3.632 \(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=232 \[ \frac {\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{5/2} d}-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}} \]

[Out]

1/4*(3*a^2-8*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(5/2)/d+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)+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)-3/4*a*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/b^2/d+1/2*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c
)^(3/2)/b/d

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Rubi [A]  time = 0.99, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3566, 3647, 3656, 6725, 63, 217, 206, 912, 93, 205, 208} \[ \frac {\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{5/2} d}-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\frac {\tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 205

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 912

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

Rule 3566

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 - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3656

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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((a + b
*ff*x)^m*(c + d*ff*x)^n*(A + C*ff^2*x^2))/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e
, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 4.09, size = 270, normalized size = 1.16 \[ \frac {\frac {\sqrt {a} \left (3 a^2-8 b^2\right ) \sqrt {\frac {b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}-\frac {4 (-1)^{3/4} b^2 \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {4 (-1)^{3/4} b^2 \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+2 b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}-3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [B]  time = 0.71, size = 946413, normalized size = 4079.37 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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 \[ \int \frac {\tan \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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