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

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

[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]])

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Rubi [A]  time = 0.150654, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {16, 3012, 2636, 2640, 2639} \[ \frac{2 b^2 (7 A+9 C) \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{2 (7 A+9 C) \sin (c+d x)}{15 d \sqrt{b \cos (c+d x)}}-\frac{2 (7 A+9 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 b d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[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 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]

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]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt{b \cos (c+d x)}} \, dx &=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]  time = 0.795612, size = 97, normalized size = 0.66 \[ \frac{6 (7 A+9 C) \sin (c+d x)-6 (7 A+9 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \tan (c+d x) \sec (c+d x) \left (5 A \sec ^2(c+d x)+7 A+9 C\right )}{45 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[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]])

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Maple [B]  time = 10.628, size = 729, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

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]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\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}}{b \cos \left (d x + c\right )}, x\right ) \end{align*}

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/(b*cos(d*x + c)), 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+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 \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} \end{align*}

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