\(\int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) [469]

Optimal result

Integrand size = 38, antiderivative size = 155 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {B \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {B \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d} \]

[Out]

-B*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a-b)^(1/2)+2*B*arctanh(b^(1/2)*tan(d*x+c
)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/b^(1/2)-B*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(
I*a+b)^(1/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {21, 3656, 924, 65, 223, 212, 926, 95, 211, 214} \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {B \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {B \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}} \]

[In]

Int[(Tan[c + d*x]^(3/2)*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(3/2),x]

[Out]

-((B*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d)) + (2*B*ArcTanh[(S
qrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[b]*d) - (B*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x
]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 924

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegr
and[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &
& NeQ[c*d^2 + a*e^2, 0] && IGtQ[m + 1/2, 0]

Rule 926

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 3656

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x]
, x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rubi steps \begin{align*} \text {integral}& = B \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = \frac {B \text {Subst}\left (\int \frac {x^{3/2}}{\sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {B \text {Subst}\left (\int \left (\frac {1}{\sqrt {x} \sqrt {a+b x}}-\frac {1}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {B \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{d}-\frac {B \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {B \text {Subst}\left (\int \left (\frac {i}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {(2 B) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {(i B) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {(i B) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(2 B) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = \frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {(i B) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i B) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = -\frac {B \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {B \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.23 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {B \left ((-1)^{3/4} \left (\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )+\frac {2 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}\right )}{d} \]

[In]

Integrate[(Tan[c + d*x]^(3/2)*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(3/2),x]

[Out]

(B*((-1)^(3/4)*(ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[-a + I*b]
+ ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[a + I*b]) + (2*Sqrt[a]*
ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]])
))/d

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.31 (sec) , antiderivative size = 943434, normalized size of antiderivative = 6086.67

\[\text {output too large to display}\]

[In]

int(tan(d*x+c)^(3/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4592 vs. \(2 (123) = 246\).

Time = 1.05 (sec) , antiderivative size = 9186, normalized size of antiderivative = 59.26 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(tan(d*x+c)^(3/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=B \int \frac {\tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]

[In]

integrate(tan(d*x+c)**(3/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))**(3/2),x)

[Out]

B*Integral(tan(c + d*x)**(3/2)/sqrt(a + b*tan(c + d*x)), x)

Maxima [F]

\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B b \tan \left (d x + c\right ) + B a\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(tan(d*x+c)^(3/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*b*tan(d*x + c) + B*a)*tan(d*x + c)^(3/2)/(b*tan(d*x + c) + a)^(3/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(tan(d*x+c)^(3/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (B\,a+B\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int((tan(c + d*x)^(3/2)*(B*a + B*b*tan(c + d*x)))/(a + b*tan(c + d*x))^(3/2),x)

[Out]

int((tan(c + d*x)^(3/2)*(B*a + B*b*tan(c + d*x)))/(a + b*tan(c + d*x))^(3/2), x)