3.2.62 \(\int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx\) [162]

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

[Out]

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

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Rubi [A]
time = 1.24, antiderivative size = 379, 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 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}} \end {gather*}

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 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)}{\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) \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 (c-i d)^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 (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) \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 )}{(c-i d)^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 )}{(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*}

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Mathematica [A]
time = 3.72, size = 403, normalized size = 1.06 \begin {gather*} \frac {3 (b c-a d)^2 \left (\frac {(i A+B-i C) (c+i d)^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 {-a+i b} \sqrt {-c+i d}}+\frac {i (A+i B-C) (c-i d)^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 {a+i b} \sqrt {c+i d}}\right )+\frac {2 (b c-a d) \left (c^2+d^2\right ) \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{(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)}}{\sqrt {c+d \tan (e+f x)}}}{3 (b c-a d)^2 \left (c^2+d^2\right )^2 f} \end {gather*}

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 + B - I*C)*(c + I*d)^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[-a + I*b]*Sqrt[-c + I*d]) + (I*(A + I*B - C)*(c - I*d)^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[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)))*S
qrt[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(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)

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

[Out]

Integral((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)

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

[Out]

\text{Hanged}

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