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

Optimal. Leaf size=210 \[ \frac{2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac{2 b (5 A+7 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{b \cos (c+d x)}}+\frac{2 b^3 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{6 b B \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.269163, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {16, 3021, 2748, 2636, 2640, 2639, 2642, 2641} \[ \frac{2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac{2 b (5 A+7 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{b \cos (c+d x)}}+\frac{2 b^3 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{6 b B \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (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^3*B*Sin[c + d*x])/(5*d*(b*Cos[c + d*x])^(5/2)) + (2*b^2*(5*A + 7*C)*Sin[c + d*x])/(21*d*(b*Cos[c
+ d*x])^(3/2)) + (6*b*B*Sin[c + d*x])/(5*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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 2640

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

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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 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]

Rubi steps

\begin{align*} \int \sqrt{b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=b^5 \int \frac{A+B \cos (c+d x)+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 (2 b^2\right ) \int \frac{\frac{7 b^2 B}{2}+\frac{1}{2} b^2 (5 A+7 C) \cos (c+d x)}{(b \cos (c+d x))^{7/2}} \, dx\\ &=\frac{2 A b^4 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\left (b^4 B\right ) \int \frac{1}{(b \cos (c+d x))^{7/2}} \, dx+\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^3 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{1}{5} \left (3 b^2 B\right ) \int \frac{1}{(b \cos (c+d x))^{3/2}} \, dx+\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^3 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{6 b B \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{1}{5} (3 B) \int \sqrt{b \cos (c+d x)} \, dx+\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^3 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{6 b B \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{\left (3 B \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 \sqrt{\cos (c+d x)}}\\ &=-\frac{6 B \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\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^3 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 b^2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{6 b B \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.916425, size = 143, normalized size = 0.68 \[ \frac{2 \sec ^3(c+d x) \sqrt{b \cos (c+d x)} \left (5 (5 A+7 C) \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{25}{2} A \sin (2 (c+d x))+15 A \tan (c+d x)+21 B \sin (c+d x)+63 B \sin (c+d x) \cos ^2(c+d x)-63 B \cos ^{\frac{5}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{35}{2} C \sin (2 (c+d x))\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*Cos[c + d*x]]*Sec[c + d*x]^3*(-63*B*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 5*(5*A + 7*C)*Cos
[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 21*B*Sin[c + d*x] + 63*B*Cos[c + d*x]^2*Sin[c + d*x] + (25*A*Sin[2
*(c + d*x)])/2 + (35*C*Sin[2*(c + d*x)])/2 + 15*A*Tan[c + d*x]))/(105*d)

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Maple [B]  time = 12.198, size = 725, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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)))-1/5*B/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)*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*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))-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*b*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2*b)^(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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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