3.7.39 \(\int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) [639]

Optimal. Leaf size=229 \[ \frac {i \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}} \]

[Out]

I*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a-b)^(1/2)-I*arctanh((I*a+b)^(1/2)*tan(d*
x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a+b)^(1/2)+2/15*(15*a^2-8*b^2)*(a+b*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)
^(1/2)-2/5*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(5/2)+8/15*b*(a+b*tan(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(3/2)

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Rubi [A]
time = 0.38, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3650, 3730, 3731, 12, 3656, 924, 95, 211, 214} \begin {gather*} \frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}+\frac {i \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

(I*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - (I*ArcTanh[(Sqrt[I
*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d) - (2*Sqrt[a + b*Tan[c + d*x]])/(5*a*d
*Tan[c + d*x]^(5/2)) + (8*b*Sqrt[a + b*Tan[c + d*x]])/(15*a^2*d*Tan[c + d*x]^(3/2)) + (2*(15*a^2 - 8*b^2)*Sqrt
[a + b*Tan[c + d*x]])/(15*a^3*d*Sqrt[Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 211

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

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 924

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegr
and[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &
& NeQ[c*d^2 + a*e^2, 0] && IGtQ[m + 1/2, 0]

Rule 3650

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 3656

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x]
, x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^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])))

Rule 3731

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*t
an[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 + a^2*C)*(a + b*Tan[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)) - a*C*(b*c*(m + 1)
+ a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 2)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \int \frac {2 b+\frac {5}{2} a \tan (c+d x)+2 b \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{5 a}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\frac {1}{4} \left (-15 a^2+8 b^2\right )+2 b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}-\frac {8 \int -\frac {15 a^3 \sqrt {\tan (c+d x)}}{8 \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^3}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}+\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}+\frac {\text {Subst}\left (\int \left (-\frac {1}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {1}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {i \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.80, size = 197, normalized size = 0.86 \begin {gather*} \frac {-\frac {15 \sqrt [4]{-1} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {15 \sqrt [4]{-1} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2+4 a b \tan (c+d x)+\left (15 a^2-8 b^2\right ) \tan ^2(c+d x)\right )}{a^3 \tan ^{\frac {5}{2}}(c+d x)}}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

((-15*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*
b] + (15*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I
*b] + (2*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2 + 4*a*b*Tan[c + d*x] + (15*a^2 - 8*b^2)*Tan[c + d*x]^2))/(a^3*Tan[c
+ d*x]^(5/2)))/(15*d)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.62, size = 947049, normalized size = 4135.59 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),x)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^(7/2)), x)

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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(1/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),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 {1}{\sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(7/2)/(a+b*tan(d*x+c))**(1/2),x)

[Out]

Integral(1/(sqrt(a + b*tan(c + d*x))*tan(c + d*x)**(7/2)), 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(1/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),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 {1}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(c + d*x)^(7/2)*(a + b*tan(c + d*x))^(1/2)),x)

[Out]

int(1/(tan(c + d*x)^(7/2)*(a + b*tan(c + d*x))^(1/2)), x)

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