3.68 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{\sqrt{b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=112 \[ \frac{2 A b^2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 (3 A+5 C) \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 b d \sqrt{\cos (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.112419, antiderivative size = 112, 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} \[ \frac{2 A b^2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 (3 A+5 C) \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 b d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 6.28102, size = 522, normalized size = 4.66 \[ b \left (\frac{\cos ^4(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (\frac{4 \sec (c) \sec (c+d x) (3 A \sin (d x)+5 C \sin (d x))}{5 d}+\frac{4 (3 A+5 C) \csc (c) \sec (c)}{5 d}+\frac{4 A \sec (c) \sin (d x) \sec ^3(c+d x)}{5 d}+\frac{4 A \tan (c) \sec ^2(c+d x)}{5 d}\right )}{(b \cos (c+d x))^{3/2} (2 A+C \cos (2 c+2 d x)+C)}-\frac{i (3 A+5 C) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos ^{\frac{7}{2}}(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (\frac{2 e^{2 i d x} \sqrt{e^{-i d x} \left (2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )\right )} \sqrt{i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{3 i d \cos (c) \left (1+e^{2 i d x}\right )-3 d \sin (c) \left (-1+e^{2 i d x}\right )}-\frac{2 \sqrt{e^{-i d x} \left (2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )\right )} \sqrt{i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{d \sin (c) \left (-1+e^{2 i d x}\right )-i d \cos (c) \left (1+e^{2 i d x}\right )}\right )}{10 (b \cos (c+d x))^{3/2} (2 A+C \cos (2 c+2 d x)+C)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

b*(((-I/10)*(3*A + 5*C)*Cos[c + d*x]^(7/2)*Csc[c/2]*Sec[c/2]*(C + A*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeo
metric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-
1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1
+ E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*
d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)
]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^
((2*I)*d*x))*Sin[c])))/((b*Cos[c + d*x])^(3/2)*(2*A + C + C*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^4*(C + A*Sec[c
+ d*x]^2)*((4*(3*A + 5*C)*Csc[c]*Sec[c])/(5*d) + (4*A*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(5*d) + (4*Sec[c]*Sec[c
+ d*x]*(3*A*Sin[d*x] + 5*C*Sin[d*x]))/(5*d) + (4*A*Sec[c + d*x]^2*Tan[c])/(5*d)))/((b*Cos[c + d*x])^(3/2)*(2*A
+ C + C*Cos[2*c + 2*d*x])))

________________________________________________________________________________________

Maple [B]  time = 9.168, size = 601, normalized size = 5.4 \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)^3/(b*cos(d*x+c))^(1/2),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

________________________________________________________________________________________

Maxima [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 )^{3}}{\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)^3/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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 )^{3}}{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)^3/(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)^3/(b*cos(d*x + c)), x)

________________________________________________________________________________________

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)**3/(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

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 )^{3}}{\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)^3/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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