\(\int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1284]

Optimal result

Integrand size = 29, antiderivative size = 218 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c-i d} f}+\frac {i (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d} f}+\frac {2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f} \]

[Out]

-I*(a-I*b)^(3/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/f/(c-I*d)^
(1/2)+I*(a+I*b)^(3/2)*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/f/(c+
I*d)^(1/2)+2*b^(3/2)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))/f/d^(1/2)

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3656, 924, 65, 223, 212, 6857, 95, 214} \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {i (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c-i d}}+\frac {i (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c+i d}} \]

[In]

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

[Out]

((-I)*(a - I*b)^(3/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]]
)])/(Sqrt[c - I*d]*f) + (I*(a + I*b)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqr
t[c + d*Tan[e + f*x]])])/(Sqrt[c + I*d]*f) + (2*b^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sq
rt[c + d*Tan[e + f*x]])])/(Sqrt[d]*f)

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 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 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]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 1.95 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {\frac {i (-a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-c+i d}}+\frac {i (a+i b)^{3/2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right ) \sqrt {c+d \tan (e+f x)}+2 b \sqrt {c+i d} \sqrt {b c-a d} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {c+i d} \sqrt {d} \sqrt {c+d \tan (e+f x)}}}{f} \]

[In]

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

[Out]

((I*(-a + I*b)^(3/2)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]
])])/Sqrt[-c + I*d] + (I*(a + I*b)^(3/2)*Sqrt[d]*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*
b]*Sqrt[c + d*Tan[e + f*x]])]*Sqrt[c + d*Tan[e + f*x]] + 2*b*Sqrt[c + I*d]*Sqrt[b*c - a*d]*ArcSinh[(Sqrt[d]*Sq
rt[a + b*Tan[e + f*x]])/Sqrt[b*c - a*d]]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(Sqrt[c + I*d]*Sqrt[d]*Sq
rt[c + d*Tan[e + f*x]]))/f

Maple [F(-1)]

Timed out.

\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\sqrt {c +d \tan \left (f x +e \right )}}d x\]

[In]

int((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x)

[Out]

int((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8410 vs. \(2 (162) = 324\).

Time = 8.60 (sec) , antiderivative size = 16847, normalized size of antiderivative = 77.28 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

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

[In]

integrate((a+b*tan(f*x+e))**(3/2)/(c+d*tan(f*x+e))**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

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

[In]

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

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

[In]

int((a + b*tan(e + f*x))^(3/2)/(c + d*tan(e + f*x))^(1/2),x)

[Out]

int((a + b*tan(e + f*x))^(3/2)/(c + d*tan(e + f*x))^(1/2), x)