Integrand size = 32, antiderivative size = 188 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 (5 c+i d) d \sqrt {a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \] Output:
-I*2^(1/2)*a^(1/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^ (1/2)/(a+I*a*tan(f*x+e))^(1/2))/(c-I*d)^(5/2)/f-2/3*d*(a+I*a*tan(f*x+e))^( 1/2)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)-2/3*(5*c+I*d)*d*(a+I*a*tan(f*x+e)) ^(1/2)/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)
Time = 1.65 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {\frac {3 i \sqrt {2} \sqrt {-a (c-i d)} (c+i d) \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^2}-\frac {2 d \sqrt {a+i a \tan (e+f x)} \left (6 c^2+i c d+d^2+(5 c+i d) d \tan (e+f x)\right )}{\left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{3 \left (c^2+d^2\right ) f} \] Input:
Integrate[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(5/2),x]
Output:
(((3*I)*Sqrt[2]*Sqrt[-(a*(c - I*d))]*(c + I*d)*ArcTan[(Sqrt[-(a*(c - I*d)) ]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/(c - I*d)^2 - (2*d*Sqrt[a + I*a*Tan[e + f*x]]*(6*c^2 + I*c*d + d^2 + (5*c + I*d )*d*Tan[e + f*x]))/((c^2 + d^2)*(c + d*Tan[e + f*x])^(3/2)))/(3*(c^2 + d^2 )*f)
Time = 0.85 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3042, 4044, 27, 3042, 4081, 27, 3042, 4027, 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.
\(\displaystyle \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4044 |
\(\displaystyle \frac {2 \int \frac {\sqrt {i \tan (e+f x) a+a} (a (3 c+i d)-2 a d \tan (e+f x))}{2 (c+d \tan (e+f x))^{3/2}}dx}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {i \tan (e+f x) a+a} (a (3 c+i d)-2 a d \tan (e+f x))}{(c+d \tan (e+f x))^{3/2}}dx}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {i \tan (e+f x) a+a} (a (3 c+i d)-2 a d \tan (e+f x))}{(c+d \tan (e+f x))^{3/2}}dx}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {\frac {2 \int \frac {3 a^2 (c+i d)^2 \sqrt {i \tan (e+f x) a+a}}{2 \sqrt {c+d \tan (e+f x)}}dx}{a \left (c^2+d^2\right )}-\frac {2 a d (5 c+i d) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3 a (c+i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 a d (5 c+i d) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 a (c+i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 a d (5 c+i d) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {-\frac {6 i a^3 (c+i d)^2 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f \left (c^2+d^2\right )}-\frac {2 a d (5 c+i d) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {3 i \sqrt {2} a^{3/2} (c+i d)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d} \left (c^2+d^2\right )}-\frac {2 a d (5 c+i d) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
Input:
Int[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(5/2),x]
Output:
(-2*d*Sqrt[a + I*a*Tan[e + f*x]])/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3 /2)) + (((-3*I)*Sqrt[2]*a^(3/2)*(c + I*d)^2*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[ c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*(c^2 + d^2)*f) - (2*a*(5*c + I*d)*d*Sqrt[a + I*a*Tan[e + f*x]])/((c ^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/(3*a*(c^2 + d^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(c^2 + d^2)*(n + 1)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d , e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2447 vs. \(2 (153 ) = 306\).
Time = 0.65 (sec) , antiderivative size = 2448, normalized size of antiderivative = 13.02
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2448\) |
default | \(\text {Expression too large to display}\) | \(2448\) |
Input:
int((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOS E)
Output:
1/6/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(-4*2^(1/2)*d^4*(-a*(I*d-c))^(1/2 )*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)-3*ln((3*a*c+I*a*t an(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*ta n(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c^3*d^2*tan(f*x+e)^3+ 9*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c)) ^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c*d^ 4*tan(f*x+e)^3-6*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/ 2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan( f*x+e)))*a*c^4*d*tan(f*x+e)^2+9*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan (f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))) ^(1/2))/(I+tan(f*x+e)))*a*c^2*d^3*tan(f*x+e)^2-9*ln((3*a*c+I*a*tan(f*x+e)* c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))* (1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c^3*d^2*tan(f*x+e)+6*ln((3*a*c+ I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c +d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c*d^4*tan(f*x+e) +12*2^(1/2)*c^3*d*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))) ^(1/2)+3*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a *(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e) ))*a*d^5*tan(f*x+e)^3+2*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1 /2)*(-a*(I*d-c))^(1/2)*d^4-9*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*t...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 968 vs. \(2 (146) = 292\).
Time = 0.12 (sec) , antiderivative size = 968, normalized size of antiderivative = 5.15 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fr icas")
Output:
-1/6*(8*sqrt(2)*((3*c^2*d - 2*I*c*d^2 + d^3)*e^(5*I*f*x + 5*I*e) + (6*c^2* d + I*c*d^2 + d^3)*e^(3*I*f*x + 3*I*e) + 3*(c^2*d + I*c*d^2)*e^(I*f*x + I* e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)) - 3*((c^6 - 2*I*c^5*d + c^4*d^2 - 4* I*c^3*d^3 - c^2*d^4 - 2*I*c*d^5 - d^6)*f*e^(4*I*f*x + 4*I*e) + 2*(c^6 + 3* c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2*I*e) + (c^6 + 2*I*c^5*d + c^4* d^2 + 4*I*c^3*d^3 - c^2*d^4 + 2*I*c*d^5 - d^6)*f)*sqrt(-2*I*a/((I*c^5 + 5* c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(((I*c^3 + 3 *c^2*d - 3*I*c*d^2 - d^3)*f*sqrt(-2*I*a/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*e^(I*f*x + I*e) + sqrt(2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^( 2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1))*e^(-I*f*x - I*e)) + 3*(( c^6 - 2*I*c^5*d + c^4*d^2 - 4*I*c^3*d^3 - c^2*d^4 - 2*I*c*d^5 - d^6)*f*e^( 4*I*f*x + 4*I*e) + 2*(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2* I*e) + (c^6 + 2*I*c^5*d + c^4*d^2 + 4*I*c^3*d^3 - c^2*d^4 + 2*I*c*d^5 - d^ 6)*f)*sqrt(-2*I*a/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^ 4 + d^5)*f^2))*log(((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*sqrt(-2*I*a/((I *c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*e^(I*f *x + I*e) + sqrt(2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I *f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I...
\[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((a+I*a*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(5/2),x)
Output:
Integral(sqrt(I*a*(tan(e + f*x) - I))/(c + d*tan(e + f*x))**(5/2), x)
Timed out. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="ma xima")
Output:
Timed out
Timed out. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="gi ac")
Output:
Timed out
Timed out. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int((a + a*tan(e + f*x)*1i)^(1/2)/(c + d*tan(e + f*x))^(5/2),x)
Output:
int((a + a*tan(e + f*x)*1i)^(1/2)/(c + d*tan(e + f*x))^(5/2), x)
\[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \] Input:
int((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x)
Output:
sqrt(a)*int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c))/(tan(e + f *x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)