\(\int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [246]

Optimal result

Integrand size = 21, antiderivative size = 136 \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{24 b \sqrt {d \tan (a+b x)}}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b} \] Output:

1/24*d^2*InverseJacobiAM(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
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.99 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96 \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {\cos (2 (a+b x)) \csc (a+b x) \left (10 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sqrt {\sec ^2(a+b x)}+(-3+10 \cos (2 (a+b x))+3 \cos (4 (a+b x))) \sqrt {\tan (a+b x)}\right ) (d \tan (a+b x))^{3/2}}{120 b \sqrt {\tan (a+b x)} \left (-1+\tan ^2(a+b x)\right )} \] Input:

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

Output:

(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))/(120* 
b*Sqrt[Tan[a + b*x]]*(-1 + Tan[a + b*x]^2))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3090, 3042, 3092, 3042, 3092, 3042, 3094, 3042, 3053, 3042, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (a+b x))^{3/2}}{\sec (a+b x)^5}dx\)

\(\Big \downarrow \) 3090

\(\displaystyle \frac {1}{10} d^2 \int \frac {\cos ^3(a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} d^2 \int \frac {1}{\sec (a+b x)^3 \sqrt {d \tan (a+b x)}}dx-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3092

\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \int \frac {\cos (a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \int \frac {1}{\sec (a+b x) \sqrt {d \tan (a+b x)}}dx+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3092

\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {1}{2} \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {1}{2} \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3094

\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {d \tan (a+b x)}}+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {d \tan (a+b x)}}+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{10} d^2 \left (\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}+\frac {5}{6} \left (\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}+\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{2 b \sqrt {d \tan (a+b x)}}\right )\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\)

Input:

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

Output:

-1/5*(d*Cos[a + b*x]^5*Sqrt[d*Tan[a + b*x]])/b + (d^2*((Cos[a + b*x]^3*Sqr 
t[d*Tan[a + b*x]])/(3*b*d) + (5*((EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x 
]*Sqrt[Sin[2*a + 2*b*x]])/(2*b*Sqrt[d*Tan[a + b*x]]) + (Cos[a + b*x]*Sqrt[ 
d*Tan[a + b*x]])/(b*d)))/6))/10
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[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 3090

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] - Simp[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 3092

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] + Simp[(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)])) && IntegersQ[2*m, 2*n]
 

rule 3094

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] 
:> Simp[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[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 3120

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]
 

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 6.79 (sec) , antiderivative size = 1526, normalized size of antiderivative = 11.22

method	result	size
default	\(\text {Expression too large to display}\)	\(1526\)

Input:

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

Output:

-1/480/b*(d*tan(b*x+a))^(1/2)*(15*2^(1/2)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b 
*x+a)+1)^2)^(1/2)*ln(2*cot(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1) 
^2)^(1/2)+2*csc(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2 
*cot(b*x+a)+2)*(-2*cos(b*x+a)^2+cot(b*x+a)+csc(b*x+a)-2*cos(b*x+a))+60*ln( 
2*cot(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+2*csc(b*x+a 
)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cot(b*x+a)+2)*(-sin( 
b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*(cos(b*x+a)^2+cos(b*x+a))+15*2^( 
1/2)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-2*cot(b*x+a)*(- 
2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*csc(b*x+a)*(-2*sin(b*x+a 
)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cot(b*x+a)+2)*(2*cos(b*x+a)^2-cot(b 
*x+a)-csc(b*x+a)+2*cos(b*x+a))+60*ln(-2*cot(b*x+a)*(-2*sin(b*x+a)*cos(b*x+ 
a)/(cos(b*x+a)+1)^2)^(1/2)-2*csc(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x 
+a)+1)^2)^(1/2)-2*cot(b*x+a)+2)*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^ 
(1/2)*(-cos(b*x+a)^2-cos(b*x+a))+30*2^(1/2)*arctan((sin(b*x+a)*(-2*sin(b*x 
+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x+a)+1)/(-1+cos(b*x+a)))*(-2* 
sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*(-2*cos(b*x+a)^2+cot(b*x+a)+ 
csc(b*x+a)-2*cos(b*x+a))+120*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/ 
2)*arctan((sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-co 
s(b*x+a)+1)/(-1+cos(b*x+a)))*(cos(b*x+a)^2+cos(b*x+a))+30*2^(1/2)*(-2*sin( 
b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*(-2*sin(b*...
 

Fricas [F]

\[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{5} \,d x } \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{5} \,d x } \] Input:

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int {\cos \left (a+b\,x\right )}^5\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\sqrt {d}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \cos \left (b x +a \right )^{5} \tan \left (b x +a \right )d x \right ) d \] Input:

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

Output:

sqrt(d)*int(sqrt(tan(a + b*x))*cos(a + b*x)**5*tan(a + b*x),x)*d