∫ A+B tan(c+dx)_ (a+b tan(c+dx))3∕2 dx

3.353 \(\int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d (a-i b)^{3/2}}+\frac {(-B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d (a+i b)^{3/2}} \]

[Out]

-(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d+(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/

(a+I*b)^(1/2))/(a+I*b)^(3/2)/d-2*(A*b-B*a)/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)

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Rubi [A] time = 0.24, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3529, 3539, 3537, 63, 208} \[ -\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d (a-i b)^{3/2}}+\frac {(-B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d (a+i b)^{3/2}} \]

Antiderivative was successfully veriﬁed.

[In]

Int[(A + B*Tan[c + d*x])/(a + b*Tan[c + d*x])^(3/2),x]

[Out]

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

+ b*Tan[c + d*x]]/Sqrt[a + I*b]])/((a + I*b)^(3/2)*d) - (2*(A*b - a*B))/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*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 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((

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

f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c

- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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]

Rubi steps

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

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Mathematica [C] time = 0.21, size = 113, normalized size = 0.82 \[ \frac {i \left (\frac {(A-i B) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \tan (c+d x)}{a-i b}\right )}{a-i b}-\frac {(A+i B) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \tan (c+d x)}{a+i b}\right )}{a+i b}\right )}{d \sqrt {a+b \tan (c+d x)}} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[(A + B*Tan[c + d*x])/(a + b*Tan[c + d*x])^(3/2),x]

[Out]

(I*(((A - I*B)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)])/(a - I*b) - ((A + I*B)*Hyperge

ometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)])/(a + I*b)))/(d*Sqrt[a + b*Tan[c + d*x]])

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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((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),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((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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