\(\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [246]
Optimal result
Integrand size = 21, antiderivative size = 136 \[
\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{24 b \sqrt {d \tan (a+b x)}}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}
\]
[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
Rubi [A] (verified)
Time = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2690, 2692, 2694, 2653, 2720}
\[
\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{24 b \sqrt {d \tan (a+b x)}}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b}
\]
[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 2653
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 2690
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 2692
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*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[2*m, 2*n]
Rule 2694
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 2720
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]
Rubi steps \begin{align*}
\text {integral}& = -\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 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,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*}
Mathematica [C] (verified)
Result contains complex when optimal does not.
Time = 2.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96
\[
\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {\cos (2 (a+b x)) \csc (a+b x) \left (10 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sqrt {\sec ^2(a+b x)}+(-3+10 \cos (2 (a+b x))+3 \cos (4 (a+b x))) \sqrt {\tan (a+b x)}\right ) (d \tan (a+b x))^{3/2}}{120 b \sqrt {\tan (a+b x)} \left (-1+\tan ^2(a+b x)\right )}
\]
[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))
Maple [C] (warning: unable to verify)
Result contains complex when optimal does not.
Time = 6.39 (sec) , antiderivative size = 1958, normalized size of antiderivative =
14.40
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\(1958\) |
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[In]
int(cos(b*x+a)^5*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
[Out]
1/240/b*sin(b*x+a)*(24*2^(1/2)*cos(b*x+a)^5*sin(b*x+a)+30*I*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot
(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))
-30*I*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticP
i((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(b*x+a)-4*2^(1/2)*cos(b*x+a)^3*sin(b*x+a)+30*I*(1+
csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((1+csc(
b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(b*x+a)-30*I*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1
+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1
/2))+50*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*Ellipti
cF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)-30*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot
(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2
))*cos(b*x+a)-30*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2
)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(b*x+a)+50*(cot(b*x+a)-csc(b*x+a))^(1/2
)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1
/2*2^(1/2))-30*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*
EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))-30*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a
)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2
*2^(1/2))-10*sin(b*x+a)*2^(1/2)*cos(b*x+a)+15*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-(cot(b*x+a)*
cos(b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*x+a)^2+csc(b*
x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-2*cos(b*x+a)-sin(b*x+a)+csc(b*x+a)+2)/(-1+cos(b*x+a)))*cos(b*x+a)-15*(-cos
(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln((2*sin(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x
+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-cot(b*x+a)*cos(b*x+a)+2*cot(b*x+a)+2*cos(b*x+a)+sin
(b*x+a)-csc(b*x+a)-2)/(-1+cos(b*x+a)))*cos(b*x+a)+30*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((s
in(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x+a)+1)/(-1+cos(b*x+a)))*cos(b*x+a)+30
*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a
)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*cos(b*x+a)+15*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(
-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*
x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-2*cos(b*x+a)-sin(b*x+a)+csc(b*x+a)+2)/(-1+cos(b*x+a)))-15*(-c
os(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln((2*sin(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b
*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-cot(b*x+a)*cos(b*x+a)+2*cot(b*x+a)+2*cos(b*x+a)+s
in(b*x+a)-csc(b*x+a)-2)/(-1+cos(b*x+a)))+30*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)
*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x+a)+1)/(-1+cos(b*x+a)))+30*(-cos(b*x+a)*sin(b*
x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*
x+a)-1)/(-1+cos(b*x+a))))*(d*tan(b*x+a))^(1/2)*d/(-1+cos(b*x+a))/(cos(b*x+a)+1)*2^(1/2)
Fricas [F]
\[
\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{5} \,d x }
\]
[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)
Sympy [F(-1)]
Timed out. \[
\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out}
\]
[In]
integrate(cos(b*x+a)**5*(d*tan(b*x+a))**(3/2),x)
[Out]
Timed out
Maxima [F]
\[
\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{5} \,d x }
\]
[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)
Giac [F(-2)]
Exception generated. \[
\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Exception raised: TypeError}
\]
[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,sageVARx):;OUTP
UT:The choice was done assuming 0=[0,0]ext_reduce Error: Bad Argument TypeThe choice was done assuming 0=[0,0]
ext_reduce
Mupad [F(-1)]
Timed out. \[
\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int {\cos \left (a+b\,x\right )}^5\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x
\]
[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)