\(\int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [113]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 b^3 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 b \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2716, 2721, 2720} \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 b^3 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 b \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}} \]

[In]

Int[Sec[c + d*x]^4/Sqrt[b*Cos[c + d*x]],x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (2*b^3*Sin[c + d*x])/(7*d*(b*C
os[c + d*x])^(7/2)) + (10*b*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*} \text {integral}& = b^4 \int \frac {1}{(b \cos (c+d x))^{9/2}} \, dx \\ & = \frac {2 b^3 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {1}{7} \left (5 b^2\right ) \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx \\ & = \frac {2 b^3 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 b \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {5}{21} \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx \\ & = \frac {2 b^3 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 b \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 \sqrt {b \cos (c+d x)}} \\ & = \frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 b^3 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {10 b \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.66 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 \left (5+3 \sec ^2(c+d x)\right ) \tan (c+d x)}{21 d \sqrt {b \cos (c+d x)}} \]

[In]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(266\) vs. \(2(107)=214\).

Time = 2.52 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.81

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}{28 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{4}}-\frac {5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}{21 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(267\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1/28*cos(1/2*d*x+1/2*c)/b*(-b*(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)^4-5/21*cos(1/2*d*x+1/2*c)/b*(-b*(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+10/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(
1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(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)))/sin(1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.21 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {-5 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} {\left (5 \, \cos \left (d x + c\right )^{2} + 3\right )} \sin \left (d x + c\right )}{21 \, b d \cos \left (d x + c\right )^{4}} \]

[In]

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

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt {b \cos {\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

Integral(sec(c + d*x)**4/sqrt(b*cos(c + d*x)), x)

Maxima [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

integrate(sec(d*x + c)^4/sqrt(b*cos(d*x + c)), x)

Giac [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

integrate(sec(d*x + c)^4/sqrt(b*cos(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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