3.1300 \(\int \frac {\sqrt {a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=283 \[ \frac {2 d \left (6 a c d-b \left (5 c^2-d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {2 d \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac {i \sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}+\frac {i \sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (c+i d)^{5/2}} \]

[Out]

-I*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))*(a-I*b)^(1/2)/(c-I*d)^(5
/2)/f+I*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))*(a+I*b)^(1/2)/(c+I*
d)^(5/2)/f+2/3*d*(6*a*c*d-b*(5*c^2-d^2))*(a+b*tan(f*x+e))^(1/2)/(-a*d+b*c)/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2
)-2/3*d*(a+b*tan(f*x+e))^(1/2)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)

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Rubi [A]  time = 1.22, antiderivative size = 283, normalized size of antiderivative = 1.00, 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 = {3568, 3649, 3616, 3615, 93, 208} \[ \frac {2 d \left (6 a c d-b \left (5 c^2-d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {2 d \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac {i \sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}+\frac {i \sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[a - I*b]*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])]
)/((c - I*d)^(5/2)*f) + (I*Sqrt[a + I*b]*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[
c + d*Tan[e + f*x]])])/((c + I*d)^(5/2)*f) - (2*d*Sqrt[a + b*Tan[e + f*x]])/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*
x])^(3/2)) + (2*d*(6*a*c*d - b*(5*c^2 - d^2))*Sqrt[a + b*Tan[e + f*x]])/(3*(b*c - a*d)*(c^2 + d^2)^2*f*Sqrt[c
+ d*Tan[e + f*x]])

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]

Rule 3568

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

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 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 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])))

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 d \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} (-3 a c-b d)-\frac {3}{2} (b c-a d) \tan (e+f x)+b d \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{3 \left (c^2+d^2\right )}\\ &=-\frac {2 d \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 d \left (6 a c d-b \left (5 c^2-d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (b c-a d) \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) \left (a c^2+2 b c d-a d^2\right )+\frac {3}{4} (b c-a d) \left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{3 (b c-a d) \left (c^2+d^2\right )^2}\\ &=-\frac {2 d \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 d \left (6 a c d-b \left (5 c^2-d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (b c-a d) \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 {a+b \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 {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=-\frac {2 d \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 d \left (6 a c d-b \left (5 c^2-d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (b c-a d) \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-i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (c-i d)^2 f}+\frac {(a+i b) \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 \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 d \left (6 a c d-b \left (5 c^2-d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (b c-a d) \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}{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 {(a+i b) \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 \sqrt {a-i b} \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 )}{(c-i d)^{5/2} f}+\frac {i \sqrt {a+i b} \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 )}{(c+i d)^{5/2} f}-\frac {2 d \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 d \left (6 a c d-b \left (5 c^2-d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 4.86, size = 266, normalized size = 0.94 \[ \frac {\frac {2 d \sqrt {a+b \tan (e+f x)} \left (d \left (6 a c d+b \left (d^2-5 c^2\right )\right ) \tan (e+f x)+a d \left (7 c^2+d^2\right )-6 b c^3\right )}{\left (c^2+d^2\right )^2 (b c-a d) (c+d \tan (e+f x))^{3/2}}-\frac {3 i \sqrt {-a+i b} \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-c+i d)^{5/2}}+\frac {3 i \sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{5/2}}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)*Sqrt[-a + I*b]*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f
*x]])])/(-c + I*d)^(5/2) + ((3*I)*Sqrt[a + I*b]*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b
]*Sqrt[c + d*Tan[e + f*x]])])/(c + I*d)^(5/2) + (2*d*Sqrt[a + b*Tan[e + f*x]]*(-6*b*c^3 + a*d*(7*c^2 + d^2) +
d*(6*a*c*d + b*(-5*c^2 + d^2))*Tan[e + f*x]))/((b*c - a*d)*(c^2 + d^2)^2*(c + d*Tan[e + f*x])^(3/2)))/(3*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((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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giac [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(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a +b \tan \left (f x +e \right )}}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((2*b*d+2*a*c)^2>0)', see `ass
ume?` for more details)Is ((2*b*d+2*a*c)^2    -4*((a*c-b*d)^2       -((-a*d)-b*c)        *(a*d+b*c)))    ^2 po
sitive or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a+b\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \tan {\left (e + f x \right )}}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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