3.13.94 \(\int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx\) [1294]

Optimal. Leaf size=301 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} (c-i d)^{3/2} f}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} (c+i d)^{3/2} f}-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.85, antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3650, 3730, 3697, 3696, 95, 214} \begin {gather*} -\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}-\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {i \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 (a-i b)^{3/2} (c-i d)^{3/2}}+\frac {i \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 (a+i b)^{3/2} (c+i d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

((-I)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/((a - I*b)^(
3/2)*(c - I*d)^(3/2)*f) + (I*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e
+ f*x]])])/((a + I*b)^(3/2)*(c + I*d)^(3/2)*f) - (2*b^2)/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*S
qrt[c + d*Tan[e + f*x]]) - (2*d*(a^2*d^2 + b^2*(c^2 + 2*d^2))*Sqrt[a + b*Tan[e + f*x]])/((a^2 + b^2)*(b*c - a*
d)^2*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*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 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 3650

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 + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*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] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3696

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[A^2/f, Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

Rule 3697

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]

Rule 3730

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] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {\frac {1}{2} \left (2 b^2 d-a (b c-a d)\right )+\frac {1}{2} b (b c-a d) \tan (e+f x)+b^2 d \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{\left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {4 \int \frac {-\frac {1}{4} (b c-a d)^2 (a c-b d)+\frac {1}{4} (b c-a d)^2 (b c+a d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right )}\\ &=-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {\int \frac {1+i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b) (c-i d)}+\frac {\int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b) (c+i d)}\\ &=-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a-i b) (c-i d) f}+\frac {\text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a+i b) (c+i d) f}\\ &=-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{i a+b-(i c+d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(a-i b) (c-i d) f}+\frac {\text {Subst}\left (\int \frac {1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(a+i b) (c+i d) f}\\ &=-\frac {i \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 )}{(a-i b)^{3/2} (c-i d)^{3/2} f}+\frac {i \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 )}{(a+i b)^{3/2} (c+i d)^{3/2} f}-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 4.95, size = 349, normalized size = 1.16 \begin {gather*} -\frac {\frac {(b c-a d)^2 \left (\frac {i (a+i b) (c+i d) \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 {-a+i b} \sqrt {-c+i d}}+\frac {(i a+b) (c-i d) \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 {a+i b} \sqrt {c+i d}}\right )}{(-b c+a d) \left (c^2+d^2\right )}+\frac {2 b^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{(b c-a d) \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{\left (a^2+b^2\right ) (b c-a d) f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((2*b*d+2*a*c)^2>0)', see `ass
ume?` for mo

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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