3.1301 \(\int \frac{1}{\sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=295 \[ -\frac{4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 d^2 \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b} (c-i d)^{5/2}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b} (c+i d)^{5/2}} \]

[Out]

((-I)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a - I*
b]*(c - I*d)^(5/2)*f) + (I*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e +
f*x]])])/(Sqrt[a + I*b]*(c + I*d)^(5/2)*f) + (2*d^2*Sqrt[a + b*Tan[e + f*x]])/(3*(b*c - a*d)*(c^2 + d^2)*f*(c
+ d*Tan[e + f*x])^(3/2)) - (4*d^2*(3*a*c*d - b*(4*c^2 + d^2))*Sqrt[a + b*Tan[e + f*x]])/(3*(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 = 1.26774, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {3569, 3649, 3616, 3615, 93, 208} \[ -\frac{4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 d^2 \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b} (c-i d)^{5/2}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b} (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a - I*
b]*(c - I*d)^(5/2)*f) + (I*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e +
f*x]])])/(Sqrt[a + I*b]*(c + I*d)^(5/2)*f) + (2*d^2*Sqrt[a + b*Tan[e + f*x]])/(3*(b*c - a*d)*(c^2 + d^2)*f*(c
+ d*Tan[e + f*x])^(3/2)) - (4*d^2*(3*a*c*d - b*(4*c^2 + d^2))*Sqrt[a + b*Tan[e + f*x]])/(3*(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 3616

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

Rule 3615

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[A^2/f, Subst[Int[((a + b*x)^m*(c + d*x)^n)/(A - B*x), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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{1}{\sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx &=\frac{2 d^2 \sqrt{a+b \tan (e+f x)}}{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{1}{2} \left (2 b d^2+3 c (b c-a d)\right )-\frac{3}{2} d (b c-a d) \tan (e+f x)+b d^2 \tan ^2(e+f x)}{\sqrt{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 \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (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} (b c-a d)^2 \left (c^2-d^2\right )-\frac{3}{2} c d (b c-a d)^2 \tan (e+f x)}{\sqrt{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 \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (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{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{\int \frac{1-i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=\frac{2 d^2 \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (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-i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (c-i d)^2 f}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=\frac{2 d^2 \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (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}{i a+b-(i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(c-i d)^2 f}+\frac{\operatorname{Subst}\left (\int \frac{1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(c+i d)^2 f}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a-i b} (c-i d)^{5/2} f}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+i b} (c+i d)^{5/2} f}+\frac{2 d^2 \sqrt{a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 (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 = 2.04511, size = 316, normalized size = 1.07 \[ \frac{\frac{4 d^2 \left (b \left (4 c^2+d^2\right )-3 a c d\right ) \sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}+\frac{2 d^2 \left (c^2+d^2\right ) (b c-a d) \sqrt{a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}}+3 i (b c-a d)^2 \left (\frac{(c-i d)^2 \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+i b} \sqrt{-c-i d}}+\frac{(c+i d)^2 \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{-a+i b} \sqrt{c-i d}}\right )}{3 f \left (c^2+d^2\right )^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*I)*(b*c - a*d)^2*(((c - I*d)^2*ArcTan[(Sqrt[-c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*
Tan[e + f*x]])])/(Sqrt[a + I*b]*Sqrt[-c - I*d]) + ((c + I*d)^2*ArcTan[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])
/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + I*b]*Sqrt[c - I*d])) + (2*d^2*(b*c - a*d)*(c^2 + d^2)*
Sqrt[a + b*Tan[e + f*x]])/(c + d*Tan[e + f*x])^(3/2) + (4*d^2*(-3*a*c*d + b*(4*c^2 + d^2))*Sqrt[a + b*Tan[e +
f*x]])/Sqrt[c + d*Tan[e + f*x]])/(3*(b*c - a*d)^2*(c^2 + d^2)^2*f)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt{a+b\tan \left ( fx+e \right ) }}} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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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))^(1/2)/(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{1}{\sqrt{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(1/(a+b*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError