Integrand size = 14, antiderivative size = 152 \[ \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}-\frac {2 b}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {4 a b}{\left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \] Output:
-I*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(5/2)/d+I*arctanh ((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(5/2)/d-2/3*b/(a^2+b^2)/d/( a+b*tan(d*x+c))^(3/2)-4*a*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )+(-i a-b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}} \] Input:
Integrate[(a + b*Tan[c + d*x])^(-5/2),x]
Output:
(I*(a + I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - I* b)] + ((-I)*a - b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/( a + I*b)])/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2))
Time = 0.79 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {3042, 3964, 3042, 4012, 3042, 4022, 3042, 4020, 25, 73, 221}
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.
\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3964 |
\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}}dx}{a^2+b^2}-\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}}dx}{a^2+b^2}-\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\frac {\int \frac {a^2-2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{a^2+b^2}-\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {a^2-2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{a^2+b^2}-\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}}{a^2+b^2}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a+i b)^2 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a-i b)^2 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a^2+b^2}}{a^2+b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a-i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (a+i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}}{a^2+b^2}}{a^2+b^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a-i b)^2 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^2 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}}{a^2+b^2}}{a^2+b^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a-i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {(a+i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}}{a^2+b^2}}{a^2+b^2}\) |
Input:
Int[(a + b*Tan[c + d*x])^(-5/2),x]
Output:
(-2*b)/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ((((a + I*b)^2*ArcTa n[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + ((a - I*b)^2*ArcTan[Tan [c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d))/(a^2 + b^2) - (4*a*b)/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))/(a^2 + b^2)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(2317\) vs. \(2(128)=256\).
Time = 0.23 (sec) , antiderivative size = 2318, normalized size of antiderivative = 15.25
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2318\) |
default | \(\text {Expression too large to display}\) | \(2318\) |
Input:
int(1/(a+b*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)-1/4/d/b/(a^2+b^2)^3*ln((a+b*tan( d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2) )*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^4-1/2/d*b/(a^2+b^2)^(7/2)*ln((a+b*tan(d* x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))* (2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3-4/d*b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2) -2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2) )/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^4+2/d*b/(a^2+b^2)^3/(2*(a^2+b^2)^(1/2)- 2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2)) /(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+1/d/b/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/ 2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/ 2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^4-1/d/b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^ (1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^ (1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^6+1/4/d/b/(a^2+b^2)^(7/2)*ln((a+b* tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^( 1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^5-3/4/d*b^3/(a^2+b^2)^(7/2)*ln((a+b* tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^( 1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+4/d*b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^( 1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^( 1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^4+1/d*b^3/(a^2+b^2)^(7/2)/(2*(a^2+b ^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2...
Leaf count of result is larger than twice the leaf count of optimal. 3296 vs. \(2 (122) = 244\).
Time = 0.14 (sec) , antiderivative size = 3296, normalized size of antiderivative = 21.68 \[ \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a+b*tan(d*x+c))**(5/2),x)
Output:
Integral((a + b*tan(c + d*x))**(-5/2), x)
Exception generated. \[ \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Exception generated. \[ \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{1,[6,14,5]%%%}+%%%{6,[6,12,5]%%%}+%%%{15,[6,10,5]%%%}+ %%%{20,[6
Time = 6.43 (sec) , antiderivative size = 3703, normalized size of antiderivative = 24.36 \[ \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:
int(1/(a + b*tan(c + d*x))^(5/2),x)
Output:
(log(16*a*b^15*d^2 - ((-1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2* 5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(((-1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(8 96*a^6*b^15*d^4 - 160*a^2*b^19*d^4 - 128*a^4*b^17*d^4 - 32*b^21*d^4 + 3136 *a^8*b^13*d^4 + 4928*a^10*b^11*d^4 + 4480*a^12*b^9*d^4 + 2432*a^14*b^7*d^4 + 736*a^16*b^5*d^4 + 96*a^18*b^3*d^4 + ((-1/(a^5*d^2 - b^5*d^2*1i + 5*a*b ^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(a + b*ta n(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18*d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 13440*a^13* b^10*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 + 64*a ^21*b^2*d^5))/2))/2 - (a + b*tan(c + d*x))^(1/2)*(320*a^4*b^14*d^3 - 16*b^ 18*d^3 + 1024*a^6*b^12*d^3 + 1440*a^8*b^10*d^3 + 1024*a^10*b^8*d^3 + 320*a ^12*b^6*d^3 - 16*a^16*b^2*d^3)))/2 + 96*a^3*b^13*d^2 + 240*a^5*b^11*d^2 + 320*a^7*b^9*d^2 + 240*a^9*b^7*d^2 + 96*a^11*b^5*d^2 + 16*a^13*b^3*d^2)*(-1 /(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10 *a^3*b^2*d^2))^(1/2))/2 - log(16*a*b^15*d^2 - ((-1/(4*(a^5*d^2 - b^5*d^2*1 i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2)))^(1/2) *(896*a^6*b^15*d^4 - 160*a^2*b^19*d^4 - 128*a^4*b^17*d^4 - 32*b^21*d^4 + 3 136*a^8*b^13*d^4 + 4928*a^10*b^11*d^4 + 4480*a^12*b^9*d^4 + 2432*a^14*b^7* d^4 + 736*a^16*b^5*d^4 + 96*a^18*b^3*d^4 - (-1/(4*(a^5*d^2 - b^5*d^2*1i...
\[ \int \frac {1}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {-2 \sqrt {a +\tan \left (d x +c \right ) b}-3 \left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) \tan \left (d x +c \right )^{2} b^{3} d -6 \left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) \tan \left (d x +c \right ) a \,b^{2} d -3 \left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) a^{2} b d}{3 b d \left (\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}\right )} \] Input:
int(1/(a+b*tan(d*x+c))^(5/2),x)
Output:
( - 2*sqrt(tan(c + d*x)*b + a) - 3*int((sqrt(tan(c + d*x)*b + a)*tan(c + d *x)**2)/(tan(c + d*x)**3*b**3 + 3*tan(c + d*x)**2*a*b**2 + 3*tan(c + d*x)* a**2*b + a**3),x)*tan(c + d*x)**2*b**3*d - 6*int((sqrt(tan(c + d*x)*b + a) *tan(c + d*x)**2)/(tan(c + d*x)**3*b**3 + 3*tan(c + d*x)**2*a*b**2 + 3*tan (c + d*x)*a**2*b + a**3),x)*tan(c + d*x)*a*b**2*d - 3*int((sqrt(tan(c + d* x)*b + a)*tan(c + d*x)**2)/(tan(c + d*x)**3*b**3 + 3*tan(c + d*x)**2*a*b** 2 + 3*tan(c + d*x)*a**2*b + a**3),x)*a**2*b*d)/(3*b*d*(tan(c + d*x)**2*b** 2 + 2*tan(c + d*x)*a*b + a**2))