Integrand size = 14, antiderivative size = 120 \[ \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}-\frac {2 b}{\left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \] Output:
-I*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d+I*arctanh ((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(3/2)/d-2*b/(a^2+b^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 = 105, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )+(-i a-b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )}{\left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \] Input:
Integrate[(a + b*Tan[c + d*x])^(-3/2),x]
Output:
(I*(a + I*b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b )] + ((-I)*a - b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)])/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])
Time = 0.56 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 3964, 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.
\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3964 |
\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a+i b) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a-i b) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a^2+b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a-i b) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (a+i b) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}}{a^2+b^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a-i b) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}}{a^2+b^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a+i b) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {(a-i b) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{a^2+b^2}\) |
Input:
Int[(a + b*Tan[c + d*x])^(-3/2),x]
Output:
(((a + I*b)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + ((a - I*b)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d))/(a^2 + b^2) - (2*b)/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])
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[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. \(1931\) vs. \(2(100)=200\).
Time = 0.24 (sec) , antiderivative size = 1932, normalized size of antiderivative = 16.10
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1932\) |
default | \(\text {Expression too large to display}\) | \(1932\) |
Input:
int(1/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/4/d/b/(a^2+b^2)^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^3-1/4/d *b/(a^2+b^2)^2*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*t an(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+1/4/d/b/(a^2+ b^2)^(5/2)*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^4-1/4/d*b^3/(a^2+ b^2)^(5/2)*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)-4/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^3-3/d*b^3/(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+1/d/b/(a^2+b^ 2)^(3/2)/(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^3+1/d*b/(a^2 +b^2)^(3/2)/(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+1/d*b/(a^ 2+b^2)^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^2-1/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^5+1/d*...
Leaf count of result is larger than twice the leaf count of optimal. 1949 vs. \(2 (94) = 188\).
Time = 0.11 (sec) , antiderivative size = 1949, normalized size of antiderivative = 16.24 \[ \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
-1/2*(((a^2*b + b^3)*d*tan(d*x + c) + (a^3 + a*b^2)*d)*sqrt(-((a^6 + 3*a^4 *b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6 *a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4) ) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log(-(3*a^2* b - b^3)*sqrt(b*tan(d*x + c) + a) + ((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d ^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + 2*(3*a^3*b^2 - a*b^4) *d)*sqrt(-((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^ 2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))) - ((a^2*b + b^3)*d*tan(d*x + c) + (a^3 + a*b^2)*d)*sqrt(-((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/(( a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^ 12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log( -(3*a^2*b - b^3)*sqrt(b*tan(d*x + c) + a) - ((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^ 8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + 2*(3*a^3*b^2 - a*b^4)*d)*sqrt(-((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^ 2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^ 4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3...
\[ \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(a+b*tan(d*x+c))**(3/2),x)
Output:
Integral((a + b*tan(c + d*x))**(-3/2), x)
Exception generated. \[ \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+b*tan(d*x+c))^(3/2),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. \[ \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+b*tan(d*x+c))^(3/2),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,[3,9,3]%%%}+%%%{4,[3,7,3]%%%}+%%%{6,[3,5,3]%%%}+%%%{ 4,[3,3,3]
Time = 3.31 (sec) , antiderivative size = 2236, normalized size of antiderivative = 18.63 \[ \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:
int(1/(a + b*tan(c + d*x))^(3/2),x)
Output:
(log(8*b^9*d^2 - ((((-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i ))^(1/2)*(64*a*b^11*d^4 + ((-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b *d^2*3i))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d ^5 + 640*a^5*b^8*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5 ))/2 + 256*a^3*b^9*d^4 + 384*a^5*b^7*d^4 + 256*a^7*b^5*d^4 + 64*a^9*b^3*d^ 4))/2 - (a + b*tan(c + d*x))^(1/2)*(16*b^10*d^3 + 32*a^2*b^8*d^3 - 32*a^6* b^4*d^3 - 16*a^8*b^2*d^3))*(-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b *d^2*3i))^(1/2))/2 + 24*a^2*b^7*d^2 + 24*a^4*b^5*d^2 + 8*a^6*b^3*d^2)*(-1/ (a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2))/2 - log(8*b^9* d^2 - ((-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)))^(1/2)* (64*a*b^11*d^4 - (-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i )))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 6 40*a^5*b^8*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5) + 25 6*a^3*b^9*d^4 + 384*a^5*b^7*d^4 + 256*a^7*b^5*d^4 + 64*a^9*b^3*d^4) + (a + b*tan(c + d*x))^(1/2)*(16*b^10*d^3 + 32*a^2*b^8*d^3 - 32*a^6*b^4*d^3 - 16 *a^8*b^2*d^3))*(-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)) )^(1/2) + 24*a^2*b^7*d^2 + 24*a^4*b^5*d^2 + 8*a^6*b^3*d^2)*(-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)))^(1/2) + atan((((-1i/(4*(a^3* d^2*1i + b^3*d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)*(64*a*b^11*d^4 + (- 1i/(4*(a^3*d^2*1i + b^3*d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)*(a + ...
\[ \int \frac {1}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {-2 \sqrt {a +\tan \left (d x +c \right ) b}-\left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}}d x \right ) \tan \left (d x +c \right ) b^{2} d -\left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}}d x \right ) a b d}{b d \left (a +\tan \left (d x +c \right ) b \right )} \] Input:
int(1/(a+b*tan(d*x+c))^(3/2),x)
Output:
( - 2*sqrt(tan(c + d*x)*b + a) - int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x )**2)/(tan(c + d*x)**2*b**2 + 2*tan(c + d*x)*a*b + a**2),x)*tan(c + d*x)*b **2*d - int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2)/(tan(c + d*x)**2*b* *2 + 2*tan(c + d*x)*a*b + a**2),x)*a*b*d)/(b*d*(tan(c + d*x)*b + a))