\(\int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [328]

Optimal result

Integrand size = 23, antiderivative size = 404 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^2-21 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {-\frac {b (-1+\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 d}+\frac {2 \sqrt {a+b} \left (-8 a^2+2 a b+21 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {-\frac {b (-1+\sec (c+d x))}{a+b}} \sqrt {\frac {b (1+\sec (c+d x))}{-a+b}}}{15 b^3 d}-\frac {8 a \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac {2 \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d} \] Output:

-2*(a+b)^(1/2)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a 
+b)/a,((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c 
))/(a-b))^(1/2)/a/d-2/15*(a-b)*(a+b)^(1/2)*(8*a^2-21*b^2)*cot(d*x+c)*Ellip 
ticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(-b*(-1+sec(d 
*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d+2/15*(a+b)^(1/2) 
*(-8*a^2+2*a*b+21*b^2)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^( 
1/2),((a+b)/(a-b))^(1/2))*(-b*(-1+sec(d*x+c))/(a+b))^(1/2)*(b*(1+sec(d*x+c 
))/(-a+b))^(1/2)/b^3/d-8/15*a*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/5* 
sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 13.71 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \sqrt {\sec (c+d x)} \left (\frac {\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-2 \left (8 a^3+8 a^2 b-21 a b^2-21 b^3\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+4 b \left (4 a^2+a b-18 b^2\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+60 b^3 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+\frac {1}{2} \left (8 a^2-21 b^2\right ) (b+a \cos (c+d x)) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+(b+a \cos (c+d x)) \sqrt {\sec (c+d x)} \left (\left (8 a^2-21 b^2\right ) \sin (c+d x)+b (-4 a+3 b \sec (c+d x)) \tan (c+d x)\right )\right )}{15 b^3 d \sqrt {a+b \sec (c+d x)}} \] Input:

Integrate[Tan[c + d*x]^4/Sqrt[a + b*Sec[c + d*x]],x]
 

Output:

(2*Sqrt[Sec[c + d*x]]*((Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(-2*(8*a^3 + 
 8*a^2*b - 21*a*b^2 - 21*b^3)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/ 
(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(a + b*Sec[c + d*x])/((a + b)* 
(1 + Sec[c + d*x]))] + 4*b*(4*a^2 + a*b - 18*b^2)*EllipticF[ArcSin[Tan[(c 
+ d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(a + b*Sec 
[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + 60*b^3*EllipticPi[-1, ArcSin[Ta 
n[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(a + 
b*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + ((8*a^2 - 21*b^2)*(b + a*C 
os[c + d*x])*Sec[(c + d*x)/2]^3*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2])) 
/2))/Sqrt[Sec[(c + d*x)/2]^2] + (b + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]]*(( 
8*a^2 - 21*b^2)*Sin[c + d*x] + b*(-4*a + 3*b*Sec[c + d*x])*Tan[c + d*x]))) 
/(15*b^3*d*Sqrt[a + b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 610, normalized size of antiderivative = 1.51, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4383, 2009}

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.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^4}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4383

\(\displaystyle \int \left (\frac {\sec ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}}-\frac {2 \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}}+\frac {1}{\sqrt {a+b \sec (c+d x)}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 (a-b) \sqrt {a+b} \left (8 a^2+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b^4 d}-\frac {2 \sqrt {a+b} \left (8 a^2-2 a b+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{15 b^3 d}+\frac {4 (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}+\frac {4 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {8 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 b^2 d}+\frac {2 \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

Input:

Int[Tan[c + d*x]^4/Sqrt[a + b*Sec[c + d*x]],x]
 

Output:

(4*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d* 
x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sq 
rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(8 
*a^2 + 9*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[ 
a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*( 
1 + Sec[c + d*x]))/(a - b))])/(15*b^4*d) + (4*Sqrt[a + b]*Cot[c + d*x]*Ell 
ipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt 
[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/ 
(b*d) - (2*Sqrt[a + b]*(8*a^2 - 2*a*b + 9*b^2)*Cot[c + d*x]*EllipticF[ArcS 
in[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Se 
c[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15*b^3*d) 
- (2*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[ 
c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + 
b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d) - (8*a*Sqrt[a + b*Sec[c 
 + d*x]]*Tan[c + d*x])/(15*b^2*d) + (2*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x 
]]*Tan[c + d*x])/(5*b*d)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4383

Int[cot[(c_.) + (d_.)*(x_)]^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_ 
), x_Symbol] :> Int[ExpandIntegrand[(a + b*Csc[c + d*x])^n, (-1 + Csc[c + d 
*x]^2)^(m/2), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0] && I 
GtQ[m/2, 0] && IntegerQ[n - 1/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(961\) vs. \(2(365)=730\).

Time = 16.98 (sec) , antiderivative size = 962, normalized size of antiderivative = 2.38

method	result	size
default	\(\text {Expression too large to display}\)	\(962\)

Input:

int(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/15/d/b^3*(a+b*sec(d*x+c))^(1/2)/(a*cos(d*x+c)^2+a*cos(d*x+c)+b*cos(d*x+c 
)+b)*(30*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d 
*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*b^3*EllipticPi(-csc(d*x+c) 
+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+8*(cos(d*x+c)^2+2*cos(d*x+c)+1)*(1/(a+ 
b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 
)*a^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(cos(d*x+c)^ 
2+2*cos(d*x+c)+1)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x 
+c)/(1+cos(d*x+c)))^(1/2)*a^2*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a 
+b))^(1/2))+21*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(1/(a+b)*(b+a*cos(d*x+c))/(1 
+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticE(-csc 
(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+21*(-cos(d*x+c)^2-2*cos(d*x+c)-1)* 
(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)) 
)^(1/2)*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(-cos( 
d*x+c)^2-2*cos(d*x+c)-1)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*( 
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2*b*EllipticF(-csc(d*x+c)+cot(d*x+c),(( 
a-b)/(a+b))^(1/2))+2*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(1/(a+b)*(b+a*cos(d*x+ 
c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^2*Elliptic 
F(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+36*(cos(d*x+c)^2+2*cos(d*x+c 
)+1)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d* 
x+c)))^(1/2)*b^3*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+...
 

Fricas [F]

\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

integral(tan(d*x + c)^4/sqrt(b*sec(d*x + c) + a), x)
 

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \] Input:

integrate(tan(d*x+c)**4/(a+b*sec(d*x+c))**(1/2),x)
 

Output:

Integral(tan(c + d*x)**4/sqrt(a + b*sec(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(tan(d*x + c)^4/sqrt(b*sec(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(tan(d*x + c)^4/sqrt(b*sec(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \] Input:

int(tan(c + d*x)^4/(a + b/cos(c + d*x))^(1/2),x)
 

Output:

int(tan(c + d*x)^4/(a + b/cos(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \tan \left (d x +c \right )^{4}}{\sec \left (d x +c \right ) b +a}d x \] Input:

int(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),x)
 

Output:

int((sqrt(sec(c + d*x)*b + a)*tan(c + d*x)**4)/(sec(c + d*x)*b + a),x)