3.1283 \(\int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=264 \[ -\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {i (a-i b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c-i d}}+\frac {i (a+i b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c+i d}} \]

[Out]

-b^(3/2)*(-5*a*d+b*c)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))/d^(3/2)/f-I*(a-I*
b)^(5/2)*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)^(5/2)*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)+b^2*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/d/f

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Rubi [A]  time = 2.45, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3566, 3655, 6725, 63, 217, 206, 93, 208} \[ -\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {i (a-i b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c-i d}}+\frac {i (a+i b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c+i d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^(5/2)/Sqrt[c + d*Tan[e + f*x]],x]

[Out]

((-I)*(a - I*b)^(5/2)*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)^(5/2)*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) - (b^(3/2)*(b*c - 5*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])
/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(d^(3/2)*f) + (b^2*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(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 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3655

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[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]

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))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\int \frac {\frac {1}{2} \left (2 a^3 d-b^2 (b c+a d)\right )+b \left (3 a^2-b^2\right ) d \tan (e+f x)-\frac {1}{2} b^2 (b c-5 a d) \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{d}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (2 a^3 d-b^2 (b c+a d)\right )+b \left (3 a^2-b^2\right ) d x-\frac {1}{2} b^2 (b c-5 a d) x^2}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d f}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\operatorname {Subst}\left (\int \left (-\frac {b^2 (b c-5 a d)}{2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {a \left (a^2-3 b^2\right ) d+b \left (3 a^2-b^2\right ) d x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{d f}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\operatorname {Subst}\left (\int \frac {a \left (a^2-3 b^2\right ) d+b \left (3 a^2-b^2\right ) d x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d f}-\frac {\left (b^2 (b c-5 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 d f}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\operatorname {Subst}\left (\int \left (\frac {i a \left (a^2-3 b^2\right ) d-b \left (3 a^2-b^2\right ) d}{2 (i-x) \sqrt {a+b x} \sqrt {c+d x}}+\frac {i a \left (a^2-3 b^2\right ) d+b \left (3 a^2-b^2\right ) d}{2 (i+x) \sqrt {a+b x} \sqrt {c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{d f}-\frac {(b (b c-5 a d)) \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 )}{d f}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {(i a-b)^3 \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 {(i a+b)^3 \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 {(b (b c-5 a d)) \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 )}{d f}\\ &=-\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {(i a-b)^3 \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+b)^3 \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)^{5/2} \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)^{5/2} \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 {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\\ \end {align*}

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Mathematica [A]  time = 3.33, size = 438, normalized size = 1.66 \[ \frac {\frac {d \left (3 a^2 b^2+a \sqrt {-b^2} \left (a^2-3 b^2\right )-b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}-c} \sqrt {a+b \tan (e+f x)}}{\sqrt {\sqrt {-b^2}-a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {\sqrt {-b^2}-a} \sqrt {\frac {\sqrt {-b^2} d}{b}-c}}+\frac {d \left (a^3 \sqrt {-b^2}-3 a^2 b^2+3 a \left (-b^2\right )^{3/2}+b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}+c} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {\frac {\sqrt {-b^2} d}{b}+c}}-\frac {b^{5/2} (b c-5 a d) \sqrt {c-\frac {a d}{b}} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right )}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}+b^3 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b d f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^(5/2)/Sqrt[c + d*Tan[e + f*x]],x]

[Out]

(((3*a^2*b^2 - b^4 + a*Sqrt[-b^2]*(a^2 - 3*b^2))*d*ArcTanh[(Sqrt[-c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e + f*x
]])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[-c + (Sqrt[-b^2]*d)/b]) + (
(-3*a^2*b^2 + b^4 + a^3*Sqrt[-b^2] + 3*a*(-b^2)^(3/2))*d*ArcTanh[(Sqrt[c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e
+ f*x]])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + (Sqrt[-b^2]*d)/b]) +
b^3*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] - (b^(5/2)*(b*c - 5*a*d)*Sqrt[c - (a*d)/b]*ArcSinh[(Sqr
t[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c - (a*d)/b])]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(Sqrt[
d]*Sqrt[c + d*Tan[e + f*x]]))/(b*d*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))^(5/2)/(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))^(5/2)/(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]=[-49,-76,20,-39]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: 93.4index.cc index_m o
perator + Error: Bad Argument Value

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{\sqrt {c +d \tan \left (f x +e \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x)

[Out]

int((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {d \tan \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^(5/2)/sqrt(d*tan(f*x + e) + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))^(5/2)/(c + d*tan(e + f*x))^(1/2),x)

[Out]

int((a + b*tan(e + f*x))^(5/2)/(c + d*tan(e + f*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(1/2),x)

[Out]

Integral((a + b*tan(e + f*x))**(5/2)/sqrt(c + d*tan(e + f*x)), x)

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