3.1.86 \(\int (b \cos (c+d x))^{3/2} \sec ^4(c+d x) \, dx\) [86]

Optimal. Leaf size=72 \[ \frac {2 b^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b^3 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}} \]

[Out]

2/3*b^3*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+2/3*b^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2716, 2721, 2720} \begin {gather*} \frac {2 b^3 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 b^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^4,x]

[Out]

(2*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*b^3*Sin[c + d*x])/(3*d*(b
*Cos[c + d*x])^(3/2))

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 51, normalized size = 0.71 \begin {gather*} \frac {2 b^2 \left (\sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\tan (c+d x)\right )}{3 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^4,x]

[Out]

(2*b^2*(Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + Tan[c + d*x]))/(3*d*Sqrt[b*Cos[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(240\) vs. \(2(88)=176\).
time = 0.09, size = 241, normalized size = 3.35

method	result	size
default	\(-\frac {2 \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) b^{2} \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(3/2)*sec(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

-2/3*(-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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*s
in(1/2*d*x+1/2*c)^2-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*b^2*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(
1/2)/(-b*(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)/sin(1/2*d*x+1/2*c)/(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((b*cos(d*x+c))^(3/2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c))^(3/2)*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 = 0.10, size = 101, normalized size = 1.40 \begin {gather*} \frac {-i \, \sqrt {2} b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} b \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/3*(-I*sqrt(2)*b^(3/2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*b
^(3/2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(b*cos(d*x + c))*b*sin
(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**(3/2)*sec(d*x+c)**4,x)

[Out]

Timed out

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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((b*cos(d*x+c))^(3/2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c))^(3/2)*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 (b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(c + d*x))^(3/2)/cos(c + d*x)^4,x)

[Out]

int((b*cos(c + d*x))^(3/2)/cos(c + d*x)^4, x)

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