3.2.25 \(\int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx\) [125]

Optimal. Leaf size=97 \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b d \sqrt {b \cos (c+d x)}}+\frac {2 b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}} \]

[Out]

2/7*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(7/2)+10/21*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+10/21*(cos(1/2*d*x+1/2*c)^2)
^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/b/d/(b*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, 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^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*b*d*Sqrt[b*Cos[c + d*x]]) + (2*b^2*Sin[c + d*x])/(7*d*(b
*Cos[c + d*x])^(7/2)) + (10*Sin[c + d*x])/(21*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 \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx &=b^3 \int \frac {1}{(b \cos (c+d x))^{9/2}} \, dx\\ &=\frac {2 b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {1}{7} (5 b) \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx\\ &=\frac {2 b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {5 \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx}{21 b}\\ &=\frac {2 b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {\left (5 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b \sqrt {b \cos (c+d x)}}\\ &=\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b d \sqrt {b \cos (c+d x)}}+\frac {2 b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 66, normalized size = 0.68 \begin {gather*} \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (5+3 \sec ^2(c+d x)\right ) \tan (c+d x)}{21 b d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(5 + 3*Sec[c + d*x]^2)*Tan[c + d*x])/(21*b*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. \(397\) vs. \(2(109)=218\).
time = 0.10, size = 398, normalized size = 4.10

method	result	size
default	\(-\frac {2 \left (-40 \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 ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+60 \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 ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+40 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-30 \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 )-16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \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 ) \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 )}}{21 b \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 )^{3} \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}\)	\(398\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*(-40*(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))
*sin(1/2*d*x+1/2*c)^6-40*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+60*(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))*sin(1/2*d*x+1/2*c)^4+40*sin(1/2*d*x+1/2*c)^4*cos(1/2
*d*x+1/2*c)-30*(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))*sin(1/2*d*x+1/2*c)^2-16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*(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*(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)^3/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(sec(d*x+c)^3/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^3/(b*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 = 0.15, size = 115, normalized size = 1.19 \begin {gather*} \frac {-5 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{4} {\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 )} {\left (5 \, \cos \left (d x + c\right )^{2} + 3\right )} \sin \left (d x + c\right )}{21 \, b^{2} d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(-5*I*sqrt(2)*sqrt(b)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt
(2)*sqrt(b)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(b*cos(d*x + c))*
(5*cos(d*x + c)^2 + 3)*sin(d*x + c))/(b^2*d*cos(d*x + c)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(b*cos(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(b*cos(c + d*x))^(3/2)), x)

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