3.13.75 \(\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{5/2}} \, dx\) [1275]

Optimal. Leaf size=277 \[ -\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} f}+\frac {i (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} f}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {4 \left (3 a b c-a^2 d+2 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 f \sqrt {a+b \tan (e+f x)}} \]

[Out]

-I*(c-I*d)^(3/2)*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)^(5
/2)/f+I*(c+I*d)^(3/2)*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)^(5/2)/f-4/3*(-a^2*d+3*a*b*c+2*b^2*d)*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)^2/f/(a+b*tan(f*x+e))^(1/2)-2/3*(-a*d+
b*c)*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)/f/(a+b*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.89, antiderivative size = 277, 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 = {3648, 3730, 3697, 3696, 95, 214} \begin {gather*} -\frac {4 \left (a^2 (-d)+3 a b c+2 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right )^2 \sqrt {a+b \tan (e+f x)}}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac {i (c-i d)^{3/2} \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 (a-i b)^{5/2}}+\frac {i (c+i d)^{3/2} \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 (a+i b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(3/2)/(a + b*Tan[e + f*x])^(5/2),x]

[Out]

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

Rule 95

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3648

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

Rule 3696

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 3697

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 3730

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

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Mathematica [A]
time = 3.35, size = 264, normalized size = 0.95 \begin {gather*} \frac {\frac {3 i (-c+i d)^{3/2} \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 )}{(-a+i b)^{5/2}}+\frac {3 i (c+i d)^{3/2} \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 )}{(a+i b)^{5/2}}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (7 a^2 b c+b^3 c-3 a^3 d+3 a b^2 d+2 b \left (3 a b c-a^2 d+2 b^2 d\right ) \tan (e+f x)\right )}{\left (a^2+b^2\right )^2 (a+b \tan (e+f x))^{3/2}}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])^(3/2)/(a + b*Tan[e + f*x])^(5/2),x]

[Out]

(((3*I)*(-c + I*d)^(3/2)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e +
f*x]])])/(-a + I*b)^(5/2) + ((3*I)*(c + I*d)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a +
I*b]*Sqrt[c + d*Tan[e + f*x]])])/(a + I*b)^(5/2) - (2*Sqrt[c + d*Tan[e + f*x]]*(7*a^2*b*c + b^3*c - 3*a^3*d +
3*a*b^2*d + 2*b*(3*a*b*c - a^2*d + 2*b^2*d)*Tan[e + f*x]))/((a^2 + b^2)^2*(a + b*Tan[e + f*x])^(3/2)))/(3*f)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(5/2),x)

[Out]

int((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(5/2),x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(a+b*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 mo

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(3/2)/(a+b*tan(f*x+e))**(5/2),x)

[Out]

Integral((c + d*tan(e + f*x))**(3/2)/(a + b*tan(e + f*x))**(5/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*tan(e + f*x))^(3/2)/(a + b*tan(e + f*x))^(5/2),x)

[Out]

int((c + d*tan(e + f*x))^(3/2)/(a + b*tan(e + f*x))^(5/2), x)

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