3.2.52 \(\int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx\) [152]

Optimal. Leaf size=375 \[ -\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} \sqrt {c-i d} f}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} \sqrt {c+i d} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (5 a^3 b B d-2 a^4 C d+b^4 (3 B c-2 A d)+a b^3 (6 A c-6 c C-B d)-a^2 b^2 (3 B c+8 A d-4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}} \]

[Out]

-(I*A+B-I*C)*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)^(5/2)/
f/(c-I*d)^(1/2)-(B-I*(A-C))*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)^(5/2)/f/(c+I*d)^(1/2)-2/3*(5*a^3*b*B*d-2*a^4*C*d+b^4*(-2*A*d+3*B*c)+a*b^3*(6*A*c-B*d-6*C*c)-a^2*b^2*(
8*A*d+3*B*c-4*C*d))*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)^2/(-a*d+b*c)^2/f/(a+b*tan(f*x+e))^(1/2)-2/3*(A*b^2-a*(B*b
-C*a))*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)/(-a*d+b*c)/f/(a+b*tan(f*x+e))^(3/2)

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Rubi [A]
time = 1.21, antiderivative size = 375, normalized size of antiderivative = 1.00, 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 = {3730, 3697, 3696, 95, 214} \begin {gather*} -\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2}}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (-2 a^4 C d+5 a^3 b B d-a^2 b^2 (8 A d+3 B c-4 C d)+a b^3 (6 A c-B d-6 c C)+b^4 (3 B c-2 A d)\right )}{3 f \left (a^2+b^2\right )^2 (b c-a d)^2 \sqrt {a+b \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)^{5/2} \sqrt {c-i d}}-\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)^{5/2} \sqrt {c+i d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/((a + b*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),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)^(5/2)*Sqrt[c - I*d]*f)) - ((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]])])/((a + I*b)^(5/2)*Sqrt[c + I*d]*f) - (2*(A*b^2 - a*(b*B - a*C))*Sqrt[c + d
*Tan[e + f*x]])/(3*(a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x])^(3/2)) - (2*(5*a^3*b*B*d - 2*a^4*C*d + b^4*(
3*B*c - 2*A*d) + a*b^3*(6*A*c - 6*c*C - B*d) - a^2*b^2*(3*B*c + 8*A*d - 4*C*d))*Sqrt[c + d*Tan[e + f*x]])/(3*(
a^2 + b^2)^2*(b*c - a*d)^2*f*Sqrt[a + b*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 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 {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (2 A b^2 d-3 a A (b c-a d)-(b B-a C) (3 b c-a d)\right )+\frac {3}{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)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (5 a^3 b B d-2 a^4 C d+b^4 (3 B c-2 A d)+a b^3 (6 A c-6 c C-B d)-a^2 b^2 (3 B c+8 A d-4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}+\frac {4 \int \frac {\frac {3}{4} \left (2 a b B+a^2 (A-C)-b^2 (A-C)\right ) (b c-a d)^2+\frac {3}{4} \left (a^2 B-b^2 B-2 a b (A-C)\right ) (b c-a d)^2 \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right )^2 (b c-a d)^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (5 a^3 b B d-2 a^4 C d+b^4 (3 B c-2 A d)+a b^3 (6 A c-6 c C-B d)-a^2 b^2 (3 B c+8 A d-4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \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)^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 (a+i b)^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (5 a^3 b B d-2 a^4 C d+b^4 (3 B c-2 A d)+a b^3 (6 A c-6 c C-B d)-a^2 b^2 (3 B c+8 A d-4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}+\frac {(A-i B-C) \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)^2 f}+\frac {(A+i B-C) \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)^2 f}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (5 a^3 b B d-2 a^4 C d+b^4 (3 B c-2 A d)+a b^3 (6 A c-6 c C-B d)-a^2 b^2 (3 B c+8 A d-4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}+\frac {(A-i B-C) \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)^2 f}+\frac {(A+i B-C) \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)^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 )}{(a-i b)^{5/2} \sqrt {c-i d} 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)^{5/2} \sqrt {c+i d} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (5 a^3 b B d-2 a^4 C d+b^4 (3 B c-2 A d)+a b^3 (6 A c-6 c C-B d)-a^2 b^2 (3 B c+8 A d-4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 4.11, size = 388, normalized size = 1.03 \begin {gather*} \frac {\frac {3 (a+i b)^2 (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} \sqrt {-c+i d}}+\frac {3 i (a-i b)^2 (A+i B-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} \sqrt {c+i d}}+\frac {2 \left (a^2+b^2\right ) \left (A b^2+a (-b B+a C)\right ) \sqrt {c+d \tan (e+f x)}}{(-b c+a d) (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (-5 a^3 b B d+2 a^4 C d+b^4 (-3 B c+2 A d)+a b^3 (-6 A c+6 c C+B d)+a^2 b^2 (3 B c+8 A d-4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{(b c-a d)^2 \sqrt {a+b \tan (e+f x)}}}{3 \left (a^2+b^2\right )^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*(a + I*b)^2*(I*A + B - I*C)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*T
an[e + f*x]])])/(Sqrt[-a + I*b]*Sqrt[-c + I*d]) + ((3*I)*(a - I*b)^2*(A + I*B - 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]*Sqrt[c + I*d]) + (2*(a^2 + b^2
)*(A*b^2 + a*(-(b*B) + a*C))*Sqrt[c + d*Tan[e + f*x]])/((-(b*c) + a*d)*(a + b*Tan[e + f*x])^(3/2)) + (2*(-5*a^
3*b*B*d + 2*a^4*C*d + b^4*(-3*B*c + 2*A*d) + a*b^3*(-6*A*c + 6*c*C + B*d) + a^2*b^2*(3*B*c + 8*A*d - 4*C*d))*S
qrt[c + d*Tan[e + f*x]])/((b*c - a*d)^2*Sqrt[a + b*Tan[e + f*x]]))/(3*(a^2 + b^2)^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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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((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),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 {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 {5}{2}} \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*tan(e + f*x) + C*tan(e + f*x)**2)/((a + b*tan(e + f*x))**(5/2)*sqrt(c + d*tan(e + f*x))), 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((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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