]))/d

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(597\) vs. \(2(122)=244\).
time = 0.77, size = 598, normalized size = 5.44

method result size
default \(-\frac {2 \sqrt {b \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 (24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+40 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 C \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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 C \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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 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 )+10 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 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 ) b +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}}{5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\) \(598\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4*(b*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/5*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)^3/(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)*(24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*A*EllipticE(c
os(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4+
40*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-20*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*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*A
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d
*x+1/2*c)^2-40*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+20*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+8*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^2-3*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))
+10*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-5*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*b+sin(1/2*d*x+1/2*c)^2*b)^(1/2)/(b*(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*cos(d*x+c)^2)*sec(d*x+c)^4*(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c))*sec(d*x + c)^4, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 136, normalized size = 1.24 \begin {gather*} \frac {\sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \sqrt {b} \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 (3 i \, A + 5 i \, C\right )} \sqrt {b} \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 (3 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \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*cos(d*x+c)^2)*sec(d*x+c)^4*(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**4*(b*cos(d*x+c))**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5008 deep

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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*cos(d*x+c)^2)*sec(d*x+c)^4*(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c))*sec(d*x + c)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(1/2))/cos(c + d*x)^4,x)

[Out]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(1/2))/cos(c + d*x)^4, x)

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