\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [70]

Optimal result

Integrand size = 33, antiderivative size = 147 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {2 (7 A+9 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b d \sqrt {\cos (c+d x)}}+\frac {2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {2 b^2 (7 A+9 C) \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {2 (7 A+9 C) \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}} \]

[Out]

2/9*A*b^4*sin(d*x+c)/d/(b*cos(d*x+c))^(9/2)+2/45*b^2*(7*A+9*C)*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/15*(7*A+9*C
)*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)-2/15*(7*A+9*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(s
in(1/2*d*x+1/2*c),2^(1/2))*(b*cos(d*x+c))^(1/2)/b/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {16, 3091, 2716, 2721, 2719} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {2 b^2 (7 A+9 C) \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {2 (7 A+9 C) \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}}-\frac {2 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{15 b d \sqrt {\cos (c+d x)}} \]

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5)/Sqrt[b*Cos[c + d*x]],x]

[Out]

(-2*(7*A + 9*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*b*d*Sqrt[Cos[c + d*x]]) + (2*A*b^4*Sin[c +
d*x])/(9*d*(b*Cos[c + d*x])^(9/2)) + (2*b^2*(7*A + 9*C)*Sin[c + d*x])/(45*d*(b*Cos[c + d*x])^(5/2)) + (2*(7*A
+ 9*C)*Sin[c + d*x])/(15*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*} \text {integral}& = b^5 \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{11/2}} \, dx \\ & = \frac {2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {1}{9} \left (b^3 (7 A+9 C)\right ) \int \frac {1}{(b \cos (c+d x))^{7/2}} \, dx \\ & = \frac {2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {2 b^2 (7 A+9 C) \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {1}{15} (b (7 A+9 C)) \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx \\ & = \frac {2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {2 b^2 (7 A+9 C) \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {2 (7 A+9 C) \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}}-\frac {(7 A+9 C) \int \sqrt {b \cos (c+d x)} \, dx}{15 b} \\ & = \frac {2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {2 b^2 (7 A+9 C) \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {2 (7 A+9 C) \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}}-\frac {\left ((7 A+9 C) \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 b \sqrt {\cos (c+d x)}} \\ & = -\frac {2 (7 A+9 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b d \sqrt {\cos (c+d x)}}+\frac {2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {2 b^2 (7 A+9 C) \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {2 (7 A+9 C) \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {-6 (7 A+9 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 (7 A+9 C) \sin (c+d x)+2 \sec (c+d x) \left (7 A+9 C+5 A \sec ^2(c+d x)\right ) \tan (c+d x)}{45 d \sqrt {b \cos (c+d x)}} \]

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5)/Sqrt[b*Cos[c + d*x]],x]

[Out]

(-6*(7*A + 9*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 6*(7*A + 9*C)*Sin[c + d*x] + 2*Sec[c + d*x]*(7*
A + 9*C + 5*A*Sec[c + d*x]^2)*Tan[c + d*x])/(45*d*Sqrt[b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(730\) vs. \(2(155)=310\).

Time = 16.57 (sec) , antiderivative size = 731, normalized size of antiderivative = 4.97

method	result	size
default	\(\text {Expression too large to display}\)	\(731\)
parts	\(\text {Expression too large to display}\)	\(782\)

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(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)*(2/5*C/b/sin(1/2*d*x+1/2*c)^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)*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(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*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(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+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(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+b*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*A*(-1/144*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)^5-7/180*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)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(
1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(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))-7/15*(s
in(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))-EllipticE(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 = 160, normalized size of antiderivative = 1.09 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {3 \, \sqrt {2} {\left (7 i \, A + 9 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-7 i \, A - 9 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, {\left (7 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (7 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{45 \, b d \cos \left (d x + c\right )^{5}} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/45*(3*sqrt(2)*(7*I*A + 9*I*C)*sqrt(b)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(
d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-7*I*A - 9*I*C)*sqrt(b)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierst
rassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(7*A + 9*C)*cos(d*x + c)^4 + (7*A + 9*C)*cos(d*x +
c)^2 + 5*A)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**5/(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{5}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^5/sqrt(b*cos(d*x + c)), x)

Giac [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{5}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^5/sqrt(b*cos(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^5\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^5*(b*cos(c + d*x))^(1/2)),x)

[Out]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^5*(b*cos(c + d*x))^(1/2)), x)