∫ 1_____________ (a+b tan(e+fx))(c+d tan(e+fx))5∕2 dx

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

1/2))/(I*a-b)/(c+I*d)^(5/2)/f-2*b^(7/2)*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/(-a*d+b*c)^(1/2))/(a^2+b^2)/(-a

*d+b*c)^(5/2)/f-2*d^2*(2*a*c*d-b*(3*c^2+d^2))/(-a*d+b*c)^2/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)+2/3*d^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 = 1.42, antiderivative size = 272, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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, -

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*}

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Mathematica [A] time = 5.22, 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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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab