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

Optimal. Leaf size=273 \[ \frac{2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{3 d^2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-c^2 d^2 (A-3 C)+A d^4-2 B c d^3+c^4 C\right )\right )}{d^2 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{(a-i b) (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}+\frac{(-b+i a) (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.797563, antiderivative size = 271, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3635, 3628, 3539, 3537, 63, 208} \[ \frac{2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{3 d^2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-c^2 d^2 (A-3 C)+A d^4-2 B c d^3+c^4 C\right )\right )}{d^2 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{(b+i a) (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}+\frac{(-b+i a) (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3635

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 - a*d)*(c^2*C - B*c*d + A*d^2)*
(c + d*Tan[e + f*x])^(n + 1))/(d^2*f*(n + 1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x
])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d +
a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] &&
NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]

Rule 3628

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

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

Mathematica [C]  time = 2.7419, size = 300, normalized size = 1.1 \[ -\frac{d (-a A d+a B c+a C d+A b c+b B d-b c C) \left (i (c+i d) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )\right )-3 d (a B+A b-b C) (c+d \tan (e+f x)) \left (i (c+i d) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )\right )+2 (c-i d) (c+i d) (-2 a C d+b B d+2 b c C)+6 C d (c-i d) (c+i d) (a+b \tan (e+f x))}{3 d^2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*(c - I*d)*(c + I*d)*(2*b*c*C + b*B*d - 2*a*C*d) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d)*(I*(c
+ I*d)*Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c - I*d)] - (I*c + d)*Hypergeometric2F1[-3/2, 1,
-1/2, (c + d*Tan[e + f*x])/(c + I*d)]) + 6*C*(c - I*d)*(c + I*d)*d*(a + b*Tan[e + f*x]) - 3*(A*b + a*B - b*C)
*d*(I*(c + I*d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)] - (I*c + d)*Hypergeometric2F1[
-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)])*(c + d*Tan[e + f*x]))/(3*d^2*(c^2 + d^2)*f*(c + d*Tan[e + f*x])
^(3/2))

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Maple [B]  time = 0.21, size = 40201, normalized size = 147.3 \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))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(5/2),x)

[Out]

result too large to display

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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))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(5/2),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))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(5/2),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 ) \left (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(c + d*tan(e + f*x))**(5/2), 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 )}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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