Integrand size = 23, antiderivative size = 160 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^4(c+d x) \, dx=\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {2 a \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {6 a^3 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {2 a^4 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}} \] Output:
2*a^(1/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-2*a*tan(d*x+ c)/d/(a+a*sec(d*x+c))^(1/2)+2/3*a^2*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2)+ 6/5*a^3*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+2/7*a^4*tan(d*x+c)^7/d/(a+a* sec(d*x+c))^(7/2)
Time = 6.52 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^4(c+d x) \, dx=\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt {a (1+\sec (c+d x))} \left (105 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {7}{2}}(c+d x)+35 \sin \left (\frac {1}{2} (c+d x)\right )-28 \sin \left (\frac {3}{2} (c+d x)\right )-23 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{105 d} \] Input:
Integrate[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^4,x]
Output:
(Sec[(c + d*x)/2]*Sec[c + d*x]^3*Sqrt[a*(1 + Sec[c + d*x])]*(105*Sqrt[2]*A rcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(7/2) + 35*Sin[(c + d*x)/2] - 28*Sin[(3*(c + d*x))/2] - 23*Sin[(7*(c + d*x))/2]))/(105*d)
Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4375, 364, 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.
| \(\displaystyle \int \tan ^4(c+d x) \sqrt {a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cot \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle -\frac {2 a^3 \int \frac {\tan ^4(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^2}{(\sec (c+d x) a+a)^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle -\frac {2 a^3 \int \left (\frac {a \tan ^6(c+d x)}{(\sec (c+d x) a+a)^3}+\frac {3 \tan ^4(c+d x)}{(\sec (c+d x) a+a)^2}+\frac {\tan ^2(c+d x)}{a (\sec (c+d x) a+a)}+\frac {1}{a^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}-\frac {1}{a^2}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^3 \left (-\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2}}+\frac {\tan (c+d x)}{a^2 \sqrt {a \sec (c+d x)+a}}-\frac {a \tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}-\frac {3 \tan ^5(c+d x)}{5 (a \sec (c+d x)+a)^{5/2}}-\frac {\tan ^3(c+d x)}{3 a (a \sec (c+d x)+a)^{3/2}}\right )}{d}\) |
Input:
Int[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^4,x]
Output:
(-2*a^3*(-(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)) - (3*Tan[c + d*x]^5)/(5*(a + a*Sec[c + d*x])^(5/2) ) - (a*Tan[c + d*x]^7)/(7*(a + a*Sec[c + d*x])^(7/2))))/d
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
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 = 0.89 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.99
| method | result | size |
| default | \(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (92 \cos \left (d x +c \right )^{3}+46 \cos \left (d x +c \right )^{2}-18 \cos \left (d x +c \right )-15\right ) \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}+\left (-105 \cos \left (d x +c \right )-105\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{105 d \left (1+\cos \left (d x +c \right )\right )}\) | \(159\) |
Input:
int((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
-2/105/d*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*((92*cos(d*x+c)^3+46*cos( d*x+c)^2-18*cos(d*x+c)-15)*tan(d*x+c)*sec(d*x+c)^2+(-105*cos(d*x+c)-105)*( -cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(2^(1/2)/(cot(d*x+c)^2-2*csc(d*x+ c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c))))
Time = 0.10 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.07 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^4(c+d x) \, dx=\left [\frac {105 \, {\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (92 \, \cos \left (d x + c\right )^{3} + 46 \, \cos \left (d x + c\right )^{2} - 18 \, \cos \left (d x + c\right ) - 15\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, -\frac {2 \, {\left (105 \, {\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (92 \, \cos \left (d x + c\right )^{3} + 46 \, \cos \left (d x + c\right )^{2} - 18 \, \cos \left (d x + c\right ) - 15\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^4,x, algorithm="fricas")
Output:
[1/105*(105*(cos(d*x + c)^4 + cos(d*x + c)^3)*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*(92*cos(d*x + c)^3 + 46*cos(d*x + c)^2 - 18*cos(d*x + c) - 15)*sqrt((a*cos(d*x + c) + a)/cos (d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3), -2/105*(10 5*(cos(d*x + c)^4 + cos(d*x + c)^3)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (92*cos(d*x + c)^3 + 46*cos(d*x + c)^2 - 18*cos(d*x + c) - 15)*sqrt((a*cos(d*x + c) + a)/cos (d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]
\[ \int \sqrt {a+a \sec (c+d x)} \tan ^4(c+d x) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \tan ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate((a+a*sec(d*x+c))**(1/2)*tan(d*x+c)**4,x)
Output:
Integral(sqrt(a*(sec(c + d*x) + 1))*tan(c + d*x)**4, x)
\[ \int \sqrt {a+a \sec (c+d x)} \tan ^4(c+d x) \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \tan \left (d x + c\right )^{4} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^4,x, algorithm="maxima")
Output:
-1/210*(105*((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 + 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)), (co s(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 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan 2(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*(d*cos(2*d*x + 2*c)^2 + d*sin(2*d*x + 2 *c)^2 + 2*d*cos(2*d*x + 2*c) + d)*integrate((((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 + si n(10*d*x + 10*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 6*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 8 *c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 6*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*d*x + 4*c)*sin(...
Time = 0.39 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.54 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^4(c+d x) \, dx=-\frac {\sqrt {2} {\left (\frac {105 \, \sqrt {2} \sqrt {-a} a \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} - \frac {4 \, {\left (105 \, a^{4} - {\left (385 \, a^{4} + {\left (43 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 203 \, a^{4}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{210 \, d} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^4,x, algorithm="giac")
Output:
-1/210*sqrt(2)*(105*sqrt(2)*sqrt(-a)*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)/a bs(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*(105*a^4 - (385*a^4 + (43*a^4*tan (1/2*d*x + 1/2*c)^2 - 203*a^4)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c )^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/ 2*d*x + 1/2*c)^2 + a)))*sgn(cos(d*x + c))/d
Timed out. \[ \int \sqrt {a+a \sec (c+d x)} \tan ^4(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^4\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \] 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)
\[ \int \sqrt {a+a \sec (c+d x)} \tan ^4(c+d x) \, dx=\sqrt {a}\, \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{4}d x \right ) \] Input:
int((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^4,x)
Output:
sqrt(a)*int(sqrt(sec(c + d*x) + 1)*tan(c + d*x)**4,x)