Integrand size = 25, antiderivative size = 186 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=-\frac {i (i a-b)^{3/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {3 a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i (i a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \] Output:
-I*(I*a-b)^(3/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1 /2))/d+3*a*b^(1/2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2) )/d+I*(I*a+b)^(3/2)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c) )^(1/2))/d+b*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d
Time = 1.61 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.18 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\frac {-\sqrt [4]{-1} (-a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} (a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {3 \sqrt {a} \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {a+b \tan (c+d x)}}{\sqrt {1+\frac {b \tan (c+d x)}{a}}}}{d} \] Input:
Integrate[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2),x]
Output:
(-((-1)^(1/4)*(-a + I*b)^(3/2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[ c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]) + (-1)^(1/4)*(a + I*b)^(3/2)*ArcTan[ ((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + b*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + (3*Sqrt[a]*Sqrt[b]*ArcSinh [(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[a + b*Tan[c + d*x]])/Sqrt[1 + (b*Tan[c + d*x])/a])/d
Time = 0.87 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4053, 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.
\(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 4053 |
\(\displaystyle \int -\frac {-3 a b \tan ^2(c+d x)-2 \left (a^2-b^2\right ) \tan (c+d x)+a b}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {1}{2} \int \frac {-3 a b \tan ^2(c+d x)-2 \left (a^2-b^2\right ) \tan (c+d x)+a b}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {1}{2} \int \frac {-3 a b \tan (c+d x)^2-2 \left (a^2-b^2\right ) \tan (c+d x)+a b}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4138 |
\(\displaystyle \frac {b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {-3 a b \tan ^2(c+d x)-2 \left (a^2-b^2\right ) \tan (c+d x)+a b}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {-3 a b \tan ^2(c+d x)-2 \left (a^2-b^2\right ) \tan (c+d x)+a b}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \left (\frac {2 \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}-\frac {3 a b}{\sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {i (-b+i a)^{3/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-3 a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-i (b+i a)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\) |
Input:
Int[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2),x]
Output:
-((I*(I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b* Tan[c + d*x]]] - 3*a*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - I*(I*a + b)^(3/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + (b*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Ta n[c + d*x]])/d
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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^n/(f*(m + n - 1))), x] + Simp[1/(m + n - 1) Int[(a + b*Ta n[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a^2*c*(m + n - 1) - b *(b*c*(m - 1) + a*d*n) + (2*a*b*c + a^2*d - b^2*d)*(m + n - 1)*Tan[e + f*x] + b*(b*c*n + a*d*(2*m + n - 2))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 , 0] && GtQ[m, 1] && GtQ[n, 0] && IntegerQ[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] :> 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 = 0.69 (sec) , antiderivative size = 1345539, normalized size of antiderivative = 7234.08
\[\text {output too large to display}\]
Input:
int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2),x)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 3600 vs. \(2 (146) = 292\).
Time = 0.90 (sec) , antiderivative size = 7202, normalized size of antiderivative = 38.72 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \] Input:
integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\tan {\left (c + d x \right )}}\, dx \] Input:
integrate(tan(d*x+c)**(1/2)*(a+b*tan(d*x+c))**(3/2),x)
Output:
Integral((a + b*tan(c + d*x))**(3/2)*sqrt(tan(c + d*x)), x)
\[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\tan \left (d x + c\right )} \,d x } \] Input:
integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((b*tan(d*x + c) + a)^(3/2)*sqrt(tan(d*x + c)), x)
\[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\tan \left (d x + c\right )} \,d x } \] Input:
integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate((b*tan(d*x + c) + a)^(3/2)*sqrt(tan(d*x + c)), x)
Timed out. \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:
int(tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(3/2),x)
Output:
int(tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(3/2), x)
\[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}d x \right ) a \] Input:
int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2),x)
Output:
int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x),x)*b + int(sq rt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a),x)*a