Integrand size = 27, antiderivative size = 248 \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i (a-i b)^4 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {i (a+i b)^4 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}-\frac {2 b^2 \left (40 a b c d-87 a^2 d^2-b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^3 f}-\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f} \] Output:
-I*(a-I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(1/2)/f +I*(a+I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(1/2)/f -2/15*b^2*(40*a*b*c*d-87*a^2*d^2-b^2*(8*c^2-15*d^2))*(c+d*tan(f*x+e))^(1/2 )/d^3/f-4/15*b^3*(-7*a*d+2*b*c)*tan(f*x+e)*(c+d*tan(f*x+e))^(1/2)/d^2/f+2/ 5*b^2*(a+b*tan(f*x+e))^2*(c+d*tan(f*x+e))^(1/2)/d/f
Time = 2.30 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {-\frac {15 i (a-i b)^4 d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {15 i (a+i b)^4 d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}+\frac {2 b^2 \left (-40 a b c d+87 a^2 d^2+b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d^2}+\frac {4 b^3 (-2 b c+7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{d}+6 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{15 d f} \] Input:
Integrate[(a + b*Tan[e + f*x])^4/Sqrt[c + d*Tan[e + f*x]],x]
Output:
(((-15*I)*(a - I*b)^4*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/S qrt[c - I*d] + ((15*I)*(a + I*b)^4*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt [c + I*d]])/Sqrt[c + I*d] + (2*b^2*(-40*a*b*c*d + 87*a^2*d^2 + b^2*(8*c^2 - 15*d^2))*Sqrt[c + d*Tan[e + f*x]])/d^2 + (4*b^3*(-2*b*c + 7*a*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/d + 6*b^2*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(15*d*f)
Time = 1.57 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4049, 27, 3042, 4120, 27, 3042, 4113, 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.
\(\displaystyle \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 4049 |
\(\displaystyle \frac {2 \int -\frac {(a+b \tan (e+f x)) \left (-5 d a^3+b^2 d a+2 b^2 (2 b c-7 a d) \tan ^2(e+f x)+4 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{5 d}+\frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\int \frac {(a+b \tan (e+f x)) \left (-5 d a^3+b^2 d a+2 b^2 (2 b c-7 a d) \tan ^2(e+f x)+4 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\int \frac {(a+b \tan (e+f x)) \left (-5 d a^3+b^2 d a+2 b^2 (2 b c-7 a d) \tan (e+f x)^2+4 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{5 d}\) |
\(\Big \downarrow \) 4120 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {2 \int \frac {15 d^2 a^4-3 b^2 d^2 a^2-40 b^3 c d a+60 b \left (a^2-b^2\right ) d^2 \tan (e+f x) a+8 b^4 c^2-b^2 \left (-\left (\left (8 c^2-15 d^2\right ) b^2\right )+40 a c d b-87 a^2 d^2\right ) \tan ^2(e+f x)}{2 \sqrt {c+d \tan (e+f x)}}dx}{3 d}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {15 d^2 a^4-3 b^2 d^2 a^2-40 b^3 c d a+60 b \left (a^2-b^2\right ) d^2 \tan (e+f x) a+8 b^4 c^2-b^2 \left (-\left (\left (8 c^2-15 d^2\right ) b^2\right )+40 a c d b-87 a^2 d^2\right ) \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {15 d^2 a^4-3 b^2 d^2 a^2-40 b^3 c d a+60 b \left (a^2-b^2\right ) d^2 \tan (e+f x) a+8 b^4 c^2-b^2 \left (-\left (\left (8 c^2-15 d^2\right ) b^2\right )+40 a c d b-87 a^2 d^2\right ) \tan (e+f x)^2}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}}{5 d}\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {15 \left (a^4-6 b^2 a^2+b^4\right ) d^2+60 a b \left (a^2-b^2\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {15 \left (a^4-6 b^2 a^2+b^4\right ) d^2+60 a b \left (a^2-b^2\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {15}{2} d^2 (a-i b)^4 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {15}{2} d^2 (a+i b)^4 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {15}{2} d^2 (a-i b)^4 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {15}{2} d^2 (a+i b)^4 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {15 i d^2 (a-i b)^4 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {15 i d^2 (a+i b)^4 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {15 i d^2 (a-i b)^4 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {15 i d^2 (a+i b)^4 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {15 d (a-i b)^4 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {15 d (a+i b)^4 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}+\frac {15 d^2 (a-i b)^4 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {15 d^2 (a+i b)^4 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{3 d}}{5 d}\) |
Input:
Int[(a + b*Tan[e + f*x])^4/Sqrt[c + d*Tan[e + f*x]],x]
Output:
(2*b^2*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f) - ((4*b^3* (2*b*c - 7*a*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) - ((15*(a - I*b)^4*d^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (15*(a + I*b)^4*d^2*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) - (2*b ^2*(40*a*b*c*d - 87*a^2*d^2 - b^2*(8*c^2 - 15*d^2))*Sqrt[c + d*Tan[e + f*x ]])/(d*f))/(3*d))/(5*d)
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[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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2)) Int[(c + d*Tan[e + f*x])^n*Si mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(5492\) vs. \(2(216)=432\).
