Integrand size = 23, antiderivative size = 222 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^6(c+d x) \, dx=-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}-\frac {2 a^2 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {2 a^3 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {2 a^4 \tan ^7(c+d x)}{d (a+a \sec (c+d x))^{7/2}}+\frac {10 a^5 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac {2 a^6 \tan ^{11}(c+d x)}{11 d (a+a \sec (c+d x))^{11/2}} \] 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) +2/5*a^3*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+2*a^4*tan(d*x+c)^7/d/(a+a*s ec(d*x+c))^(7/2)+10/9*a^5*tan(d*x+c)^9/d/(a+a*sec(d*x+c))^(9/2)+2/11*a^6*t an(d*x+c)^11/d/(a+a*sec(d*x+c))^(11/2)
Time = 7.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.60 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^6(c+d x) \, dx=-\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt {a (1+\sec (c+d x))} \left (3960 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {11}{2}}(c+d x)+792 \sin \left (\frac {1}{2} (c+d x)\right )-1386 \sin \left (\frac {3}{2} (c+d x)\right )+495 \sin \left (\frac {5}{2} (c+d x)\right )-616 \sin \left (\frac {7}{2} (c+d x)\right )-247 \sin \left (\frac {11}{2} (c+d x)\right )\right )}{3960 d} \] Input:
Integrate[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^6,x]
Output:
-1/3960*(Sec[(c + d*x)/2]*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*(3960* Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(11/2) + 792*Sin[(c + d*x)/2] - 1386*Sin[(3*(c + d*x))/2] + 495*Sin[(5*(c + d*x))/2] - 616*Sin [(7*(c + d*x))/2] - 247*Sin[(11*(c + d*x))/2]))/d
Time = 0.29 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.92, 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.
\(\displaystyle \int \tan ^6(c+d x) \sqrt {a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cot \left (c+d x+\frac {\pi }{2}\right )^6 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 4375 |
\(\displaystyle -\frac {2 a^4 \int \frac {\tan ^6(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^3}{(\sec (c+d x) a+a)^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle -\frac {2 a^4 \int \left (\frac {a^2 \tan ^{10}(c+d x)}{(\sec (c+d x) a+a)^5}+\frac {5 a \tan ^8(c+d x)}{(\sec (c+d x) a+a)^4}+\frac {7 \tan ^6(c+d x)}{(\sec (c+d x) a+a)^3}+\frac {\tan ^4(c+d x)}{a (\sec (c+d x) a+a)^2}-\frac {\tan ^2(c+d x)}{a^2 (\sec (c+d x) a+a)}-\frac {1}{a^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}+\frac {1}{a^3}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^4 \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{7/2}}-\frac {\tan (c+d x)}{a^3 \sqrt {a \sec (c+d x)+a}}-\frac {a^2 \tan ^{11}(c+d x)}{11 (a \sec (c+d x)+a)^{11/2}}+\frac {\tan ^3(c+d x)}{3 a^2 (a \sec (c+d x)+a)^{3/2}}-\frac {5 a \tan ^9(c+d x)}{9 (a \sec (c+d x)+a)^{9/2}}-\frac {\tan ^7(c+d x)}{(a \sec (c+d x)+a)^{7/2}}-\frac {\tan ^5(c+d x)}{5 a (a \sec (c+d x)+a)^{5/2}}\right )}{d}\) |
Input:
Int[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^6,x]
Output:
(-2*a^4*(ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]/a^(7/2) - Tan[c + d*x]/(a^3*Sqrt[a + a*Sec[c + d*x]]) + Tan[c + d*x]^3/(3*a^2*(a + a*Sec[c + d*x])^(3/2)) - Tan[c + d*x]^5/(5*a*(a + a*Sec[c + d*x])^(5/2)) - Tan[c + d*x]^7/(a + a*Sec[c + d*x])^(7/2) - (5*a*Tan[c + d*x]^9)/(9*(a + a*Sec[c + d*x])^(9/2)) - (a^2*Tan[c + d*x]^11)/(11*(a + a*Sec[c + d*x])^(1 1/2))))/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 = 0.96 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (494 \cos \left (d x +c \right )^{5}+247 \cos \left (d x +c \right )^{4}-186 \cos \left (d x +c \right )^{3}-155 \cos \left (d x +c \right )^{2}+50 \cos \left (d x +c \right )+45\right ) \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}+495 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{495 d \left (1+\cos \left (d x +c \right )\right )}\) | \(178\) |
Input:
