\(\int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [337]

Optimal result

Integrand size = 23, antiderivative size = 530 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {2 \left (8 a^4-11 a^2 b^2+3 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {-\frac {b (-1+\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a b^4 \sqrt {a+b} d}+\frac {2 (2 a+b) \left (4 a^2+a b-3 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {-\frac {b (-1+\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a b^3 \sqrt {a+b} d}-\frac {4 a \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 b^2 \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a^2 \sec (c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (4 a^2-b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d} \] 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^2/d+2/3*(8*a^4-11*a^2*b^2+3*b^4)*cot(d*x+c)*EllipticE((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)/a/b^4/(a+b)^(1/2)/d+2/3*(2*a+ 
b)*(4*a^2+a*b-3*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)/a/b^3/(a+b)^(1/2)/d-4*a*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x 
+c))^(1/2)+2*b^2*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)-2*a^2*sec 
(d*x+c)*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)+2/3*(4*a^2-b^2)*(a 
+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/(a^2-b^2)/d
 

Mathematica [A] (warning: unable to verify)

Time = 16.37 (sec) , antiderivative size = 531, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 (b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 \left (8 a^3+8 a^2 b-3 a b^2-3 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-4 a b (4 a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+12 b^3 \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\frac {1}{2} \left (8 a^2-3 b^2\right ) (b+a \cos (c+d x)) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{3 a b^3 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{3/2}}+\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \left (\frac {2 \left (-8 a^2+3 b^2\right ) \sin (c+d x)}{3 a b^3}-\frac {2 \left (-a^2 \sin (c+d x)+b^2 \sin (c+d x)\right )}{a b^2 (b+a \cos (c+d x))}+\frac {2 \tan (c+d x)}{3 b^2}\right )}{d (a+b \sec (c+d x))^{3/2}} \] Input:

Integrate[Tan[c + d*x]^4/(a + b*Sec[c + d*x])^(3/2),x]
 

Output:

(2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + 
 d*x]]*(2*(8*a^3 + 8*a^2*b - 3*a*b^2 - 3*b^3)*Sqrt[Cos[c + d*x]/(1 + Cos[c 
 + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Elliptic 
E[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 4*a*b*(4*a + b)*Sqrt[Cos[c 
+ d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + 
 d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 12*b^3*Sqr 
t[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + 
 Cos[c + d*x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] 
 + ((8*a^2 - 3*b^2)*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^3*(-Sin[(c + d*x 
)/2] + Sin[(3*(c + d*x))/2]))/2))/(3*a*b^3*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + 
 b*Sec[c + d*x])^(3/2)) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((2*(-8*a 
^2 + 3*b^2)*Sin[c + d*x])/(3*a*b^3) - (2*(-(a^2*Sin[c + d*x]) + b^2*Sin[c 
+ d*x]))/(a*b^2*(b + a*Cos[c + d*x])) + (2*Tan[c + d*x])/(3*b^2)))/(d*(a + 
 b*Sec[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.77 (sec) , antiderivative size = 907, normalized size of antiderivative = 1.71, 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.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4383

\(\displaystyle \int \left (\frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}}-\frac {2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}}+\frac {1}{(a+b \sec (c+d x))^{3/2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sec (c+d x) \tan (c+d x) a^2}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {4 \tan (c+d x) a}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (8 a^2-5 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 (\sec (c+d x)+1)}{a-b}} a}{3 b^4 \sqrt {a+b} d}-\frac {4 \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 (\sec (c+d x)+1)}{a-b}} a}{b^2 \sqrt {a+b} d}+\frac {2 \left (4 a^2-b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {2 (2 a+b) (4 a+b) \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 (\sec (c+d x)+1)}{a-b}}}{3 b^3 \sqrt {a+b} d}-\frac {4 \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 (\sec (c+d x)+1)}{a-b}}}{b \sqrt {a+b} d}+\frac {2 b^2 \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)} a}+\frac {2 \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 (\sec (c+d x)+1)}{a-b}}}{\sqrt {a+b} d a}-\frac {2 \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 (\sec (c+d x)+1)}{a-b}}}{\sqrt {a+b} d a}-\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 (\sec (c+d x)+1)}{a-b}}}{d a^2}\)

Input:

Int[Tan[c + d*x]^4/(a + b*Sec[c + d*x])^(3/2),x]
 

Output:

(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))])/(a*Sqrt[a + b]*d) - (4*a*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))])/(b^2*Sqrt[a + 
b]*d) + (2*a*(8*a^2 - 5*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))])/(3*b^4*Sqrt[a + b]*d) - (2*Co 
t[c + d*x]*EllipticF[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*Sqrt[a + b]*d) - (4*Cot[c + d*x]*EllipticF[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*Sqrt[a + b]*d) + ( 
2*(2*a + b)*(4*a + b)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x 
]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqr 
t[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^3*Sqrt[a + b]*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^2*d) - (4*a*Tan[c + d*x])/((a^2 - b^2)*d 
*Sqrt[a + b*Sec[c + d*x]]) + (2*b^2*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt...
 

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 [A] (verified)

Time = 15.06 (sec) , antiderivative size = 848, normalized size of antiderivative = 1.60

method	result	size
default	\(\text {Expression too large to display}\)	\(848\)

Input:

int(tan(d*x+c)^4/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3/d/a/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)*(6*(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))+3*(-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))+3*(-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)*(co 
s(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*EllipticF( 
-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*a^3*cos(d*x+c)*sin(d*x+c)+4* 
sin(d*x+c)*(1-cos(d*x+c))*a^2*b+(-3*cos(d*x+c)^2-cos(d*x+c)-1)*a*b^2*tan(d 
*x+c)+3*b^3*cos(d*x+c)*sin(d*x+c))
 

Fricas [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(tan(d*x+c)**4/(a+b*sec(d*x+c))**(3/2),x)
 

Output:

Integral(tan(c + d*x)**4/(a + b*sec(c + d*x))**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate(tan(d*x + c)^4/(b*sec(d*x + c) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(tan(c + d*x)^4/(a + b/cos(c + d*x))^(3/2),x)
 

Output:

int(tan(c + d*x)^4/(a + b/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \tan \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \] Input:

int(tan(d*x+c)^4/(a+b*sec(d*x+c))^(3/2),x)
 

Output:

int((sqrt(sec(c + d*x)*b + a)*tan(c + d*x)**4)/(sec(c + d*x)**2*b**2 + 2*s 
ec(c + d*x)*a*b + a**2),x)