Integrand size = 33, antiderivative size = 233 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d} \]
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Time = 0.76 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3688, 3728, 3711, 3609, 3620, 3618, 65, 214} \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {\sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d} \]
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Rule 65
Rule 214
Rule 3609
Rule 3618
Rule 3620
Rule 3688
Rule 3711
Rule 3728
Rubi steps \begin{align*} \text {integral}& = \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac {2 \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \left (-2 a B-\frac {7}{2} b B \tan (c+d x)+\frac {1}{2} (7 A b-4 a B) \tan ^2(c+d x)\right ) \, dx}{7 b} \\ & = \frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac {4 \int \sqrt {a+b \tan (c+d x)} \left (-\frac {1}{2} a (7 A b-4 a B)-\frac {35}{4} A b^2 \tan (c+d x)-\frac {1}{4} \left (14 a A b-8 a^2 B+35 b^2 B\right ) \tan ^2(c+d x)\right ) \, dx}{35 b^2} \\ & = -\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac {4 \int \sqrt {a+b \tan (c+d x)} \left (\frac {35 b^2 B}{4}-\frac {35}{4} A b^2 \tan (c+d x)\right ) \, dx}{35 b^2} \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac {4 \int \frac {\frac {35}{4} b^2 (A b+a B)-\frac {35}{4} b^2 (a A-b B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx}{35 b^2} \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac {1}{2} ((i a+b) (A-i B)) \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} ((i a-b) (A+i B)) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {((a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac {((a+i b) (A+i B)) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d} \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {((i a+b) (A-i B)) \text {Subst}\left (\int \frac {1}{-1-\frac {i a}{b}+\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {((i a-b) (A+i B)) \text {Subst}\left (\int \frac {1}{-1+\frac {i a}{b}-\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d} \\ & = \frac {\sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d} \\ \end{align*}
Time = 2.97 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.91 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-14 a^2 A b-105 A b^3+8 a^3 B-35 a b^2 B-b \left (-7 a A b+4 a^2 B+35 b^2 B\right ) \tan (c+d x)+3 b^2 (7 A b+a B) \tan ^2(c+d x)+15 b^3 B \tan ^3(c+d x)\right )}{105 b^3 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(936\) vs. \(2(201)=402\).
Time = 0.40 (sec) , antiderivative size = 937, normalized size of antiderivative = 4.02
method | result | size |
parts | \(\frac {2 A \left (\frac {\left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {a \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\sqrt {a +b \tan \left (d x +c \right )}\, b^{2}-b^{2} \left (-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )\right )}{d \,b^{2}}+B \left (\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d \,b^{3}}-\frac {4 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}} a}{5 d \,b^{3}}+\frac {2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d \,b^{3}}-\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d b}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d b}\right )\) | \(937\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1092\) |
default | \(\text {Expression too large to display}\) | \(1092\) |
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Leaf count of result is larger than twice the leaf count of optimal. 1312 vs. \(2 (195) = 390\).
Time = 0.27 (sec) , antiderivative size = 1312, normalized size of antiderivative = 5.63 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
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\[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\, dx \]
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\[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{3} \,d x } \]
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Timed out. \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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Time = 66.52 (sec) , antiderivative size = 1093, normalized size of antiderivative = 4.69 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {2\,A\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^2\,d}-\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (2\,a\,\left (\frac {2\,B\,\left (a^2+b^2\right )}{b^3\,d}-\frac {4\,B\,a^2}{b^3\,d}\right )+\frac {8\,B\,a^3}{b^3\,d}-\frac {4\,B\,a\,\left (a^2+b^2\right )}{b^3\,d}\right )-\left (\frac {2\,A\,\left (a^2+b^2\right )}{b^2\,d}-\frac {2\,A\,a^2}{b^2\,d}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}-\left (\frac {2\,B\,\left (a^2+b^2\right )}{3\,b^3\,d}-\frac {4\,B\,a^2}{3\,b^3\,d}\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}+\frac {2\,B\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{7/2}}{7\,b^3\,d}-\frac {2\,A\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b^2\,d}-\frac {4\,B\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^3\,d}+\mathrm {atan}\left (\frac {d^3\,\left (\frac {16\,\left (B^2\,b^4-B^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {-\frac {\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2}{4\,d^4}}\,1{}\mathrm {i}}{8\,\left (B^3\,a^2\,b^3+B^3\,b^5\right )}\right )\,\sqrt {-\frac {\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {d^3\,\left (\frac {16\,\left (B^2\,b^4-B^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}-\frac {16\,a\,b^2\,\left (\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2}{4\,d^4}}\,1{}\mathrm {i}}{8\,\left (B^3\,a^2\,b^3+B^3\,b^5\right )}\right )\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {A^2\,b^4\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}}{4\,d^4}+\frac {A^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,A\,b^4\,\sqrt {-A^4\,b^2\,d^4}}{d^3}+\frac {16\,A\,a^2\,b^2\,\sqrt {-A^4\,b^2\,d^4}}{d^3}}+\frac {a\,b^2\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}}{4\,d^4}+\frac {A^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-A^4\,b^2\,d^4}\,32{}\mathrm {i}}{\frac {16\,A\,b^4\,\sqrt {-A^4\,b^2\,d^4}}{d}+\frac {16\,A\,a^2\,b^2\,\sqrt {-A^4\,b^2\,d^4}}{d}}\right )\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {A^2\,b^4\,\sqrt {\frac {A^2\,a}{4\,d^2}-\frac {\sqrt {-A^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,A\,b^4\,\sqrt {-A^4\,b^2\,d^4}}{d^3}+\frac {16\,A\,a^2\,b^2\,\sqrt {-A^4\,b^2\,d^4}}{d^3}}-\frac {a\,b^2\,\sqrt {\frac {A^2\,a}{4\,d^2}-\frac {\sqrt {-A^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-A^4\,b^2\,d^4}\,32{}\mathrm {i}}{\frac {16\,A\,b^4\,\sqrt {-A^4\,b^2\,d^4}}{d}+\frac {16\,A\,a^2\,b^2\,\sqrt {-A^4\,b^2\,d^4}}{d}}\right )\,\sqrt {-\frac {\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i} \]
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