Integrand size = 23, antiderivative size = 159 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\frac {\sqrt {a-i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d}-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d} \] Output:
(a-I*b)^(1/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+(a+I*b)^(1/2 )*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d-2*(a+b*tan(d*x+c))^(1/2) /d-4/15*a*(a+b*tan(d*x+c))^(3/2)/b^2/d+2/5*tan(d*x+c)*(a+b*tan(d*x+c))^(3/ 2)/b/d
Time = 0.63 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.88 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\frac {15 \sqrt {a-i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+15 \sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-2 a^2-15 b^2+a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)\right )}{b^2}}{15 d} \] Input:
Integrate[Tan[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]],x]
Output:
(15*Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + 15*Sqr t[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + (2*Sqrt[a + b *Tan[c + d*x]]*(-2*a^2 - 15*b^2 + a*b*Tan[c + d*x] + 3*b^2*Tan[c + d*x]^2) )/b^2)/(15*d)
Time = 0.95 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4049, 27, 3042, 4113, 27, 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.
| \(\displaystyle \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^3 \sqrt {a+b \tan (c+d x)}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {2 \int -\frac {1}{2} \sqrt {a+b \tan (c+d x)} \left (2 a \tan ^2(c+d x)+5 b \tan (c+d x)+2 a\right )dx}{5 b}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\int \sqrt {a+b \tan (c+d x)} \left (2 a \tan ^2(c+d x)+5 b \tan (c+d x)+2 a\right )dx}{5 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\int \sqrt {a+b \tan (c+d x)} \left (2 a \tan (c+d x)^2+5 b \tan (c+d x)+2 a\right )dx}{5 b}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\int 5 b \tan (c+d x) \sqrt {a+b \tan (c+d x)}dx+\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {5 b \int \tan (c+d x) \sqrt {a+b \tan (c+d x)}dx+\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {5 b \int \tan (c+d x) \sqrt {a+b \tan (c+d x)}dx+\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {5 b \left (\int \frac {a \tan (c+d x)-b}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {5 b \left (\int \frac {a \tan (c+d x)-b}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}+5 b \left (\frac {1}{2} (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\right )}{5 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}+5 b \left (\frac {1}{2} (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\right )}{5 b}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}+5 b \left (-\frac {i (b+i a) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (-b+i a) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\right )}{5 b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}+5 b \left (\frac {i (b+i a) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (-b+i a) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\right )}{5 b}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}+5 b \left (\frac {(-b+i a) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {(b+i a) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\right )}{5 b}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{3/2}}{3 b d}+5 b \left (-\frac {(b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {(-b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\right )}{5 b}\) |
Input:
Int[Tan[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]],x]
Output:
(2*Tan[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/(5*b*d) - ((4*a*(a + b*Tan[c + d*x])^(3/2))/(3*b*d) + 5*b*(-(((I*a + b)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b ]])/(Sqrt[a - I*b]*d)) + ((I*a - b)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(S qrt[a + I*b]*d) + (2*Sqrt[a + b*Tan[c + d*x]])/d))/(5*b)
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_))^(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]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) )
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(369\) vs. \(2(133)=266\).
Time = 1.70 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.33
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 b^{2} \sqrt {a +b \tan \left (d x +c \right )}+2 b^{2} \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{b^{2} d}\) | \(370\) |
| default | \(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 b^{2} \sqrt {a +b \tan \left (d x +c \right )}+2 b^{2} \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{b^{2} d}\) | \(370\) |
Input:
int(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/d/b^2*(1/5*(a+b*tan(d*x+c))^(5/2)-1/3*a*(a+b*tan(d*x+c))^(3/2)-b^2*(a+b* tan(d*x+c))^(1/2)+b^2*(1/8*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c)+a +(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+1/2 *((a^2+b^2)^(1/2)-a)/(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))-1/ 8*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*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))+1/2*(a-(a^2+b^2)^(1/2))/(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))))
Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (129) = 258\).
