3.51 \(\int (b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

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

[Out]

(-2*b*(3*A + 5*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*A*b^4*Sin[c +
d*x])/(5*d*(b*Cos[c + d*x])^(5/2)) + (2*b^2*(3*A + 5*C)*Sin[c + d*x])/(5*d*Sqrt[b*Cos[c + d*x]])

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Rubi [A]  time = 0.126973, antiderivative size = 113, normalized size of antiderivative = 1., 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, 2640, 2639} \[ \frac{2 b^2 (3 A+5 C) \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}+\frac{2 A b^4 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}-\frac{2 b (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

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

Mathematica [A]  time = 0.228633, size = 84, normalized size = 0.74 \[ -\frac{\sec ^3(c+d x) (b \cos (c+d x))^{3/2} \left (-(3 A+5 C) \sin (2 (c+d x))+2 (3 A+5 C) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )-2 A \tan (c+d x)\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

-((b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^3*(2*(3*A + 5*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] - (3*A + 5
*C)*Sin[2*(c + d*x)] - 2*A*Tan[c + d*x]))/(5*d)

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Maple [B]  time = 8.875, size = 599, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

2/5*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b/sin(1/2*d*x+1/2*c)^3/(8*sin(1/2*d*x+1/2*c)^6-1
2*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(12*A*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*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+20*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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)^4-40*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*si
n(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*A*cos(1/2*d*x+1/2*c)*sin(1/2*
d*x+1/2*c)^4-20*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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+40*C*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*A*(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*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
2+5*C*(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))-10*C
*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)/(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} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(3/2)*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 b \cos \left (d x + c\right )^{3} + A b \cos \left (d x + c\right )\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((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

integral((C*b*cos(d*x + c)^3 + A*b*cos(d*x + c))*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((b*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**5,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} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(3/2)*sec(d*x + c)^5, x)