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

Optimal. Leaf size=287 \[ \frac{2 \sqrt{c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{15 d^3 f}-\frac{(a-i b)^2 (B+i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}}+\frac{(a+i b)^2 (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}}-\frac{2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f} \]

[Out]

-(((a - I*b)^2*(B + I*(A - C))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f)) + ((a + I*b
)^2*(I*A - B - I*C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) + (2*(12*a^2*C*d^2 - 10
*a*b*d*(2*c*C - 3*B*d) + b^2*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2))*Sqrt[c + d*Tan[e + f*x]])/(15*d^3*f) - (2*
b*(4*b*c*C - 5*b*B*d - 4*a*C*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(15*d^2*f) + (2*C*(a + b*Tan[e + f*x])^
2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f)

________________________________________________________________________________________

Rubi [A]  time = 1.0014, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.149, Rules used = {3647, 3637, 3630, 3539, 3537, 63, 208} \[ \frac{2 \sqrt{c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{15 d^3 f}-\frac{(a-i b)^2 (B+i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}}+\frac{(a+i b)^2 (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}}-\frac{2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - I*b)^2*(B + I*(A - C))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f)) + ((a + I*b
)^2*(I*A - B - I*C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) + (2*(12*a^2*C*d^2 - 10
*a*b*d*(2*c*C - 3*B*d) + b^2*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2))*Sqrt[c + d*Tan[e + f*x]])/(15*d^3*f) - (2*
b*(4*b*c*C - 5*b*B*d - 4*a*C*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(15*d^2*f) + (2*C*(a + b*Tan[e + f*x])^
2*Sqrt[c + d*Tan[e + f*x]])/(5*d*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 3637

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

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -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]

Rubi steps

\begin{align*} \int \frac{(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}+\frac{2 \int \frac{(a+b \tan (e+f x)) \left (\frac{1}{2} (-4 b c C+a (5 A-C) d)+\frac{5}{2} (A b+a B-b C) d \tan (e+f x)-\frac{1}{2} (4 b c C-5 b B d-4 a C d) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{5 d}\\ &=-\frac{2 b (4 b c C-5 b B d-4 a C d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}-\frac{4 \int \frac{\frac{1}{4} \left (20 a b c C d-3 a^2 (5 A-C) d^2-4 b^2 \left (2 c^2 C-\frac{5 B c d}{2}\right )\right )-\frac{15}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)-\frac{1}{4} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{15 d^2}\\ &=\frac{2 \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{2 b (4 b c C-5 b B d-4 a C d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}-\frac{4 \int \frac{\frac{15}{4} \left (2 a b B-a^2 (A-C)+b^2 (A-C)\right ) d^2-\frac{15}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{15 d^2}\\ &=\frac{2 \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{2 b (4 b c C-5 b B d-4 a C d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}+\frac{1}{2} \left ((a-i b)^2 (A-i B-C)\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{1}{2} \left ((a+i b)^2 (A+i B-C)\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{2 \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{2 b (4 b c C-5 b B d-4 a C d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}-\frac{\left (i (a+i b)^2 (A+i B-C)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 f}+\frac{\left ((a-i b)^2 (i A+B-i C)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 f}\\ &=\frac{2 \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{2 b (4 b c C-5 b B d-4 a C d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}-\frac{\left ((a-i b)^2 (A-i B-C)\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 )}{d f}-\frac{\left ((a+i b)^2 (A+i B-C)\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 )}{d f}\\ &=-\frac{(a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d} f}-\frac{(a+i b)^2 (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d} f}+\frac{2 \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{2 b (4 b c C-5 b B d-4 a C d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}\\ \end{align*}

Mathematica [A]  time = 5.99704, size = 275, normalized size = 0.96 \[ \frac{\frac{2 \sqrt{c+d \tan (e+f x)} \left (12 a^2 C d^2+10 a b d (3 B d-2 c C)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d^2}-\frac{15 d (a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}+\frac{15 i d (a+i b)^2 (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}+\frac{2 b \tan (e+f x) (4 a C d+5 b B d-4 b c C) \sqrt{c+d \tan (e+f x)}}{d}+6 C (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{15 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-15*(a - I*b)^2*(I*A + B - I*C)*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt[c - I*d] + ((15*I)*(
a + I*b)^2*(A + I*B - C)*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d] + (2*(12*a^2*C*d^2 +
10*a*b*d*(-2*c*C + 3*B*d) + b^2*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2))*Sqrt[c + d*Tan[e + f*x]])/d^2 + (2*b*(
-4*b*c*C + 5*b*B*d + 4*a*C*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/d + 6*C*(a + b*Tan[e + f*x])^2*Sqrt[c + d
*Tan[e + f*x]])/(15*d*f)

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Maple [B]  time = 0.183, size = 18289, normalized size = 63.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 (b \tan \left (f x + e\right ) + a\right )}^{2}}{\sqrt{d \tan \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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