Time = 0.10 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.10 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\frac {15 \, b^{2} d \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} \log \left (d \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) - 15 \, b^{2} d \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} \log \left (-d \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) + 15 \, b^{2} d \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} \log \left (d \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) - 15 \, b^{2} d \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} \log \left (-d \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) + 4 \, {\left (3 \, b^{2} \tan \left (d x + c\right )^{2} + a b \tan \left (d x + c\right ) - 2 \, a^{2} - 15 \, b^{2}\right )} \sqrt {b \tan \left (d x + c\right ) + a}}{30 \, b^{2} d} \] Input:
integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
Output:
1/30*(15*b^2*d*sqrt((d^2*sqrt(-b^2/d^4) + a)/d^2)*log(d*sqrt((d^2*sqrt(-b^ 2/d^4) + a)/d^2) + sqrt(b*tan(d*x + c) + a)) - 15*b^2*d*sqrt((d^2*sqrt(-b^ 2/d^4) + a)/d^2)*log(-d*sqrt((d^2*sqrt(-b^2/d^4) + a)/d^2) + sqrt(b*tan(d* x + c) + a)) + 15*b^2*d*sqrt(-(d^2*sqrt(-b^2/d^4) - a)/d^2)*log(d*sqrt(-(d ^2*sqrt(-b^2/d^4) - a)/d^2) + sqrt(b*tan(d*x + c) + a)) - 15*b^2*d*sqrt(-( d^2*sqrt(-b^2/d^4) - a)/d^2)*log(-d*sqrt(-(d^2*sqrt(-b^2/d^4) - a)/d^2) + sqrt(b*tan(d*x + c) + a)) + 4*(3*b^2*tan(d*x + c)^2 + a*b*tan(d*x + c) - 2 *a^2 - 15*b^2)*sqrt(b*tan(d*x + c) + a))/(b^2*d)
\[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(tan(d*x+c)**3*(a+b*tan(d*x+c))**(1/2),x)
Output:
Integral(sqrt(a + b*tan(c + d*x))*tan(c + d*x)**3, x)
\[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int { \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{3} \,d x } \] Input:
integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^3, x)
Exception generated. \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)^3*(a+b*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:Unable to divide, perhaps due to ro unding error%%%{%%%{1,[0,19,7]%%%}+%%%{8,[0,17,7]%%%}+%%%{28,[0,15,7]%%%}+ %%%{56,[0
Time = 5.83 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.23 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\left (\frac {2\,a^2}{b^2\,d}-\frac {2\,\left (a^2+b^2\right )}{b^2\,d}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+\frac {2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^2\,d}-\frac {2\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b^2\,d}+\mathrm {atan}\left (\frac {b^4\,\sqrt {\frac {a}{4\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}+\frac {32\,a\,b^3\,\sqrt {\frac {a}{4\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}\right )\,\sqrt {\frac {a-b\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {b^4\,\sqrt {\frac {a}{4\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}-\frac {32\,a\,b^3\,\sqrt {\frac {a}{4\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}\right )\,\sqrt {\frac {a+b\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i} \] Input:
int(tan(c + d*x)^3*(a + b*tan(c + d*x))^(1/2),x)
Output:
((2*a^2)/(b^2*d) - (2*(a^2 + b^2))/(b^2*d))*(a + b*tan(c + d*x))^(1/2) + a tan((b^4*(a/(4*d^2) - (b*1i)/(4*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*32i )/((b^5*16i)/d + (a^2*b^3*16i)/d) + (32*a*b^3*(a/(4*d^2) - (b*1i)/(4*d^2)) ^(1/2)*(a + b*tan(c + d*x))^(1/2))/((b^5*16i)/d + (a^2*b^3*16i)/d))*((a - b*1i)/(4*d^2))^(1/2)*2i - atan((b^4*(a/(4*d^2) + (b*1i)/(4*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*32i)/((b^5*16i)/d + (a^2*b^3*16i)/d) - (32*a*b^3*( a/(4*d^2) + (b*1i)/(4*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((b^5*16i)/d + (a^2*b^3*16i)/d))*((a + b*1i)/(4*d^2))^(1/2)*2i + (2*(a + b*tan(c + d*x ))^(5/2))/(5*b^2*d) - (2*a*(a + b*tan(c + d*x))^(3/2))/(3*b^2*d)
\[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{3}d x \] Input:
int(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2),x)
Output:
int(sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**3,x)