Integrand size = 23, antiderivative size = 125 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {2 a^2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}} \] Output:
2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1/2)/d-2*tan(d*x+c) /d/(a+a*sec(d*x+c))^(1/2)+2/3*a*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2)+2/5* a^2*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.83 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.90 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {16 \sqrt {2} \left (\frac {1}{1+\sec (c+d x)}\right )^{9/2} \left (-\frac {\cos (c+d x) (9+5 \cos (c+d x)) \csc ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (30 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \cos ^2(c+d x)+(-29+22 \cos (c+d x)-23 \cos (2 (c+d x))) \sqrt {1-\sec (c+d x)}\right )}{480 \sqrt {1-\sec (c+d x)}}-\frac {4}{9} \operatorname {Hypergeometric2F1}\left (2,\frac {9}{2},\frac {11}{2},-2 \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^5(c+d x)}{5 d \sqrt {a (1+\sec (c+d x))} \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{7/2}} \] Input:
Integrate[Tan[c + d*x]^4/Sqrt[a + a*Sec[c + d*x]],x]
Output:
(16*Sqrt[2]*((1 + Sec[c + d*x])^(-1))^(9/2)*(-1/480*(Cos[c + d*x]*(9 + 5*C os[c + d*x])*Csc[(c + d*x)/2]^6*Sec[(c + d*x)/2]^2*(30*ArcTanh[Sqrt[1 - Se c[c + d*x]]]*Cos[c + d*x]^2 + (-29 + 22*Cos[c + d*x] - 23*Cos[2*(c + d*x)] )*Sqrt[1 - Sec[c + d*x]]))/Sqrt[1 - Sec[c + d*x]] - (4*Hypergeometric2F1[2 , 9/2, 11/2, -2*Sec[c + d*x]*Sin[(c + d*x)/2]^2]*Sec[c + d*x]*Tan[(c + d*x )/2]^2)/9)*Tan[c + d*x]^5)/(5*d*Sqrt[a*(1 + Sec[c + d*x])]*(1 - Tan[(c + d *x)/2]^2)^(7/2))
Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4375, 363, 254, 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.
| \(\displaystyle \int \frac {\tan ^4(c+d x)}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^4}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle -\frac {2 a^2 \int \frac {\tan ^4(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}{(\sec (c+d x) a+a)^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle -\frac {2 a^2 \left (\int \frac {\tan ^4(c+d x)}{(\sec (c+d x) a+a)^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\frac {\tan ^5(c+d x)}{5 (a \sec (c+d x)+a)^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle -\frac {2 a^2 \left (\int \left (\frac {\tan ^2(c+d x)}{a (\sec (c+d x) a+a)}+\frac {1}{a^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}-\frac {1}{a^2}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\frac {\tan ^5(c+d x)}{5 (a \sec (c+d x)+a)^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^2 \left (-\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2}}+\frac {\tan (c+d x)}{a^2 \sqrt {a \sec (c+d x)+a}}-\frac {\tan ^5(c+d x)}{5 (a \sec (c+d x)+a)^{5/2}}-\frac {\tan ^3(c+d x)}{3 a (a \sec (c+d x)+a)^{3/2}}\right )}{d}\) |
Input:
Int[Tan[c + d*x]^4/Sqrt[a + a*Sec[c + d*x]],x]
Output:
(-2*a^2*(-(ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]/a^(5/2) ) + Tan[c + d*x]/(a^2*Sqrt[a + a*Sec[c + d*x]]) - Tan[c + d*x]^3/(3*a*(a + a*Sec[c + d*x])^(3/2)) - Tan[c + d*x]^5/(5*(a + a*Sec[c + d*x])^(5/2))))/ d
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d) Subst[Int[x^m*((2 + a*x^2 )^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] ]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I ntegerQ[n - 1/2]
Time = 0.93 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (17 \sin \left (d x +c \right )+\tan \left (d x +c \right )-3 \sec \left (d x +c \right ) \tan \left (d x +c \right )+15 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{15 d a \left (1+\cos \left (d x +c \right )\right )}\) | \(144\) |
Input:
int(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/15/d/a*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*(17*sin(d*x+c)+tan(d*x+c )-3*sec(d*x+c)*tan(d*x+c)+15*(1+cos(d*x+c))*(-cos(d*x+c)/(1+cos(d*x+c)))^( 1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x+c)*c ot(d*x+c)+csc(d*x+c)^2-1)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.49 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {15 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (17 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (17 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:
integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
Output:
[-1/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d*x + c )^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin( d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(17*cos(d*x + c)^2 + cos(d*x + c) - 3)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/ (a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2), -2/15*(15*(cos(d*x + c)^3 + cos (d*x + c)^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d* x + c)/(sqrt(a)*sin(d*x + c))) + (17*cos(d*x + c)^2 + cos(d*x + c) - 3)*sq rt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)]
\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \] Input:
integrate(tan(d*x+c)**4/(a+a*sec(d*x+c))**(1/2),x)
Output:
Integral(tan(c + d*x)**4/sqrt(a*(sec(c + d*x) + 1)), x)
\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
Output:
1/30*(20*(3*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d *x + 2*c), cos(2*d*x + 2*c) + 1)) - 4*(15*cos(4*d*x + 4*c) + 20*cos(2*d*x + 2*c) + 17)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 15 *((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arcta n2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4 )*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2 *c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2 (sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - (cos(2*d*x + 2*c)^2 + sin (2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arctan2((cos(2*d*x + 2*c)^2 + si n(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d* x + 2*c) + 1)) - 1) + 2*(a*d*cos(2*d*x + 2*c)^2 + a*d*sin(2*d*x + 2*c)^2 + 2*a*d*cos(2*d*x + 2*c) + a*d)*integrate(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(10*d*x + 10*c)*cos(2*d*x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(1 0*d*x + 10*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 6*s in(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + s in(2*d*x + 2*c)^2)*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))...
Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (109) = 218\).
Time = 0.65 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.82 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} \sqrt {-a} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} + \frac {4 \, {\left ({\left (13 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 40 \, a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )}}{30 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:
integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
Output:
-1/30*sqrt(2)*(15*sqrt(2)*sqrt(-a)*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c ) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2 *(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*abs(a) - 6*a))/abs(a) + 4*((13*a^2*tan(1/2*d*x + 1/2*c)^2 - 40* a^2)*tan(1/2*d*x + 1/2*c)^2 + 15*a^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(d*sgn(cos(d*x + c)))
Timed out. \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int(tan(c + d*x)^4/(a + a/cos(c + d*x))^(1/2),x)
Output:
int(tan(c + d*x)^4/(a + a/cos(c + d*x))^(1/2), x)
\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{4}}{\sec \left (d x +c \right )+1}d x \right )}{a} \] Input:
int(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x)
Output:
(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*tan(c + d*x)**4)/(sec(c + d*x) + 1),x ))/a