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

Optimal. Leaf size=113 \[ \frac {2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {2 b (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}} \]

[Out]

2/7*A*b^4*sin(d*x+c)/d/(b*cos(d*x+c))^(7/2)+2/21*b^2*(5*A+7*C)*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+2/21*b*(5*A+7
*C)*(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)

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Rubi [A]  time = 0.11, antiderivative size = 113, 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, 3012, 2636, 2642, 2641} \[ \frac {2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^4*Sin[c +
d*x])/(7*d*(b*Cos[c + d*x])^(7/2)) + (2*b^2*(5*A + 7*C)*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 2636

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 3012

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 ^5(c+d x) \, dx &=b^5 \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{9/2}} \, dx\\ &=\frac {2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {1}{7} \left (b^3 (5 A+7 C)\right ) \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx\\ &=\frac {2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {1}{21} (b (5 A+7 C)) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx\\ &=\frac {2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {\left (b (5 A+7 C) \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {b \cos (c+d x)}}\\ &=\frac {2 b (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 83, normalized size = 0.73 \[ \frac {\sec ^3(c+d x) \sqrt {b \cos (c+d x)} \left ((5 A+7 C) \sin (2 (c+d x))+2 (5 A+7 C) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 A \tan (c+d x)\right )}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*Cos[c + d*x]]*Sec[c + d*x]^3*(2*(5*A + 7*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + (5*A + 7*C)
*Sin[2*(c + d*x)] + 6*A*Tan[c + d*x]))/(21*d)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{5}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)*sqrt(b*cos(d*x + c))*sec(d*x + c)^5, x)

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maple [B]  time = 2.96, size = 411, normalized size = 3.64 \[ -\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 )}\, b \left (A \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 )}}{56 b \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )}}{42 b \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {5 \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}\, \EllipticF \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 )+C \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 )}}{6 b \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\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 )}}\right )\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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*(A*(-1/56*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^4-5/42*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+5/21*(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)))+C*(-1/6*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)/(-1/2+
cos(1/2*d*x+1/2*c)^2)^2+1/3*(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)/(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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)*sqrt(b*cos(d*x + c))*sec(d*x + c)^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^5} \,d x \]

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)^5,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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