Time = 0.58 (sec) , antiderivative size = 5493, normalized size of antiderivative = 22.15
method | result | size |
parts | \(\text {Expression too large to display}\) | \(5493\) |
derivativedivides | \(\text {Expression too large to display}\) | \(10033\) |
default | \(\text {Expression too large to display}\) | \(10033\) |
Input:
int((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 4951 vs. \(2 (208) = 416\).
Time = 0.61 (sec) , antiderivative size = 4951, normalized size of antiderivative = 19.96 \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:
integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{4}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \] Input:
integrate((a+b*tan(f*x+e))**4/(c+d*tan(f*x+e))**(1/2),x)
Output:
Integral((a + b*tan(e + f*x))**4/sqrt(c + d*tan(e + f*x)), x)
Timed out. \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \] Input:
integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
Timed out
Exception generated. \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(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 = 14.31 (sec) , antiderivative size = 3771, normalized size of antiderivative = 15.21 \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:
int((a + b*tan(e + f*x))^4/(c + d*tan(e + f*x))^(1/2),x)
Output:
atan(((((32*(a^4*d^3*f^2 + b^4*d^3*f^2 - 6*a^2*b^2*d^3*f^2 + 4*a*b^3*c*d^2 *f^2 - 4*a^3*b*c*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a* b^7*8i - a^7*b*8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^ 5*b^3*56i - 28*a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2))*(-(a*b^7*8i - a^7*b *8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^5*b^3*56i - 28 *a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2) - (16*(c + d*tan(e + f*x))^(1/2)*( a^8*d^2 + b^8*d^2 - 28*a^2*b^6*d^2 + 70*a^4*b^4*d^2 - 28*a^6*b^2*d^2))/f^2 )*(-(a*b^7*8i - a^7*b*8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b ^4 + a^5*b^3*56i - 28*a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2)*1i - (((32*(a ^4*d^3*f^2 + b^4*d^3*f^2 - 6*a^2*b^2*d^3*f^2 + 4*a*b^3*c*d^2*f^2 - 4*a^3*b *c*d^2*f^2))/f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a*b^7*8i - a^7*b *8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^5*b^3*56i - 28 *a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2))*(-(a*b^7*8i - a^7*b*8i + a^8 + b^ 8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^5*b^3*56i - 28*a^6*b^2)/(4*( c*f^2 - d*f^2*1i)))^(1/2) + (16*(c + d*tan(e + f*x))^(1/2)*(a^8*d^2 + b^8* d^2 - 28*a^2*b^6*d^2 + 70*a^4*b^4*d^2 - 28*a^6*b^2*d^2))/f^2)*(-(a*b^7*8i - a^7*b*8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^5*b^3*5 6i - 28*a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2)*1i)/((((32*(a^4*d^3*f^2 + b ^4*d^3*f^2 - 6*a^2*b^2*d^3*f^2 + 4*a*b^3*c*d^2*f^2 - 4*a^3*b*c*d^2*f^2))/f ^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a*b^7*8i - a^7*b*8i + a^8 +...
\[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {2 \sqrt {d \tan \left (f x +e \right )+c}\, a^{4}+\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{4}}{d \tan \left (f x +e \right )+c}d x \right ) b^{4} d f +4 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{3}}{d \tan \left (f x +e \right )+c}d x \right ) a \,b^{3} d f -\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{d \tan \left (f x +e \right )+c}d x \right ) a^{4} d f +6 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{d \tan \left (f x +e \right )+c}d x \right ) a^{2} b^{2} d f +4 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{d \tan \left (f x +e \right )+c}d x \right ) a^{3} b d f}{d f} \] Input:
int((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x)
Output:
(2*sqrt(tan(e + f*x)*d + c)*a**4 + int((sqrt(tan(e + f*x)*d + c)*tan(e + f *x)**4)/(tan(e + f*x)*d + c),x)*b**4*d*f + 4*int((sqrt(tan(e + f*x)*d + c) *tan(e + f*x)**3)/(tan(e + f*x)*d + c),x)*a*b**3*d*f - int((sqrt(tan(e + f *x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*d + c),x)*a**4*d*f + 6*int((sqrt (tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*d + c),x)*a**2*b**2*d* f + 4*int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)*d + c),x)* a**3*b*d*f)/(d*f)