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

Optimal. Leaf size=379 \[ \frac{2 \left (A d^2-B c d+c^2 C\right ) \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac{2 \sqrt{a+b \tan (e+f x)} \left (b \left (4 c^2 d^2 (2 A-C)+2 A d^4-5 B c^3 d+B c d^3+2 c^4 C\right )-3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right )}{3 f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{(B+i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b} (c-i d)^{5/2}}+\frac{(i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b} (c+i d)^{5/2}} \]

[Out]

-(((B + I*(A - C))*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]*(c - I*d)^(5/2)*f)) + ((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]])])/(Sqrt[a + I*b]*(c + I*d)^(5/2)*f) + (2*(c^2*C - B*c*d + A*d^2)*Sqrt[a + b
*Tan[e + f*x]])/(3*(b*c - a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (2*(b*(2*c^4*C - 5*B*c^3*d + 4*c^2*
(2*A - C)*d^2 + B*c*d^3 + 2*A*d^4) - 3*a*d^2*(2*c*(A - C)*d - B*(c^2 - d^2)))*Sqrt[a + b*Tan[e + f*x]])/(3*(b*
c - a*d)^2*(c^2 + d^2)^2*f*Sqrt[c + d*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.81413, antiderivative size = 379, 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} \[ \frac{2 \left (A d^2-B c d+c^2 C\right ) \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac{2 \sqrt{a+b \tan (e+f x)} \left (b \left (4 c^2 d^2 (2 A-C)+2 A d^4-5 B c^3 d+B c d^3+2 c^4 C\right )-3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right )}{3 f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{(B+i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b} (c-i d)^{5/2}}+\frac{(i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b} (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((B + I*(A - C))*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]*(c - I*d)^(5/2)*f)) + ((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]])])/(Sqrt[a + I*b]*(c + I*d)^(5/2)*f) + (2*(c^2*C - B*c*d + A*d^2)*Sqrt[a + b
*Tan[e + f*x]])/(3*(b*c - a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (2*(b*(2*c^4*C - 5*B*c^3*d + 4*c^2*
(2*A - C)*d^2 + B*c*d^3 + 2*A*d^4) - 3*a*d^2*(2*c*(A - C)*d - B*(c^2 - d^2)))*Sqrt[a + b*Tan[e + f*x]])/(3*(b*
c - a*d)^2*(c^2 + d^2)^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)}{\sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{\frac{1}{2} \left (2 A b d^2+3 A c (b c-a d)-(b c-3 a d) (c C-B d)\right )+\frac{3}{2} (b c-a d) (B c-(A-C) d) \tan (e+f x)+b \left (c^2 C-B c d+A d^2\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{3 (b c-a d) \left (c^2+d^2\right )}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (2 c^4 C-5 B c^3 d+4 c^2 (2 A-C) d^2+B c d^3+2 A d^4\right )-3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{4 \int \frac{-\frac{3}{4} (b c-a d)^2 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-\frac{3}{4} (b c-a d)^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{3 (b c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (2 c^4 C-5 B c^3 d+4 c^2 (2 A-C) d^2+B c d^3+2 A d^4\right )-3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d)^2 \left (c^2+d^2\right )^2 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 (c-i d)^2}+\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 (c+i d)^2}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (2 c^4 C-5 B c^3 d+4 c^2 (2 A-C) d^2+B c d^3+2 A d^4\right )-3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d)^2 \left (c^2+d^2\right )^2 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 (c-i d)^2 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 (c+i d)^2 f}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (2 c^4 C-5 B c^3 d+4 c^2 (2 A-C) d^2+B c d^3+2 A d^4\right )-3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d)^2 \left (c^2+d^2\right )^2 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 )}{(c-i d)^2 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 )}{(c+i d)^2 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 )}{\sqrt{a-i b} (c-i d)^{5/2} 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 )}{\sqrt{a+i b} (c+i d)^{5/2} f}+\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (2 c^4 C-5 B c^3 d+4 c^2 (2 A-C) d^2+B c d^3+2 A d^4\right )-3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 5.36485, size = 403, normalized size = 1.06 \[ \frac{\frac{2 \left (c^2+d^2\right ) (b c-a d) \left (A d^2-B c d+c^2 C\right ) \sqrt{a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}}+\frac{2 \sqrt{a+b \tan (e+f x)} \left (3 a d^2 \left (2 c d (C-A)+B \left (c^2-d^2\right )\right )+b \left (4 c^2 d^2 (2 A-C)+2 A d^4-5 B c^3 d+B c d^3+2 c^4 C\right )\right )}{\sqrt{c+d \tan (e+f x)}}+3 (b c-a d)^2 \left (\frac{i (c-i d)^2 (A+i B-C) \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+i b} \sqrt{-c-i d}}+\frac{(c+i d)^2 (i A+B-i C) \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{-a+i b} \sqrt{c-i d}}\right )}{3 f \left (c^2+d^2\right )^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*c - a*d)^2*((I*(A + I*B - C)*(c - I*d)^2*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]) + ((I*A + B - I*C)*(c + I*d)^2*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])) + (2*
(b*c - a*d)*(c^2 + d^2)*(c^2*C - B*c*d + A*d^2)*Sqrt[a + b*Tan[e + f*x]])/(c + d*Tan[e + f*x])^(3/2) + (2*(b*(
2*c^4*C - 5*B*c^3*d + 4*c^2*(2*A - C)*d^2 + B*c*d^3 + 2*A*d^4) + 3*a*d^2*(2*c*(-A + C)*d + B*(c^2 - d^2)))*Sqr
t[a + b*Tan[e + f*x]])/Sqrt[c + d*Tan[e + f*x]])/(3*(b*c - a*d)^2*(c^2 + d^2)^2*f)

________________________________________________________________________________________

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}){\frac{1}{\sqrt{a+b\tan \left ( fx+e \right ) }}} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{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))^(1/2)/(c+d*tan(f*x+e))^(5/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [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))**(1/2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError