Integrand size = 25, antiderivative size = 150 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=-\frac {(i a+b) (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(i a-b) (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 (b c+a d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f} \]
-(I*a+b)*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+(I* a-b)*(c+I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(a*d+ b*c)*(c+d*tan(f*x+e))^(1/2)/f+2/3*b*(c+d*tan(f*x+e))^(3/2)/f
Time = 0.64 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.93 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\frac {-3 i (a-i b) (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+3 i (a+i b) (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+2 \sqrt {c+d \tan (e+f x)} (4 b c+3 a d+b d \tan (e+f x))}{3 f} \]
((-3*I)*(a - I*b)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + (3*I)*(a + I*b)*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]] /Sqrt[c + I*d]] + 2*Sqrt[c + d*Tan[e + f*x]]*(4*b*c + 3*a*d + b*d*Tan[e + f*x]))/(3*f)
Time = 0.75 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4011, 3042, 4011, 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.
\(\displaystyle \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \sqrt {c+d \tan (e+f x)} (a c-b d+(b c+a d) \tan (e+f x))dx+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {c+d \tan (e+f x)} (a c-b d+(b c+a d) \tan (e+f x))dx+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \frac {-2 b c d+a \left (c^2-d^2\right )+\left (2 a c d+b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {-2 b c d+a \left (c^2-d^2\right )+\left (2 a c d+b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {1}{2} (a-i b) (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (a-i b) (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {i (a-i b) (c-i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b) (c+i d)^2 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i (a-i b) (c-i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i (a+i b) (c+i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a-i b) (c-i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a+i b) (c+i d)^2 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a-i b) (c-i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b) (c+i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f}\) |
((a - I*b)*(c - I*d)^(3/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((a + I *b)*(c + I*d)^(3/2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f + (2*(b*c + a*d) *Sqrt[c + d*Tan[e + f*x]])/f + (2*b*(c + d*Tan[e + f*x])^(3/2))/(3*f)
3.13.37.3.1 Defintions of rubi rules used
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[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
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. \(1656\) vs. \(2(126)=252\).
Time = 0.84 (sec) , antiderivative size = 1657, normalized size of antiderivative = 11.05
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1657\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1665\) |
default | \(\text {Expression too large to display}\) | \(1665\) |
a*(2/f*d*(c+d*tan(f*x+e))^(1/2)-1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e)) ^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2 *c)^(1/2)*(c^2+d^2)^(1/2)*c+1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/ 2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^ (1/2)*c^2-1/4/f*d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1 /2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+2/f*d/(2*(c^ 2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2 )+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-1/f*d/(2*(c^2+d^2)^(1/2)-2* c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/( 2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)+1/4/f/d*ln((c+d*tan(f*x+e))^ (1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^ 2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c-1/4/f/d*ln((c+d*tan(f*x+e))^(1/2 )*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^ 2)^(1/2)+2*c)^(1/2)*c^2+1/4/f*d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/ 2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2 )-2/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2 )-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+1/f*d/(2*(c^2 +d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f* x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2))+b*(2/3/f*(c+d *tan(f*x+e))^(3/2)+2/f*c*(c+d*tan(f*x+e))^(1/2)+1/4/f*ln(d*tan(f*x+e)+c...
Leaf count of result is larger than twice the leaf count of optimal. 3058 vs. \(2 (121) = 242\).
Time = 0.48 (sec) , antiderivative size = 3058, normalized size of antiderivative = 20.39 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \]
1/6*(3*f*sqrt((6*a*b*c^2*d - 2*a*b*d^3 - (a^2 - b^2)*c^3 + 3*(a^2 - b^2)*c *d^2 + f^2*sqrt(-(4*a^2*b^2*c^6 + 12*(a^3*b - a*b^3)*c^5*d + 3*(3*a^4 - 14 *a^2*b^2 + 3*b^4)*c^4*d^2 - 40*(a^3*b - a*b^3)*c^3*d^3 - 6*(a^4 - 8*a^2*b^ 2 + b^4)*c^2*d^4 + 12*(a^3*b - a*b^3)*c*d^5 + (a^4 - 2*a^2*b^2 + b^4)*d^6) /f^4))/f^2)*log((2*(a^3*b + a*b^3)*c^5 + 3*(a^4 - b^4)*c^4*d - 4*(a^3*b + a*b^3)*c^3*d^2 + 2*(a^4 - b^4)*c^2*d^3 - 6*(a^3*b + a*b^3)*c*d^4 - (a^4 - b^4)*d^5)*sqrt(d*tan(f*x + e) + c) + ((a*c - b*d)*f^3*sqrt(-(4*a^2*b^2*c^6 + 12*(a^3*b - a*b^3)*c^5*d + 3*(3*a^4 - 14*a^2*b^2 + 3*b^4)*c^4*d^2 - 40* (a^3*b - a*b^3)*c^3*d^3 - 6*(a^4 - 8*a^2*b^2 + b^4)*c^2*d^4 + 12*(a^3*b - a*b^3)*c*d^5 + (a^4 - 2*a^2*b^2 + b^4)*d^6)/f^4) - (2*a*b^2*c^4 + (5*a^2*b - 3*b^3)*c^3*d + 3*(a^3 - 3*a*b^2)*c^2*d^2 - (7*a^2*b - b^3)*c*d^3 - (a^3 - a*b^2)*d^4)*f)*sqrt((6*a*b*c^2*d - 2*a*b*d^3 - (a^2 - b^2)*c^3 + 3*(a^2 - b^2)*c*d^2 + f^2*sqrt(-(4*a^2*b^2*c^6 + 12*(a^3*b - a*b^3)*c^5*d + 3*(3 *a^4 - 14*a^2*b^2 + 3*b^4)*c^4*d^2 - 40*(a^3*b - a*b^3)*c^3*d^3 - 6*(a^4 - 8*a^2*b^2 + b^4)*c^2*d^4 + 12*(a^3*b - a*b^3)*c*d^5 + (a^4 - 2*a^2*b^2 + b^4)*d^6)/f^4))/f^2)) - 3*f*sqrt((6*a*b*c^2*d - 2*a*b*d^3 - (a^2 - b^2)*c^ 3 + 3*(a^2 - b^2)*c*d^2 + f^2*sqrt(-(4*a^2*b^2*c^6 + 12*(a^3*b - a*b^3)*c^ 5*d + 3*(3*a^4 - 14*a^2*b^2 + 3*b^4)*c^4*d^2 - 40*(a^3*b - a*b^3)*c^3*d^3 - 6*(a^4 - 8*a^2*b^2 + b^4)*c^2*d^4 + 12*(a^3*b - a*b^3)*c*d^5 + (a^4 - 2* a^2*b^2 + b^4)*d^6)/f^4))/f^2)*log((2*(a^3*b + a*b^3)*c^5 + 3*(a^4 - b^...
\[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right ) \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \]
Exception generated. \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(d-c>0)', see `assume?` for more details)Is
Timed out. \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\text {Timed out} \]
Time = 20.88 (sec) , antiderivative size = 2823, normalized size of antiderivative = 18.82 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \]
log(((((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^ 2)/f^4)^(1/2)*((16*c*d^2*(((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3* f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(b*c^2 + b*d^2 - f*(((-b^4*d^2*f^4*(3*c^ 2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*b^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6 *c^2*d^2))/f^2))/2 - (8*b^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*((6*b^4*c^ 2*d^4*f^4 - b^4*d^6*f^4 - 9*b^4*c^4*d^2*f^4)^(1/2)/(4*f^4) + (b^2*c^3)/(4* f^2) - (3*b^2*c*d^2)/(4*f^2))^(1/2) - log(((-((-b^4*d^2*f^4*(3*c^2 - d^2)^ 2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)*((16*c*d^2*(-((-b^4*d ^2*f^4*(3*c^2 - d^2)^2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)* (b*c^2 + b*d^2 + f*(-((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*b^2*d^2* (c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (8*b^3*d^2*( c^2 - d^2)*(c^2 + d^2)^2)/f^3)*(-((6*b^4*c^2*d^4*f^4 - b^4*d^6*f^4 - 9*b^4 *c^4*d^2*f^4)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/(4*f^4))^(1/2) - log( ((((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f ^4)^(1/2)*((16*c*d^2*(((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(b*c^2 + b*d^2 + f*(((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f *x))^(1/2)))/f - (16*b^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*...