Integrand size = 26, antiderivative size = 126 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 \sqrt {a+i a \tan (c+d x)}}{a d}-\frac {5 (a+i a \tan (c+d x))^{3/2}}{3 a^2 d} \] Output:
1/2*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(1/2)/ d-tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/2)+4*(a+I*a*tan(d*x+c))^(1/2)/a/d-5 /3*(a+I*a*tan(d*x+c))^(3/2)/a^2/d
Time = 0.46 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {7+2 i \tan (c+d x)+2 \tan ^2(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Tan[c + d*x]^3/Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(Sqrt[2]*Sqrt[a]*d) + (7 + (2*I)*Tan[c + d*x] + 2*Tan[c + d*x]^2)/(3*d*Sqrt[a + I*a*Tan[c + d* x]])
Time = 0.63 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4041, 27, 3042, 4075, 3042, 4010, 3042, 3961, 219}
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.
\(\displaystyle \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^3}{\sqrt {a+i a \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4041 |
\(\displaystyle -\frac {\int -\frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} (4 a-5 i a \tan (c+d x))dx}{a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} (4 a-5 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} (4 a-5 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 4075 |
\(\displaystyle \frac {\int \sqrt {i \tan (c+d x) a+a} (4 \tan (c+d x) a+5 i a)dx-\frac {10 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {i \tan (c+d x) a+a} (4 \tan (c+d x) a+5 i a)dx-\frac {10 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 4010 |
\(\displaystyle \frac {i a \int \sqrt {i \tan (c+d x) a+a}dx-\frac {10 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {8 a \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i a \int \sqrt {i \tan (c+d x) a+a}dx-\frac {10 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {8 a \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {\frac {2 a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {8 a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {10 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {10 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
Input:
Int[Tan[c + d*x]^3/Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
-(Tan[c + d*x]^2/(d*Sqrt[a + I*a*Tan[c + d*x]])) + ((Sqrt[2]*a^(3/2)*ArcTa nh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (8*a*Sqrt[a + I*a*Ta n[c + d*x]])/d - (10*(a + I*a*Tan[c + d*x])^(3/2))/(3*d))/(2*a^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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Simp [(b*c + a*d)/b Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e , f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && !LtQ[m, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[m] || 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_)]), x_Symbol] :> Simp[B *d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1]
Time = 1.44 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}+\frac {a^{2}}{\sqrt {a +i a \tan \left (d x +c \right )}}}{a^{2} d}\) | \(92\) |
default | \(\frac {-\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}+\frac {a^{2}}{\sqrt {a +i a \tan \left (d x +c \right )}}}{a^{2} d}\) | \(92\) |
Input:
int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/d/a^2*(-1/3*(a+I*a*tan(d*x+c))^(3/2)+a*(a+I*a*tan(d*x+c))^(1/2)+1/4*a^(3 /2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+1/2*a^2/ (a+I*a*tan(d*x+c))^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (102) = 204\).
Time = 0.11 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.29 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (-4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (7 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )}}{12 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \] Input:
integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
Output:
1/12*(3*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(1/(a* d^2))*log(4*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3*sqrt(2)*(a *d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(-4*((a*d *e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^ 2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(7*e^(4*I*d*x + 4*I*c) + 18*e^(2*I*d*x + 2*I*c) + 3))/(a*d* e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))
\[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:
integrate(tan(d*x+c)**3/(a+I*a*tan(d*x+c))**(1/2),x)
Output:
Integral(tan(c + d*x)**3/sqrt(I*a*(tan(c + d*x) - I)), x)
Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 24 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3} - \frac {12 \, a^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{12 \, a^{4} d} \] Input:
integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
Output:
-1/12*(3*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a ))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 8*(I*a*tan(d*x + c) + a)^(3/2)*a^2 - 24*sqrt(I*a*tan(d*x + c) + a)*a^3 - 12*a^4/sqrt(I*a*tan(d* x + c) + a))/(a^4*d)
Exception generated. \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone
Time = 1.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {1}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \] Input:
int(tan(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(1/2),x)
Output:
1/(d*(a + a*tan(c + d*x)*1i)^(1/2)) + (2*(a + a*tan(c + d*x)*1i)^(1/2))/(a *d) - (2*(a + a*tan(c + d*x)*1i)^(3/2))/(3*a^2*d) - (2^(1/2)*atan((2^(1/2) *(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*1i)/(2*a^(1/2)*d)
\[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{2}+1}d x \right ) i +\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{2}+1}d x \right )}{a} \] Input:
int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x)
Output:
(sqrt(a)*( - int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**4)/(tan(c + d*x)* *2 + 1),x)*i + int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3)/(tan(c + d*x )**2 + 1),x)))/a