3.1263 \(\int \frac{1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=272 \[ -\frac{2 b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right ) (b c-a d)^{5/2}}-\frac{2 d^2 \left (2 a c d-b \left (3 c^2+d^2\right )\right )}{f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 d^2}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a) (c-i d)^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a) (c+i d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 1.42329, antiderivative size = 272, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {3569, 3649, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{2 b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right ) (b c-a d)^{5/2}}-\frac{2 d^2 \left (2 a c d-b \left (3 c^2+d^2\right )\right )}{f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 d^2}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a) (c-i d)^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a) (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

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

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3649

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*T
an[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])))

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

Mathematica [A]  time = 4.874, size = 323, normalized size = 1.19 \[ \frac{\frac{3 \left (-\frac{2 b^{7/2} \left (c^2+d^2\right )^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{\sqrt{b c-a d}}+\frac{(b-i a) (c+i d)^2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}+\frac{(b+i a) (c-i d)^2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}\right )}{\left (a^2+b^2\right ) \left (c^2+d^2\right ) (a d-b c)}-\frac{6 d^2 \left (b \left (3 c^2+d^2\right )-2 a c d\right )}{\left (c^2+d^2\right ) (b c-a d) \sqrt{c+d \tan (e+f x)}}-\frac{2 d^2}{(c+d \tan (e+f x))^{3/2}}}{3 f \left (c^2+d^2\right ) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

((3*((((-I)*a + b)*(c + I*d)^2*(b*c - a*d)^2*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt[c - I*d] +
((I*a + b)*(c - I*d)^2*(b*c - a*d)^2*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d] - (2*b^(7/
2)*(c^2 + d^2)^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/Sqrt[b*c - a*d]))/((a^2 + b^2)*(
-(b*c) + a*d)*(c^2 + d^2)) - (2*d^2)/(c + d*Tan[e + f*x])^(3/2) - (6*d^2*(-2*a*c*d + b*(3*c^2 + d^2)))/((b*c -
a*d)*(c^2 + d^2)*Sqrt[c + d*Tan[e + f*x]]))/(3*(-(b*c) + a*d)*(c^2 + d^2)*f)

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Maple [B]  time = 0.106, size = 13982, normalized size = 51.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/(a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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