3.1284 \(\int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=218 \[ \frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {i (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c-i d}}+\frac {i (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c+i d}} \]

[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)

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Rubi [A]  time = 1.10, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3575, 910, 63, 217, 206, 6725, 93, 208} \[ \frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {i (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c-i d}}+\frac {i (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c+i d}} \]

Antiderivative was successfully verified.

[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 63

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 93

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 206

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

Rule 208

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

Rule 217

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 910

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 3575

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 6725

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*} \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx &=\frac {\operatorname {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 {\operatorname {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 {\operatorname {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 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {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) \operatorname {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 ) \operatorname {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 ) \operatorname {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) \operatorname {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} \tanh ^{-1}\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 ) \operatorname {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 ) \operatorname {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} \tanh ^{-1}\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} \tanh ^{-1}\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} \tanh ^{-1}\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*}

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Mathematica [A]  time = 2.04, size = 294, normalized size = 1.35 \[ \frac {\frac {i (-a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-c+i d}}+\frac {i \sqrt {d} (a+i b)^{3/2} \sqrt {c+d \tan (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )+2 b \sqrt {c+i d} \sqrt {b c-a d} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {d} \sqrt {c+i d} \sqrt {c+d \tan (e+f x)}}}{f} \]

Antiderivative was successfully verified.

[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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [a,b,c,d]=[-48,85,18,67]Warning, choosing root of [1,0,%%%{-2,[2,0,2,0]%%%}+%%%{-2,[2,0,0,2]%%%}
+%%%{4,[1,7,1,3]%%%}+%%%{-4,[0,8,0,4]%%%}+%%%{-2,[0,2,2,0]%%%}+%%%{-2,[0,2,0,2]%%%},%%%{-8,[2,7,2,3]%%%}+%%%{-
8,[2,7,0,5]%%%}+%%%{-8,[0,9,2,3]%%%}+%%%{-8,[0,9,0,5]%%%},%%%{1,[4,0,4,0]%%%}+%%%{2,[4,0,2,2]%%%}+%%%{1,[4,0,0
,4]%%%}+%%%{4,[3,7,3,3]%%%}+%%%{4,[3,7,1,5]%%%}+%%%{-4,[2,14,0,8]%%%}+%%%{-4,[2,8,2,4]%%%}+%%%{-4,[2,8,0,6]%%%
}+%%%{2,[2,2,4,0]%%%}+%%%{4,[2,2,2,2]%%%}+%%%{2,[2,2,0,4]%%%}+%%%{-8,[1,15,1,7]%%%}+%%%{4,[1,9,3,3]%%%}+%%%{4,
[1,9,1,5]%%%}+%%%{-4,[0,16,2,6]%%%}+%%%{-4,[0,10,2,4]%%%}+%%%{-4,[0,10,0,6]%%%}+%%%{1,[0,4,4,0]%%%}+%%%{2,[0,4
,2,2]%%%}+%%%{1,[0,4,0,4]%%%}] at parameters values [-49,-86,-64,-30]Evaluation time: 90.2index.cc index_m ope
rator + Error: Bad Argument Value

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\sqrt {c +d \tan \left (f x +e \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {d \tan \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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