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

Optimal. Leaf size=336 \[ -\frac{2 (b c-a d)^{5/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{7/2} f \left (a^2+b^2\right )}+\frac{2 \sqrt{c+d \tan (e+f x)} \left ((b c-a d) (-a C d+b B d+b c C)+b^2 d (d (A-C)+B c)\right )}{b^3 f}-\frac{(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)}+\frac{(c+i d)^{5/2} (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)}+\frac{2 (-a C d+b B d+b c C) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f} \]

[Out]

-(((I*A + B - I*C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)*f)) + ((I*A - B
- I*C)*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)*f) - (2*(A*b^2 - a*(b*B -
a*C))*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(b^(7/2)*(a^2 + b^2)*f) +
(2*(b^2*d*(B*c + (A - C)*d) + (b*c - a*d)*(b*c*C + b*B*d - a*C*d))*Sqrt[c + d*Tan[e + f*x]])/(b^3*f) + (2*(b*
c*C + b*B*d - a*C*d)*(c + d*Tan[e + f*x])^(3/2))/(3*b^2*f) + (2*C*(c + d*Tan[e + f*x])^(5/2))/(5*b*f)

________________________________________________________________________________________

Rubi [A]  time = 2.81173, antiderivative size = 336, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.149, Rules used = {3647, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{2 (b c-a d)^{5/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{7/2} f \left (a^2+b^2\right )}+\frac{2 \sqrt{c+d \tan (e+f x)} \left ((b c-a d) (-a C d+b B d+b c C)+b^2 d (d (A-C)+B c)\right )}{b^3 f}-\frac{(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)}+\frac{(c+i d)^{5/2} (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)}+\frac{2 (-a C d+b B d+b c C) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A + B - I*C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)*f)) + ((I*A - B
- I*C)*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)*f) - (2*(A*b^2 - a*(b*B -
a*C))*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(b^(7/2)*(a^2 + b^2)*f) +
(2*(b^2*d*(B*c + (A - C)*d) + (b*c - a*d)*(b*c*C + b*B*d - a*C*d))*Sqrt[c + d*Tan[e + f*x]])/(b^3*f) + (2*(b*
c*C + b*B*d - a*C*d)*(c + d*Tan[e + f*x])^(3/2))/(3*b^2*f) + (2*C*(c + d*Tan[e + f*x])^(5/2))/(5*b*f)

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*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] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rule 3539

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

Rule 3537

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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]

Rule 3634

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

Rubi steps

\begin{align*} \int \frac{(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx &=\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac{2 \int \frac{(c+d \tan (e+f x))^{3/2} \left (\frac{5}{2} (A b c-a C d)+\frac{5}{2} b (B c+(A-C) d) \tan (e+f x)+\frac{5}{2} (b c C+b B d-a C d) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{5 b}\\ &=\frac{2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac{4 \int \frac{\sqrt{c+d \tan (e+f x)} \left (\frac{15}{4} \left (A b^2 c^2+a d (a C d-b (2 c C+B d))\right )+\frac{15}{4} b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)+\frac{15}{4} \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{15 b^2}\\ &=\frac{2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 f}+\frac{2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac{8 \int \frac{\frac{15}{8} \left (A b^2 \left (b c^3-a d^3\right )-a d \left (a^2 C d^2-a b d (3 c C+B d)+b^2 \left (3 c^2 C+3 B c d-C d^2\right )\right )\right )+\frac{15}{8} b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) \tan (e+f x)+\frac{15}{8} \left ((b c-a d) \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right )+b^3 d \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{15 b^3}\\ &=\frac{2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 f}+\frac{2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac{8 \int \frac{-\frac{15}{8} b^3 \left (a \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )-b \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right )+\frac{15}{8} b^3 \left (a A d \left (3 c^2-d^2\right )-A b \left (c^3-3 c d^2\right )+b \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3\right )-a \left (C d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{15 b^3 \left (a^2+b^2\right )}+\frac{\left (\left (A b^2-a (b B-a C)\right ) (b c-a d)^3\right ) \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{b^3 \left (a^2+b^2\right )}\\ &=\frac{2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 f}+\frac{2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac{\left ((A-i B-C) (c-i d)^3\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)}+\frac{\left ((A+i B-C) (c+i d)^3\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)}+\frac{\left (\left (A b^2-a (b B-a C)\right ) (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{b^3 \left (a^2+b^2\right ) f}\\ &=\frac{2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 f}+\frac{2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f}+\frac{\left ((i A+B-i C) (c-i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (a-i b) f}-\frac{\left (i (A+i B-C) (c+i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (a+i b) f}+\frac{\left (2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{b^3 \left (a^2+b^2\right ) d f}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right ) f}+\frac{2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 f}+\frac{2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f}-\frac{\left ((A-i B-C) (c-i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(a-i b) d f}-\frac{\left ((A+i B-C) (c+i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(a+i b) d f}\\ &=-\frac{(i A+B-i C) (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(a-i b) f}-\frac{(A+i B-C) (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(i a-b) f}-\frac{2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right ) f}+\frac{2 \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 f}+\frac{2 (b c C+b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 b^2 f}+\frac{2 C (c+d \tan (e+f x))^{5/2}}{5 b f}\\ \end{align*}

Mathematica [A]  time = 5.32831, size = 322, normalized size = 0.96 \[ \frac{\frac{15 \left (b^{7/2} (b-i a) (c-i d)^{5/2} (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )+b^{7/2} (b+i a) (c+i d)^{5/2} (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )-2 (b c-a d)^{5/2} \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )\right )}{b^{5/2} \left (a^2+b^2\right )}+\frac{30 \sqrt{c+d \tan (e+f x)} \left ((b c-a d) (-a C d+b B d+b c C)+b^2 d (d (A-C)+B c)\right )}{b^2}+\frac{10 (-a C d+b B d+b c C) (c+d \tan (e+f x))^{3/2}}{b}+6 C (c+d \tan (e+f x))^{5/2}}{15 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*(b^(7/2)*((-I)*a + b)*(A - I*B - C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + b^(
7/2)*(I*a + b)*(A + I*B - C)*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] - 2*(A*b^2 + a*(-
(b*B) + a*C))*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]]))/(b^(5/2)*(a^2 +
b^2)) + (30*(b^2*d*(B*c + (A - C)*d) + (b*c - a*d)*(b*c*C + b*B*d - a*C*d))*Sqrt[c + d*Tan[e + f*x]])/b^2 + (1
0*(b*c*C + b*B*d - a*C*d)*(c + d*Tan[e + f*x])^(3/2))/b + 6*C*(c + d*Tan[e + f*x])^(5/2))/(15*b*f)

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Maple [B]  time = 0.214, size = 8698, normalized size = 25.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{b \tan \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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