3.268 \(\int \frac{\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=142 \[ -\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{77 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{20 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

[Out]

(-2*Cos[a + b*x]^5)/(b*d*Sqrt[d*Tan[a + b*x]]) - (77*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a +
b*x]])/(20*b*d^2*Sqrt[Sin[2*a + 2*b*x]]) - (77*Cos[a + b*x]^3*(d*Tan[a + b*x])^(3/2))/(30*b*d^3) - (11*Cos[a +
b*x]^5*(d*Tan[a + b*x])^(3/2))/(5*b*d^3)

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Rubi [A]  time = 0.187095, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2609, 2612, 2615, 2572, 2639} \[ -\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{77 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{20 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5/(d*Tan[a + b*x])^(3/2),x]

[Out]

(-2*Cos[a + b*x]^5)/(b*d*Sqrt[d*Tan[a + b*x]]) - (77*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a +
b*x]])/(20*b*d^2*Sqrt[Sin[2*a + 2*b*x]]) - (77*Cos[a + b*x]^3*(d*Tan[a + b*x])^(3/2))/(30*b*d^3) - (11*Cos[a +
b*x]^5*(d*Tan[a + b*x])^(3/2))/(5*b*d^3)

Rule 2609

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*(n + 1)), x] - Dist[(m + n + 1)/(b^2*(n + 1)), Int[(a*Sec[e + f*x])^m*(
b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && IntegersQ[2*m, 2*n]

Rule 2612

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integer
sQ[2*m, 2*n]

Rule 2615

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[(Sqrt[Cos[e + f*x]]*Sqrt[b*
Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
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{\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{11 \int \cos ^5(a+b x) \sqrt{d \tan (a+b x)} \, dx}{d^2}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{77 \int \cos ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx}{10 d^2}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{77 \int \cos (a+b x) \sqrt{d \tan (a+b x)} \, dx}{20 d^2}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{\left (77 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{20 d^2 \sqrt{\sin (a+b x)}}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{\left (77 \cos (a+b x) \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{20 d^2 \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{77 \cos (a+b x) E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{d \tan (a+b x)}}{20 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}\\ \end{align*}

Mathematica [C]  time = 0.838343, size = 89, normalized size = 0.63 \[ \frac{\sin (a+b x) \left (-308 \tan ^2(a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )+34 \cos (2 (a+b x))+3 \cos (4 (a+b x))-277\right )}{120 b (d \tan (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^5/(d*Tan[a + b*x])^(3/2),x]

[Out]

(Sin[a + b*x]*(-277 + 34*Cos[2*(a + b*x)] + 3*Cos[4*(a + b*x)] - 308*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a +
b*x]^2]*Sqrt[Sec[a + b*x]^2]*Tan[a + b*x]^2))/(120*b*(d*Tan[a + b*x])^(3/2))

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Maple [B]  time = 0.165, size = 536, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^5/(d*tan(b*x+a))^(3/2),x)

[Out]

1/120/b*2^(1/2)*(12*cos(b*x+a)^6*2^(1/2)+22*cos(b*x+a)^4*2^(1/2)+462*cos(b*x+a)*EllipticE((-(cos(b*x+a)-1-sin(
b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a)
)^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)-231*cos(b*x+a)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin
(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(
cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+462*EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(
1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a
))/sin(b*x+a))^(1/2)-231*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/
sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+7
7*cos(b*x+a)^2*2^(1/2)-231*cos(b*x+a)*2^(1/2))*sin(b*x+a)/cos(b*x+a)^2/(d*sin(b*x+a)/cos(b*x+a))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*cos(b*x + a)^5/(d^2*tan(b*x + a)^2), 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(cos(b*x+a)**5/(d*tan(b*x+a))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)