int((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^6,x,method=_RETURNVERBOSE)
Output:
2/495/d*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*((494*cos(d*x+c)^5+247*cos (d*x+c)^4-186*cos(d*x+c)^3-155*cos(d*x+c)^2+50*cos(d*x+c)+45)*tan(d*x+c)*s ec(d*x+c)^4+495*(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)*cot(d*x+c)+csc (d*x+c)^2-1)^(1/2)))
Time = 0.11 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.67 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^6(c+d x) \, dx=\left [\frac {495 \, {\left (\cos \left (d x + c\right )^{6} + \cos \left (d x + c\right )^{5}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (494 \, \cos \left (d x + c\right )^{5} + 247 \, \cos \left (d x + c\right )^{4} - 186 \, \cos \left (d x + c\right )^{3} - 155 \, \cos \left (d x + c\right )^{2} + 50 \, \cos \left (d x + c\right ) + 45\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{495 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}, \frac {2 \, {\left (495 \, {\left (\cos \left (d x + c\right )^{6} + \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (494 \, \cos \left (d x + c\right )^{5} + 247 \, \cos \left (d x + c\right )^{4} - 186 \, \cos \left (d x + c\right )^{3} - 155 \, \cos \left (d x + c\right )^{2} + 50 \, \cos \left (d x + c\right ) + 45\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{495 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^6,x, algorithm="fricas")
Output:
[1/495*(495*(cos(d*x + c)^6 + cos(d*x + c)^5)*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*(494*cos(d*x + c)^ 5 + 247*cos(d*x + c)^4 - 186*cos(d*x + c)^3 - 155*cos(d*x + c)^2 + 50*cos( d*x + c) + 45)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*co s(d*x + c)^6 + d*cos(d*x + c)^5), 2/495*(495*(cos(d*x + c)^6 + cos(d*x + c )^5)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/( sqrt(a)*sin(d*x + c))) + (494*cos(d*x + c)^5 + 247*cos(d*x + c)^4 - 186*co s(d*x + c)^3 - 155*cos(d*x + c)^2 + 50*cos(d*x + c) + 45)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5 )]
\[ \int \sqrt {a+a \sec (c+d x)} \tan ^6(c+d x) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \tan ^{6}{\left (c + d x \right )}\, dx \] Input:
integrate((a+a*sec(d*x+c))**(1/2)*tan(d*x+c)**6,x)
Output:
Integral(sqrt(a*(sec(c + d*x) + 1))*tan(c + d*x)**6, x)
Timed out. \[ \int \sqrt {a+a \sec (c+d x)} \tan ^6(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^6,x, algorithm="maxima")
Output:
Timed out
Time = 0.53 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.28 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^6(c+d x) \, dx=\frac {\sqrt {2} {\left (\frac {495 \, \sqrt {2} \sqrt {-a} a \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} - \frac {4 \, {\left (495 \, a^{6} - {\left (2805 \, a^{6} - {\left (6666 \, a^{6} - {\left (4158 \, a^{6} + {\left (221 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1463 \, a^{6}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{990 \, d} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^6,x, algorithm="giac")
Output:
1/990*sqrt(2)*(495*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)/ab s(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*(495*a^6 - (2805*a^6 - (6666*a^6 - (4158*a^6 + (221*a^6*tan(1/2*d*x + 1/2*c)^2 - 1463*a^6)*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)^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)^5*sqrt(-a*tan(1/ 2*d*x + 1/2*c)^2 + a)))*sgn(cos(d*x + c))/d
Timed out. \[ \int \sqrt {a+a \sec (c+d x)} \tan ^6(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^6\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \] Input:
int(tan(c + d*x)^6*(a + a/cos(c + d*x))^(1/2),x)
Output:
int(tan(c + d*x)^6*(a + a/cos(c + d*x))^(1/2), x)
\[ \int \sqrt {a+a \sec (c+d x)} \tan ^6(c+d x) \, dx=\sqrt {a}\, \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{6}d x \right ) \] Input:
int((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^6,x)
Output:
sqrt(a)*int(sqrt(sec(c + d*x) + 1)*tan(c + d*x)**6,x)