3.157 \(\int \frac{A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=383 \[ -\frac{2 d \sqrt{a+b \tan (e+f x)} \left (A \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right )+a^2 \left (-B c d+2 c^2 C+C d^2\right )-a b B \left (c^2+d^2\right )+b^2 c (c C-B d)\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A b^2-a (b B-a C)\right )}{f \left (a^2+b^2\right ) (b c-a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{3/2} (c-i d)^{3/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{3/2} (c+i d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 1.87757, antiderivative size = 382, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 49, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.102, Rules used = {3649, 3616, 3615, 93, 208} \[ -\frac{2 d \sqrt{a+b \tan (e+f x)} \left (a^2 A d^2+a^2 \left (-B c d+2 c^2 C+C d^2\right )-a b B \left (c^2+d^2\right )+A b^2 \left (c^2+2 d^2\right )+b^2 c (c C-B d)\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A b^2-a (b B-a C)\right )}{f \left (a^2+b^2\right ) (b c-a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{3/2} (c-i d)^{3/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{3/2} (c+i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A + B - I*C)*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)) - ((B - I*(A - C))*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqr
t[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/((a + I*b)^(3/2)*(c + I*d)^(3/2)*f) - (2*(A*b^2 - a*(b*B - a*C)))/((a^2
+ b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) - (2*d*(a^2*A*d^2 + b^2*c*(c*C - B*d)
- a*b*B*(c^2 + d^2) + A*b^2*(c^2 + 2*d^2) + a^2*(2*c^2*C - B*c*d + C*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 3649

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*T
an[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])))

Rule 3616

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 3615

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 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 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]

Rubi steps

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

Mathematica [A]  time = 6.67112, size = 484, normalized size = 1.26 \[ -\frac{2 \left (A b^2-a (b B-a C)\right )}{f \left (a^2+b^2\right ) (b c-a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (-\frac{2 \sqrt{a+b \tan (e+f x)} \left (\frac{1}{2} d^2 \left (-a A (b c-a d)-(b B-a C) (a d+b c)+2 A b^2 d\right )-c \left (\frac{1}{2} d (b c-a d) (-a B+A b-b C)-c d \left (A b^2-a (b B-a C)\right )\right )\right )}{f \left (c^2+d^2\right ) (a d-b c) \sqrt{c+d \tan (e+f x)}}+\frac{(b c-a d)^2 \left (\frac{(b+i a) (c-i d) (A+i B-C) \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+i b} \sqrt{-c-i d}}+\frac{(a+i b) (c+i d) (i A+B-i C) \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{-a+i b} \sqrt{c-i d}}\right )}{2 f \left (c^2+d^2\right ) (a d-b c)}\right )}{\left (a^2+b^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*b^2 - a*(b*B - a*C)))/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) - (
2*(((b*c - a*d)^2*(((I*a + b)*(A + I*B - C)*(c - I*d)*ArcTan[(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]) + ((a + I*b)*(I*A + B - I*C)*(c + I*d)*ArcT
an[(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])))/(2*(-(b*c) + a*d)*(c^2 + d^2)*f) - (2*(-(c*(-(c*(A*b^2 - a*(b*B - a*C))*d) + ((A*b - a*B - b*C)*d*(
b*c - a*d))/2)) + (d^2*(2*A*b^2*d - a*A*(b*c - a*d) - (b*B - a*C)*(b*c + a*d)))/2)*Sqrt[a + b*Tan[e + f*x]])/(
(-(b*c) + a*d)*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])))/((a^2 + b^2)*(b*c - a*d))

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{(A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2}) \left ( a+b\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(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., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}}{\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError