Integrand size = 43, antiderivative size = 253 \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} (2 a-3 i b) B \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 a d}+\frac {2 b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(2 a+3 i b) (i a+b)^{5/2} B \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 a d}-\frac {2 \left (a^2+3 b^2\right ) B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)} \]
1/2*(I*a-b)^(5/2)*(2*a-3*I*b)*B*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b *tan(d*x+c))^(1/2))/a/d+2*b^(5/2)*B*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b* tan(d*x+c))^(1/2))/d-1/2*(2*a+3*I*b)*(I*a+b)^(5/2)*B*arctanh((I*a+b)^(1/2) *tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/a/d-2*(a^2+3*b^2)*B*(a+b*tan(d*x +c))^(1/2)/d/tan(d*x+c)^(1/2)-b*B*(a+b*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(3/2 )
Time = 4.86 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.41 \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {B \cos (c+d x) (3 b+2 a \tan (c+d x)) \left (4 \sqrt {a} b^{5/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}-\sqrt {1+\frac {b \tan (c+d x)}{a}} \left (\sqrt [4]{-1} (-a+i b)^{5/2} (2 a+3 i b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {3}{2}}(c+d x)+\sqrt [4]{-1} (a+i b)^{5/2} (2 a-3 i b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {3}{2}}(c+d x)+2 a \sqrt {a+b \tan (c+d x)} \left (a b+\left (2 a^2+7 b^2\right ) \tan (c+d x)\right )\right )\right )}{2 a d (3 b \cos (c+d x)+2 a \sin (c+d x)) \tan ^{\frac {3}{2}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]
Integrate[((a + b*Tan[c + d*x])^(5/2)*((3*b*B)/(2*a) + B*Tan[c + d*x]))/Ta n[c + d*x]^(5/2),x]
(B*Cos[c + d*x]*(3*b + 2*a*Tan[c + d*x])*(4*Sqrt[a]*b^(5/2)*ArcSinh[(Sqrt[ b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x] ] - Sqrt[1 + (b*Tan[c + d*x])/a]*((-1)^(1/4)*(-a + I*b)^(5/2)*(2*a + (3*I) *b)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(3/2) + (-1)^(1/4)*(a + I*b)^(5/2)*(2*a - (3*I)*b)* ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d* x]]]*Tan[c + d*x]^(3/2) + 2*a*Sqrt[a + b*Tan[c + d*x]]*(a*b + (2*a^2 + 7*b ^2)*Tan[c + d*x]))))/(2*a*d*(3*b*Cos[c + d*x] + 2*a*Sin[c + d*x])*Tan[c + d*x]^(3/2)*Sqrt[1 + (b*Tan[c + d*x])/a])
Time = 1.57 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 4088, 27, 3042, 4128, 27, 3042, 4138, 2035, 2257, 2009}
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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan (c+d x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2}{3} \int \frac {3 \sqrt {a+b \tan (c+d x)} \left (2 b^2 B \tan ^2(c+d x)+b \left (\frac {3 b^2}{a}+a\right ) B \tan (c+d x)+2 \left (a^2+3 b^2\right ) B\right )}{4 \tan ^{\frac {3}{2}}(c+d x)}dx-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b \tan (c+d x)} \left (2 b^2 B \tan ^2(c+d x)+b \left (\frac {3 b^2}{a}+a\right ) B \tan (c+d x)+2 \left (a^2+3 b^2\right ) B\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b \tan (c+d x)} \left (2 b^2 B \tan (c+d x)^2+b \left (\frac {3 b^2}{a}+a\right ) B \tan (c+d x)+2 \left (a^2+3 b^2\right ) B\right )}{\tan (c+d x)^{3/2}}dx-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4128 |
| \(\displaystyle \frac {1}{2} \left (2 \int \frac {2 B \tan ^2(c+d x) b^3+3 \left (a^2+3 b^2\right ) B b-\frac {\left (2 a^4+3 b^2 a^2-3 b^4\right ) B \tan (c+d x)}{a}}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {4 B \left (a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 B \tan ^2(c+d x) b^3+3 \left (a^2+3 b^2\right ) B b-\frac {\left (2 a^4+3 b^2 a^2-3 b^4\right ) B \tan (c+d x)}{a}}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {4 B \left (a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 B \tan (c+d x)^2 b^3+3 \left (a^2+3 b^2\right ) B b-\frac {\left (2 a^4+3 b^2 a^2-3 b^4\right ) B \tan (c+d x)}{a}}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {4 B \left (a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {2 B \tan ^2(c+d x) b^3+3 \left (a^2+3 b^2\right ) B b-\frac {\left (2 a^4+3 b^2 a^2-3 b^4\right ) B \tan (c+d x)}{a}}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {4 B \left (a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 \int \frac {2 B \tan ^2(c+d x) b^3+3 \left (a^2+3 b^2\right ) B b-\frac {\left (2 a^4+3 b^2 a^2-3 b^4\right ) B \tan (c+d x)}{a}}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-\frac {4 B \left (a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 \int \left (\frac {2 B b^3}{\sqrt {a+b \tan (c+d x)}}+\frac {a b \left (3 a^2+7 b^2\right ) B-\left (2 a^4+3 b^2 a^2-3 b^4\right ) B \tan (c+d x)}{a \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}-\frac {4 B \left (a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{2} \left (-\frac {4 B \left (a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {2 \left (\frac {B (2 a-3 i b) (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 a}+2 b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {B (2 a+3 i b) (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 a}\right )}{d}\right )\) |
-((b*B*(a + b*Tan[c + d*x])^(3/2))/(d*Tan[c + d*x]^(3/2))) + ((2*(((I*a - b)^(5/2)*(2*a - (3*I)*b)*B*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[ a + b*Tan[c + d*x]]])/(2*a) + 2*b^(5/2)*B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d* x]])/Sqrt[a + b*Tan[c + d*x]]] - ((2*a + (3*I)*b)*(I*a + b)^(5/2)*B*ArcTan h[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*a)))/d - (4*(a^2 + 3*b^2)*B*Sqrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/2
3.5.51.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x ])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* (b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & & LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim p[1/(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* (n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b *(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ [a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f , A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 5.86 (sec) , antiderivative size = 1491744, normalized size of antiderivative = 5896.22
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 8171 vs. \(2 (205) = 410\).
Time = 2.54 (sec) , antiderivative size = 16341, normalized size of antiderivative = 64.59 \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]
integrate((a+b*tan(d*x+c))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2) ,x, algorithm="fricas")
\[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {B \left (\int \frac {2 a^{3} \sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {3 b^{3} \sqrt {a + b \tan {\left (c + d x \right )}}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx + \int \frac {6 a b^{2} \sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int 2 a b^{2} \sqrt {a + b \tan {\left (c + d x \right )}} \sqrt {\tan {\left (c + d x \right )}}\, dx + \int \frac {3 a^{2} b \sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {4 a^{2} b \sqrt {a + b \tan {\left (c + d x \right )}}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx\right )}{2 a} \]
B*(Integral(2*a**3*sqrt(a + b*tan(c + d*x))/tan(c + d*x)**(3/2), x) + Inte gral(3*b**3*sqrt(a + b*tan(c + d*x))/sqrt(tan(c + d*x)), x) + Integral(6*a *b**2*sqrt(a + b*tan(c + d*x))/tan(c + d*x)**(3/2), x) + Integral(2*a*b**2 *sqrt(a + b*tan(c + d*x))*sqrt(tan(c + d*x)), x) + Integral(3*a**2*b*sqrt( a + b*tan(c + d*x))/tan(c + d*x)**(5/2), x) + Integral(4*a**2*b*sqrt(a + b *tan(c + d*x))/sqrt(tan(c + d*x)), x))/(2*a)
\[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (2 \, B \tan \left (d x + c\right ) + \frac {3 \, B b}{a}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{2 \, \tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((a+b*tan(d*x+c))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2) ,x, algorithm="maxima")
1/2*integrate((2*B*tan(d*x + c) + 3*B*b/a)*(b*tan(d*x + c) + a)^(5/2)/tan( d*x + c)^(5/2), x)
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
integrate((a+b*tan(d*x+c))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2) ,x, algorithm="giac")
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (B\,\mathrm {tan}\left (c+d\,x\right )+\frac {3\,B\,b}{2\,a}\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]