3.246 \(\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx\)
Optimal. Leaf size=136 \[ \frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{24 b \sqrt {d \tan (a+b x)}}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b} \]
[Out]
-1/24*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sec(b*x+a)*sin(2*
b*x+2*a)^(1/2)/b/(d*tan(b*x+a))^(1/2)+1/12*d*cos(b*x+a)*(d*tan(b*x+a))^(1/2)/b+1/30*d*cos(b*x+a)^3*(d*tan(b*x+
a))^(1/2)/b-1/5*d*cos(b*x+a)^5*(d*tan(b*x+a))^(1/2)/b
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Rubi [A] time = 0.17, antiderivative size = 136, normalized size of antiderivative = 1.00,
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 =
{2610, 2612, 2614, 2573, 2641} \[ \frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{24 b \sqrt {d \tan (a+b x)}}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b} \]
Antiderivative was successfully verified.
[In]
Int[Cos[a + b*x]^5*(d*Tan[a + b*x])^(3/2),x]
[Out]
(d^2*EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(24*b*Sqrt[d*Tan[a + b*x]]) + (d*Cos[a
+ b*x]*Sqrt[d*Tan[a + b*x]])/(12*b) + (d*Cos[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/(30*b) - (d*Cos[a + b*x]^5*Sqrt[
d*Tan[a + b*x]])/(5*b)
Rule 2573
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]
Rule 2610
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(n - 1))/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && 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 2614
Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, 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 \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx &=-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}+\frac {1}{10} d^2 \int \frac {\cos ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}+\frac {1}{12} d^2 \int \frac {\cos (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}+\frac {1}{24} d^2 \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}+\frac {\left (d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{24 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}+\frac {\left (d^2 \sec (a+b x) \sqrt {\sin (2 a+2 b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx}{24 \sqrt {d \tan (a+b x)}}\\ &=\frac {d^2 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{24 b \sqrt {d \tan (a+b x)}}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\\ \end {align*}
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Mathematica [C] time = 2.47, size = 131, normalized size = 0.96 \[ \frac {\cos (2 (a+b x)) \csc (a+b x) (d \tan (a+b x))^{3/2} \left ((10 \cos (2 (a+b x))+3 \cos (4 (a+b x))-3) \sqrt {\tan (a+b x)}+10 \sqrt [4]{-1} \sqrt {\sec ^2(a+b x)} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right )\right |-1\right )\right )}{120 b \sqrt {\tan (a+b x)} \left (\tan ^2(a+b x)-1\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[a + b*x]^5*(d*Tan[a + b*x])^(3/2),x]
[Out]
(Cos[2*(a + b*x)]*Csc[a + b*x]*(10*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Sqrt[Sec
[a + b*x]^2] + (-3 + 10*Cos[2*(a + b*x)] + 3*Cos[4*(a + b*x)])*Sqrt[Tan[a + b*x]])*(d*Tan[a + b*x])^(3/2))/(12
0*b*Sqrt[Tan[a + b*x]]*(-1 + Tan[a + b*x]^2))
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fricas [F] time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (b x + a\right )} d \cos \left (b x + a\right )^{5} \tan \left (b x + a\right ), x\right ) \]
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))*d*cos(b*x + a)^5*tan(b*x + a), x)
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)ext_reduce Error: Bad Argument Typeext_reduce Error: Bad Argument Type
Evaluation time: 16.39Done
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maple [A] time = 0.51, size = 250, normalized size = 1.84 \[ -\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (12 \sqrt {2}\, \left (\cos ^{6}\left (b x +a \right )\right )-12 \sqrt {2}\, \left (\cos ^{5}\left (b x +a \right )\right )-2 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}+5 \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-5 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+5 \cos \left (b x +a \right ) \sqrt {2}\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{120 b \sin \left (b x +a \right )^{5}} \]
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*(-1+cos(b*x+a))*(12*cos(b*x+a)^6*2^(1/2)-12*2^(1/2)*cos(b*x+a)^5-2*cos(b*x+a)^4*2^(1/2)+5*sin(b*x+a)*
((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(-sin(b*x+a)-1+cos(b*x+a))/
sin(b*x+a))^(1/2)*EllipticF((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+2*cos(b*x+a)^3*2^(1/2)
-5*cos(b*x+a)^2*2^(1/2)+5*cos(b*x+a)*2^(1/2))*cos(b*x+a)*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(3/2)/sin(
b*x+a)^5*2^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{5}\,{d x} \]
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((d*tan(b*x + a))^(3/2)*cos(b*x + a)^5, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (a+b\,x\right )}^5\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(a + b*x)^5*(d*tan(a + b*x))^(3/2),x)
[Out]
int(cos(a + b*x)^5*(d*tan(a + b*x))^(3/2), x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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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