3.1262 \(\int \frac{a+b \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=186 \[ \frac{2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{(b+i a) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}+\frac{(-b+i a) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.384112, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3529, 3539, 3537, 63, 208} \[ \frac{2 (b c-a d)}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 a c d-b \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{(b+i a) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}+\frac{(-b+i a) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.138301, size = 115, normalized size = 0.62 \[ -\frac{i \left (\frac{(a+i b) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )}{c+i d}-\frac{(a-i b) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )}{c-i d}\right )}{3 f (c+d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.072, size = 12836, normalized size = 69. \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))/(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((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((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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \tan{\left (e + f x \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))/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Integral((a + b*tan(e + f*x))/(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{b \tan \left (f x + e\right ) + a}{{\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))/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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