Optimal. Leaf size=318 \[ \frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {(-b+i a)^{5/2} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {(b+i a)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 1.56, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3565, 3649, 3616, 3615, 93, 203, 206} \[ \frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (-483 a^2 b^2+315 a^4-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {(-b+i a)^{5/2} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {(b+i a)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 93
Rule 203
Rule 206
Rule 3565
Rule 3615
Rule 3616
Rule 3649
Rubi steps
\begin {align*} \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2}{9} \int \frac {\frac {19 a^2 b}{2}-\frac {9}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac {1}{2} b \left (8 a^2-9 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {9}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {4 \int \frac {\frac {3}{4} a^2 \left (21 a^2-25 b^2\right )+\frac {63}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {57}{2} a^2 b^2 \tan ^2(c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{63 a}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 \int \frac {-\frac {3}{8} a^2 b \left (231 a^2-5 b^2\right )+\frac {315}{8} a^3 \left (a^2-3 b^2\right ) \tan (c+d x)+\frac {3}{2} a^2 b \left (21 a^2-25 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{315 a^2}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {16 \int \frac {-\frac {3}{16} a^2 \left (315 a^4-483 a^2 b^2-10 b^4\right )-\frac {945}{16} a^3 b \left (3 a^2-b^2\right ) \tan (c+d x)-\frac {3}{8} a^2 b^2 \left (231 a^2-5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{945 a^3}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {32 \int \frac {\frac {945}{32} a^4 b \left (3 a^2-b^2\right )-\frac {945}{32} a^5 \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{945 a^4}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {1}{2} (i a-b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (i a+b)^3 \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {(i a-b)^3 \operatorname {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {(i a+b)^3 \operatorname {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {(i a-b)^3 \operatorname {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^3 \operatorname {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {(i a-b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 4.22, size = 300, normalized size = 0.94 \[ \frac {\frac {\sec ^4(c+d x) \sqrt {a+b \tan (c+d x)} \left (-987 a^4+272 a^3 b \sin (2 (c+d x))-326 a^3 b \sin (4 (c+d x))+1374 a^2 b^2+4 \left (280 a^4-483 a^2 b^2-10 b^4\right ) \cos (2 (c+d x))+\left (-413 a^4+558 a^2 b^2+10 b^4\right ) \cos (4 (c+d x))-10 a b^3 \sin (2 (c+d x))+5 a b^3 \sin (4 (c+d x))+30 b^4\right )}{a^2 \tan ^{\frac {9}{2}}(c+d x)}-1260 \sqrt [4]{-1} (-a+i b)^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-1260 \sqrt [4]{-1} (a+i b)^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{1260 d} \]
Antiderivative was successfully verified.
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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.
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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.
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maple [B] time = 0.96, size = 1347695, normalized size = 4238.03 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{11/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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